5.9: The Divergence Theorem
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Learning Objectives
- Explain the meaning of the divergence theorem.
- Use the divergence theorem to calculate the flux of a vector field.
- Apply the divergence theorem to an electrostatic field.
We have examined several versions of the Fundamental Theorem of Calculus in higher dimensions that relate the integral around an oriented boundary of a domain to a “derivative” of that entity on the oriented domain. In this section, we state the divergence theorem, which is the final theorem of this type that we will study. The divergence theorem has many uses in physics; in particular, the divergence theorem is used in the field of partial differential equations to derive equations modeling heat flow and conservation of mass. We use the theorem to calculate flux integrals and apply it to electrostatic fields.
Overview of Theorems
Before examining the divergence theorem, it is helpful to begin with an overview of the versions of the Fundamental Theorem of Calculus we have discussed:
- The Fundamental Theorem of Calculus: \[\int_a^b f' (x) \, dx = f(b) - f(a). \nonumber \] This theorem relates the integral of derivative \(f'\) over line segment \([a,b]\) along the \(x\)-axis to a difference of \(f\) evaluated on the boundary.
- The Fundamental Theorem for Line Integrals: \[\int_C \vecs \nabla f \cdot d\vecs r = f(P_1) - f(P_0), \nonumber \] where \(P_0\) is the initial point of \(C\) and \(P_1\) is the terminal point of \(C\). The Fundamental Theorem for Line Integrals allows path \(C\) to be a path in a plane or in space, not just a line segment on the \(x\)-axis. If we think of the gradient as a derivative, then this theorem relates an integral of derivative \(\vecs \nabla f\) over path \(C\) to a difference of \(f\) evaluated on the boundary of \(C\).
- Green’s theorem, circulation form: \[\iint_D (Q_x - P_y)\,dA = \int_C \vecs F \cdot d\vecs r. \nonumber \] Since \(Q_x - P_y = \text{curl } \vecs F \cdot \mathbf{\hat k}\) and curl is a derivative of sorts, Green’s theorem relates the integral of derivative curl \(\vecs F\) over planar region \(D\) to an integral of \(\vecs F\) over the boundary of \(D\).
- Green’s theorem, flux form: \[\iint_D (P_x + Q_y)\,dA = \int_C \vecs F \cdot \vecs N \, dS. \nonumber \] Since \(P_x + Q_y = \text{div }\vecs F\) and divergence is a derivative of sorts, the flux form of Green’s theorem relates the integral of derivative div \(\vecs F\) over planar region \(D\) to an integral of \(\vecs F\) over the boundary of \(D\).
- Stokes’ theorem: \[\iint_S curl \, \vecs F \cdot d\vecs S = \int_C \vecs F \cdot d\vecs r. \nonumber \] If we think of the curl as a derivative of sorts, then Stokes’ theorem relates the integral of derivative curl \(\vecs F\) over surface \(S\) (not necessarily planar) to an integral of \(\vecs F\) over the boundary of \(S\).
Stating the Divergence Theorem
The divergence theorem follows the general pattern of these other theorems. If we think of divergence as a derivative of sorts, then the divergence theorem relates a triple integral of derivative div \(\vecs F\) over a solid to a flux integral of \(\vecs F\) over the boundary of the solid. More specifically, the divergence theorem relates a flux integral of vector field \(\vecs F\) over a closed surface \(S\) to a triple integral of the divergence of \(\vecs F\) over the solid enclosed by \(S\).
The Divergence Theorem
Let \(S\) be a piecewise, smooth closed surface that encloses solid \(E\) in space. Assume that \(S\) is oriented outward, and let \(\vecs F\) be a vector field with continuous partial derivatives on an open region containing \(E\) (Figure \(\PageIndex{1}\)). Then
\[\iiint_E \text{div }\vecs F \, dV = \iint_S \vecs F \cdot d\vecs S. \label{divtheorem} \]
Recall that the flux form of Green’s theorem states that
\[ \iint_D \text{div }\vecs F \, dA = \int_C \vecs F \cdot \vecs N \, dS. \nonumber \]
Therefore, the divergence theorem is a version of Green’s theorem in one higher dimension.
The proof of the divergence theorem is beyond the scope of this text. However, we look at an informal proof that gives a general feel for why the theorem is true, but does not prove the theorem with full rigor. This explanation follows the informal explanation given for why Stokes’ theorem is true.
Proof
Let \(B\) be a small box with sides parallel to the coordinate planes inside \(E\) (Figure \(\PageIndex{2a}\)). Let the center of \(B\) have coordinates \((x,y,z)\) and suppose the edge lengths are \(\Delta x, \, \Delta y\), and \(\Delta z\). (Figure \(\PageIndex{1b}\)). The normal vector out of the top of the box is \(\mathbf{\hat k}\) and the normal vector out of the bottom of the box is \(-\mathbf{\hat k}\). The dot product of \(\vecs F = \langle P, Q, R \rangle\) with \(\mathbf{\hat k}\) is \(R\) and the dot product with \(-\mathbf{\hat k}\) is \(-R\). The area of the top of the box (and the bottom of the box) \(\Delta S\) is \(\Delta x \Delta y\).
The flux out of the top of the box can be approximated by \(R \left(x,\, y,\, z + \frac{\Delta z}{2}\right) \,\Delta x \,\Delta y\) (Figure \(\PageIndex{2c}\)) and the flux out of the bottom of the box is \(- R \left(x,\, y,\, z - \frac{\Delta z}{2}\right) \,\Delta x \,\Delta y\). If we denote the difference between these values as \(\Delta R\), then the net flux in the vertical direction can be approximated by \(\Delta R\, \Delta x \,\Delta y\). However,
\[\Delta R \,\Delta x \,\Delta y = \left(\frac{\Delta R}{\Delta z}\right) \,\Delta x \,\Delta y \Delta z \approx \left(\frac{\partial R}{\partial z}\right) \,\Delta V.\nonumber \]
Therefore, the net flux in the vertical direction can be approximated by \(\left(\frac{\partial R}{\partial z}\right)\Delta V\). Similarly, the net flux in the \(x\)-direction can be approximated by \(\left(\frac{\partial P}{\partial x}\right)\,\Delta V\) and the net flux in the \(y\)-direction can be approximated by \(\left(\frac{\partial Q}{\partial y}\right)\,\Delta V\). Adding the fluxes in all three directions gives an approximation of the total flux out of the box:
\[\text{Total flux }\approx \left(\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \right) \Delta V = \text{div }\vecs F \,\Delta V. \nonumber \]
This approximation becomes arbitrarily close to the value of the total flux as the volume of the box shrinks to zero.
The sum of \(\text{div }\vecs F \,\Delta V\) over all the small boxes approximating \(E\) is approximately \(\iiint_E \text{div }\vecs F \,dV\). On the other hand, the sum of \(\text{div }\vecs F \,\Delta V\) over all the small boxes approximating \(E\) is the sum of the fluxes over all these boxes. Just as in the informal proof of Stokes’ theorem, adding these fluxes over all the boxes results in the cancelation of a lot of the terms. If an approximating box shares a face with another approximating box, then the flux over one face is the negative of the flux over the shared face of the adjacent box. These two integrals cancel out. When adding up all the fluxes, the only flux integrals that survive are the integrals over the faces approximating the boundary of \(E\). As the volumes of the approximating boxes shrink to zero, this approximation becomes arbitrarily close to the flux over \(S\).
\(\Box\)
Example \(\PageIndex{1}\): Verifying the Divergence Theorem
Verify the divergence theorem for vector field \(\vecs F = \langle x - y, \, x + z, \, z - y \rangle\) and surface \(S\) that consists of cone \(x^2 + y^2 = z^2, \, 0 \leq z \leq 1\), and the circular top of the cone (see the following figure). Assume this surface is positively oriented.
Solution
Let \(E\) be the solid cone enclosed by \(S\). To verify the theorem for this example, we show that
\[\iiint_E \text{div } \vecs F \,dV = \iint_S \vecs F \cdot d\vecs S\nonumber \]
by calculating each integral separately.
To compute the triple integral, note that \(\text{div } \vecs F = P_x + Q_y + R_z = 2\), and therefore the triple integral is
\[ \begin{align*} \iiint_E \text{div } \vecs F \, dV &= 2 \iiint_E dV \\[4pt] &= 2 \, (volume \, of \, E). \end{align*}\]
The volume of a right circular cone is given by \(\pi r^2 \frac{h}{3}\). In this case, \(h = r = 1\). Therefore,
\[\iiint_E \text{div } \vecs F \,dV = 2 \, (volume \, of \, E) = \frac{2\pi}{3}.\nonumber \]
To compute the flux integral, first note that \(S\) is piecewise smooth; \(S\) can be written as a union of smooth surfaces. Therefore, we break the flux integral into two pieces: one flux integral across the circular top of the cone and one flux integral across the remaining portion of the cone. Call the circular top \(S_1\) and the portion under the top \(S_2\). We start by calculating the flux across the circular top of the cone.    Since the surface is positively oriented, we use vector \(\vecs t_v \times \vecs t_u = \langle u \, \cos v, \, u \, \sin v, \, -u \rangle\) in the flux integral.
Exercise \(\PageIndex{1}\)
Verify the divergence theorem for vector field \(\vecs F (x,y,z) = \langle x + y + z, \, y, \, 2x - y \rangle\) and surface \(S\) given by the cylinder \(x^2 + y^2 = 1, \, 0 \leq z \leq 3\) plus the circular top and bottom of the cylinder. Assume that \(S\) is positively oriented.
- Hint
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Calculate both the flux integral and the triple integral with the divergence theorem and verify they are equal.
- Answer
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Both integrals equal \(6\pi\).
Recall that the divergence of continuous field \(\vecs F\) at point \(P\) is a measure of the “outflowing-ness” of the field at \(P\). If \(\vecs F\) represents the velocity field of a fluid, then the divergence can be thought of as the rate per unit volume of the fluid flowing out less the rate per unit volume flowing in. The divergence theorem confirms this interpretation. To see this, let \(P\) be a point and let \(B_{\tau}\) be a ball of small radius \(r\) centered at \(P\) (Figure \(\PageIndex{3}\)). Let \(S_{\tau}\) be the boundary sphere of \(B_{\tau}\). Since the radius is small and \(\vecs F\) is continuous, \(\text{div }\vecs F(Q) \approx \text{div }\vecs F(P)\) for all other points \(Q\) in the ball. Therefore, the flux across \(S_{\tau}\) can be approximated using the divergence theorem:
Exercise \(\PageIndex{2}\)
Use the divergence theorem to calculate flux integral \[\iint_S \vecs F \cdot d\vecs S,\nonumber \] where \(S\) is the boundary of the box given by \(0 \leq x \leq 2, \, 0 \leq y \leq 4, \, 0 \leq z \leq 1\) and \(\vecs F = \langle x^2 + yz, \, y - z, \, 2x + 2y + 2z \rangle \) (see the following figure).
- Hint
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Calculate the corresponding triple integral.
- Answer
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40
Example \(\PageIndex{3}\): Applying the Divergence Theorem
Let \(\vecs v = \left\langle - \frac{y}{z}, \, \frac{x}{z}, \, 0 \right\rangle\) be the velocity field of a fluid. Let \(C\) be the solid cube given by \(1 \leq x \leq 4, \, 2 \leq y \leq 5, \, 1 \leq z \leq 4\), and let \(S\) be the boundary of this cube (see the following figure). Find the flow rate of the fluid across \(S\).
Solution
The flow rate of the fluid across \(S\) is \(\iint_S \vecs v \cdot d\vecs S\). Before calculating this flux integral, let’s discuss what the value of the integral should be. Based on Figure \(\PageIndex{4}\), we see that if we place this cube in the fluid (as long as the cube doesn’t encompass the origin), then the rate of fluid entering the cube is the same as the rate of fluid exiting the cube. The field is rotational in nature and, for a given circle parallel to the \(xy\)-plane that has a center on the z -axis, the vectors along that circle are all the same magnitude. That is how we can see that the flow rate is the same entering and exiting the cube. The flow into the cube cancels with the flow out of the cube, and therefore the flow rate of the fluid across the cube should be zero.
To verify this intuition, we need to calculate the flux integral. Calculating the flux integral directly requires breaking the flux integral into six separate flux integrals, one for each face of the cube. We also need to find tangent vectors, compute their cross product. However, using the divergence theorem makes this calculation go much more quickly:
\[ \begin{align*} \iint_S \vecs v \cdot d\vecs S &= \iiint_C \text{div }\vecs v \, dV \\[4pt]
&= \iiint_C 0 \, dV = 0.\end{align*}\]
Therefore the flux is zero, as expected.
Exercise \(\PageIndex{3}\)
Let \(\vecs v = \left\langle \frac{x}{z}, \, \frac{y}{z}, \, 0 \right\rangle\) be the velocity field of a fluid. Let \(C\) be the solid cube given by \(1 \leq x \leq 4, \, 2 \leq y \leq 5, \, 1 \leq z \leq 4\), and let \(S\) be the boundary of this cube (see the following figure). Find the flow rate of the fluid across \(S\).
- Hint
-
Use the divergence theorem and calculate a triple integral
- Answer
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\(9 \, \ln (16)\)
Example illustrates a remarkable consequence of the divergence theorem. Let \(S\) be a piecewise, smooth closed surface and let \(\vecs F\) be a vector field defined on an open region containing the surface enclosed by \(S\). If \(\vecs F\) has the form \(F = \langle f (y,z), \, g(x,z), \, h(x,y)\rangle\), then the divergence of \(\vecs F\) is zero. By the divergence theorem, the flux of \(\vecs F\) across \(S\) is also zero. This makes certain flux integrals incredibly easy to calculate. For example, suppose we wanted to calculate the flux integral \(\iint_S \vecs F \cdot d\vecs S\) where \(S\) is a cube and
\[\vecs F = \langle \sin (y) \, e^{yz}, \, x^2z^2, \, \cos (xy) \, e^{\sin x} \rangle. \nonumber \]
Calculating the flux integral directly would be difficult, if not impossible, using techniques we studied previously. At the very least, we would have to break the flux integral into six integrals, one for each face of the cube. But, because the divergence of this field is zero, the divergence theorem immediately shows that the flux integral is zero.
We can now use the divergence theorem to justify the physical interpretation of divergence that we discussed earlier. Recall that if \(\vecs F\) is a continuous three-dimensional vector field and \(P\) is a point in the domain of \(\vecs F\), then the divergence of \(\vecs F\) at \(P\) is a measure of the “outflowing-ness” of \(\vecs F\) at \(P\). If \(\vecs F\) represents the velocity field of a fluid, then the divergence of \(\vecs F\) at \(P\) is a measure of the net flow rate out of point \(P\) (the flow of fluid out of \(P\) less the flow of fluid in to \(P\)). To see how the divergence theorem justifies this interpretation, let \(B_{\tau}\) be a ball of very small radius r with center \(P\), and assume that \(B_{\tau}\) is in the domain of \(\vecs F\). Furthermore, assume that \(B_{\tau}\) has a positive, outward orientation. Since the radius of \(B_{\tau}\) is small and \(\vecs F\) is continuous, the divergence of \(\vecs F\) is approximately constant on \(B_{\tau}\). That is, ifv \(P'\) is any point in \(B_{\tau}\), then \(\text{div } \vecs F(P) \approx \text{div } \vecs F(P')\). Let \(S_{\tau}\) denote the boundary sphere of \(B_{\tau}\). We can approximate the flux across \(S_{\tau}\) using the divergence theorem as follows:
\[\begin{align*} \iint_{S_{\tau}} \vecs F \cdot d\vecs S &= \iiint_{B_{\tau}} \text{div }\vecs F \, dV \\[4pt]
&\approx \iiint_{B_{\tau}} \text{div } \vecs F (P) \, dV \\[4pt]
&= \text{div } \vecs F (P) \, V(B_{\tau}). \end{align*}\]
As we shrink the radius \(r\) to zero via a limit, the quantity \(\text{div }\vecs F (P) \, V(B_{\tau})\) gets arbitrarily close to the flux. Therefore,
\[\text{div }\vecs F(P) = \lim_{\tau \rightarrow 0} \frac{1}{V(B_{\tau})} \iint_{S_{\tau}} \vecs F \cdot d\vecs S \nonumber \]
and we can consider the divergence at \(P\) as measuring the net rate of outward flux per unit volume at \(P\). Since “outflowing-ness” is an informal term for the net rate of outward flux per unit volume, we have justified the physical interpretation of divergence we discussed earlier, and we have used the divergence theorem to give this justification.
Application to Electrostatic Fields
The divergence theorem has many applications in physics and engineering. It allows us to write many physical laws in both an integral form and a differential form (in much the same way that Stokes’ theorem allowed us to translate between an integral and differential form of Faraday’s law). Areas of study such as fluid dynamics, electromagnetism, and quantum mechanics have equations that describe the conservation of mass, momentum, or energy, and the divergence theorem allows us to give these equations in both integral and differential forms.
One of the most common applications of the divergence theorem is to electrostatic fields . An important result in this subject is Gauss’ law . This law states that if \(S\) is a closed surface in electrostatic field \(\vecs E\), then the flux of \(\vecs E\) across \(S\) is the total charge enclosed by \(S\) (divided by an electric constant). We now use the divergence theorem to justify the special case of this law in which the electrostatic field is generated by a stationary point charge at the origin.
If \((x,y,z)\) is a point in space, then the distance from the point to the origin is \(r = \sqrt{x^2 + y^2 + z^2}\). Let \(\vecs F_{\tau}\) denote radial vector field \(\vecs F_{\tau} = \dfrac{1}{\tau^2} \left\langle \dfrac{x}{\tau}, \, \dfrac{y}{\tau}, \, \dfrac{z}{\tau}\right\rangle \).The vector at a given position in space points in the direction of unit radial vector \(\left\langle \dfrac{x}{\tau}, \, \dfrac{y}{\tau}, \, \dfrac{z}{\tau}\right\rangle \) and is scaled by the quantity \(1/\tau^2\). Therefore, the magnitude of a vector at a given point is inversely proportional to the square of the vector’s distance from the origin. Suppose we have a stationary charge of \(q\) Coulombs at the origin, existing in a vacuum. The charge generates electrostatic field \(\vecs E\) given by
\[\vecs E = \dfrac{q}{4\pi \epsilon_0}\vecs F_{\tau}, \nonumber \]
where the approximation \(\epsilon_0 = 8.854 \times 10^{-12}\) farad (F)/m is an electric constant. (The constant \(\epsilon_0\) is a measure of the resistance encountered when forming an electric field in a vacuum.) Notice that \(\vecs E\) is a radial vector field similar to the gravitational field described in [link] . The difference is that this field points outward whereas the gravitational field points inward. Because
\[\vecs E = \dfrac{q}{4\pi \epsilon_0}\vecs F_{\tau} = \dfrac{q}{4\pi \epsilon_0}\left(\dfrac{1}{\tau^2} \left\langle \dfrac{x}{\tau}, \, \dfrac{y}{\tau}, \, \dfrac{z}{\tau}\right\rangle\right), \nonumber \]
we say that electrostatic fields obey an inverse-square law . That is, the electrostatic force at a given point is inversely proportional to the square of the distance from the source of the charge (which in this case is at the origin). Given this vector field, we show that the flux across closed surface \(S\) is zero if the charge is outside of \(S\), and that the flux is \(q/epsilon_0\) if the charge is inside of \(S\). In other words, the flux across S is the charge inside the surface divided by constant \(\epsilon_0\). This is a special case of Gauss’ law, and here we use the divergence theorem to justify this special case.
To show that the flux across \(S\) is the charge inside the surface divided by constant \(\epsilon_0\), we need two intermediate steps. First we show that the divergence of \(\vecs F_{\tau}\) is zero and then we show that the flux of \(\vecs F_{\tau}\) across any smooth surface \(S\) is either zero or \(4\pi\). We can then justify this special case of Gauss’ law.
Example \(\PageIndex{4}\): The Divergence of \(F_{\tau}\) is Zero
Verify that the divergence of \(\vecs F_{\tau}\) is zero where \(\vecs F_{\tau}\) is defined (away from the origin).
Solution
Since \(\tau = \sqrt{x^2 + y^2 + z^2}\), the quotient rule gives us
\[ \begin{align*} \dfrac{\partial}{\partial x} \left( \dfrac{x}{\tau^3} \right) &= \dfrac{\partial}{\partial x} \left( \dfrac{x}{(x^2+y^2+z^2)^{3/2}} \right) \\[4pt]
&= \dfrac{(x^2+y^2+z^2)^{3/2} - x\left[\dfrac{3}{2} (x^2+y^2+z^2)^{1/2}2x\right]}{(x^2+y^2+z^2)^3} \\[4pt]
&= \dfrac{\tau^3 -3x^2\tau}{\tau^6} = \dfrac{\tau^2 - 3x^2}{\tau^5}. \end{align*}\]
Similarly,
\[\dfrac{\partial}{\partial y} \left( \dfrac{y}{\tau^3} \right) = \dfrac{\tau^2 - 3y^2}{\tau^5} \, and \, \dfrac{\partial}{\partial z} \left( \dfrac{z}{\tau^3} \right) = \dfrac{\tau^2 - 3z^2}{\tau^5}. \nonumber \]
Therefore,
\[ \begin{align*} \text{div } \vecs F_{\tau} &= \dfrac{\tau^2 - 3x^2}{\tau^5} + \dfrac{\tau^2 - 3y^2}{\tau^5} + \dfrac{\tau^2 - 3z^2}{\tau^5} \\[4pt]
&= \dfrac{3\tau^2 - 3(x^2+y^2+z^2)}{\tau^5} \\[4pt]
&= \dfrac{3\tau^2 - 3\tau^2}{\tau^5} = 0. \end{align*}\]
Notice that since the divergence of \(\vecs F_{\tau}\) is zero and \(\vecs E\) is \(\vecs F_{\tau}\) scaled by a constant, the divergence of electrostatic field \(\vecs E\) is also zero (except at the origin).
Flux across a Smooth Surface
Let \(S\) be a connected, piecewise smooth closed surface and let \(\vecs F_{\tau} = \dfrac{1}{\tau^2} \left\langle \dfrac{x}{\tau}, \, \dfrac{y}{\tau}, \, \dfrac{z}{\tau}\right \rangle\). Then,
\[\iint_S \vecs F_{\tau} \cdot d\vecs S = \begin{cases}0, & \text{if }S\text{ does not encompass the origin} \\ 4\pi, & \text{if }S\text{ encompasses the origin.} \end{cases} \nonumber \]
In other words, this theorem says that the flux of \(\vecs F_{\tau}\) across any piecewise smooth closed surface \(S\) depends only on whether the origin is inside of \(S\).
Proof
The logic of this proof follows the logic of [link] , only we use the divergence theorem rather than Green’s theorem.
First, suppose that \(S\) does not encompass the origin. In this case, the solid enclosed by \(S\) is in the domain of \(\vecs F_{\tau}\), and since the divergence of \(\vecs F_{\tau}\) is zero, we can immediately apply the divergence theorem and find that \[\iint_S \vecs F \cdot d\vecs S \nonumber \] is zero.
Now suppose that \(S\) does encompass the origin. We cannot just use the divergence theorem to calculate the flux, because the field is not defined at the origin. Let \(S_a\) be a sphere of radius a inside of \(S\) centered at the origin. The outward normal vector field on the sphere, in spherical coordinates, is
\[\vecs t_{\phi} \times \vecs t_{\theta} = \langle a^2 \cos \theta \, \sin^2 \phi, \, a^2 \sin \theta \, \sin^2 \phi, \, a^2 \sin \phi \, \cos \phi \rangle \nonumber \]
(see [link] ). Therefore, on the surface of the sphere, the dot product \(\vecs F_{\tau} \cdot \vecs N\) (in spherical coordinates) is
\[ \begin{align*} \vecs F_{\tau} \cdot \vecs N &= \left \langle \dfrac{\sin \phi \, \cos \theta}{a^2}, \, \dfrac{\sin \phi \, \sin \theta}{a^2}, \, \dfrac{\cos \phi}{a^2} \right \rangle \cdot \langle a^2 \cos \theta \, \sin^2 \phi, a^2 \sin \theta \, \sin^2 \phi, \, a^2 \sin \phi \, \cos \phi \rangle \\[4pt]
&= \sin \phi ( \langle \sin \phi \, \cos \theta, \, \sin \phi \, \sin \theta, \, \cos \phi \rangle \cdot \langle \sin \phi \, \cos \theta, \sin \phi \, \sin \theta, \, \cos \phi \rangle ) \\[4pt]
&= \sin \phi. \end{align*}\]
The flux of \(\vecs F_{\tau}\) across \(S_a\) is
\[\iint_{S_a} \vecs F_{\tau} \cdot \vecs N dS = \int_0^{2\pi} \int_0^{\pi} \sin \phi \, d\phi \, d\theta = 4\pi. \nonumber \]
Now, remember that we are interested in the flux across \(S\), not necessarily the flux across \(S_a\). To calculate the flux across \(S\), let \(E\) be the solid between surfaces \(S_a\) and \(S\). Then, the boundary of \(E\) consists of \(S_a\) and \(S\). Denote this boundary by \(S - S_a\) to indicate that \(S\) is oriented outward but now \(S_a\) is oriented inward. We would like to apply the divergence theorem to solid \(E\). Notice that the divergence theorem, as stated, can’t handle a solid such as \(E\) because \(E\) has a hole. However, the divergence theorem can be extended to handle solids with holes, just as Green’s theorem can be extended to handle regions with holes. This allows us to use the divergence theorem in the following way. By the divergence theorem,
\[ \begin{align*} \iint_{S-S_a} \vecs F_{\tau} \cdot d\vecs S &= \iint_S \vecs F_{\tau} \cdot d\vecs S - \iint_{S_a} \vecs F_{\tau} \cdot d\vecs S \\[4pt]
&= \iiint_E \text{div } \vecs F_{\tau} \, dV \\[4pt]
&= \iiint_E 0 \, dV = 0. \end{align*}\]
Therefore,
\[\iint_S \vecs F_{\tau} \cdot d\vecs S = \iint_{S_a} \vecs F_{\tau} \cdot d\vecs S = 4\pi, \nonumber \]
and we have our desired result.
\(\Box\)
Now we return to calculating the flux across a smooth surface in the context of electrostatic field \(\vecs E = \dfrac{q}{4\pi \epsilon_0} \vecs F_{\tau} \) of a point charge at the origin. Let \(S\) be a piecewise smooth closed surface that encompasses the origin. Then
\[ \begin{align*} \iint_S \vecs E \cdot d\vecs S &= \iint_S \dfrac{q}{4\pi \epsilon_0} \vecs F_{\tau} \cdot d\vecs S\\[4pt]
&= \dfrac{q}{4\pi \epsilon_0} \iint_S \vecs F_{\tau} \cdot d\vecs S \\[4pt]
&= \dfrac{q}{\epsilon_0}. \end{align*}\]
If \(S\) does not encompass the origin, then
\[\iint_S \vecs E \cdot d\vecs S = \dfrac{q}{4\pi \epsilon_0} \iint_S \vecs F_{\tau} \cdot d\vecs S = 0. \nonumber \]
Therefore, we have justified the claim that we set out to justify: the flux across closed surface \(S\) is zero if the charge is outside of \(S\), and the flux is \(q/\epsilon_0\) if the charge is inside of \(S\).
This analysis works only if there is a single point charge at the origin. In this case, Gauss’ law says that the flux of \(\vecs E\) across \(S\) is the total charge enclosed by \(S\). Gauss’ law can be extended to handle multiple charged solids in space, not just a single point charge at the origin. The logic is similar to the previous analysis, but beyond the scope of this text. In full generality, Gauss’ law states that if \(S\) is a piecewise smooth closed surface and \(Q\) is the total amount of charge inside of \(S\), then the flux of \(\vecs E\) across \(S\) is \(Q/\epsilon_0\).
Example \(\PageIndex{5}\): Using Gauss’ law
Suppose we have four stationary point charges in space, all with a charge of 0.002 Coulombs (C). The charges are located at \((0,0,1), \, (1,1,4), (-1,0,0)\), and \((-2,-2,2)\). Let \(\vecs E\) denote the electrostatic field generated by these point charges. If \(S\) is the sphere of radius \(2\) oriented outward and centered at the origin, then find
\[\iint_S \vecs E \cdot d\vecs S. \nonumber \]
Solution
According to Gauss’ law, the flux of \(\vecs E\) across \(S\) is the total charge inside of \(S\) divided by the electric constant. Since \(S\) has radius \(2\), notice that only two of the charges are inside of \(S\): the charge at \((0,1,1)\) and the charge at \((-1,0,0)\). Therefore, the total charge encompassed by \(S\) is \(0.004\) and, by Gauss’ law,
\[\iint_S \vecs E \cdot d\vecs S = \dfrac{0.004}{8.854 \times 10^{-12}} \approx 4.418 \times 10^9 \, V - m. \nonumber \]
Exercise \(\PageIndex{4}\)
Work the previous example for surface \(S\) that is a sphere of radius 4 centered at the origin, oriented outward.
- Hint
-
Use Gauss’ law.
- Answer
-
\(\approx 6.777 \times 10^9\)
Key Concepts
- The divergence theorem relates a surface integral across closed surface \(S\) to a triple integral over the solid enclosed by \(S\). The divergence theorem is a higher dimensional version of the flux form of Green’s theorem, and is therefore a higher dimensional version of the Fundamental Theorem of Calculus.
- The divergence theorem can be used to transform a difficult flux integral into an easier triple integral and vice versa.
- The divergence theorem can be used to derive Gauss’ law, a fundamental law in electrostatics.
Key Equations
- Divergence theorem \[\iiint_E \text{div } \vecs F \, dV = \iint_S \vecs F \cdot d\vecs S \nonumber \]
Glossary
- divergence theorem
- a theorem used to transform a difficult flux integral into an easier triple integral and vice versa
- Gauss’ law
- if S is a piecewise, smooth closed surface in a vacuum and \(Q\) is the total stationary charge inside of \(S\), then the flux of electrostatic field \(\vecs E\) across \(S\) is \(Q/\epsilon_0\)
- inverse-square law
- the electrostatic force at a given point is inversely proportional to the square of the distance from the source of the charge
\[\iint_{S_{\tau}} \vecs F \cdot d\vecs S = \iiint_{B_{\tau}} \text{div }\vecs F \,dV \approx \iiint_{B_{\tau}} \text{div }\vecs F(P) \,dV.\nonumber \]
Since \( div F(P)\) is a constant,
\[\iiint_{B_{\tau}} \text{div }\vecs F(P) \,dV = \text{div }\vecs F(P) \, V(B_{\tau}).\nonumber \]
Therefore, flux \[\iint_{S_{\tau}} \vecs F \cdot d\vecs S \nonumber \] can be approximated by \(\vecs F(P) \, V(B_{\tau})\). This approximation gets better as the radius shrinks to zero, and therefore
\[\text{div } \vecs F(P) = \lim_{\tau \rightarrow 0} \frac{1}{V(B_{\tau})} \iint_{S_{\tau}} \vecs F \cdot d\vecs S.\nonumber \]
This equation says that the divergence at \(P\) is the net rate of outward flux of the fluid per unit volume.
Using the Divergence Theorem
The divergence theorem translates between the flux integral of closed surface \(S\) and a triple integral over the solid enclosed by \(S\). Therefore, the theorem allows us to compute flux integrals or triple integrals that would ordinarily be difficult to compute by translating the flux integral into a triple integral and vice versa.
Example \(\PageIndex{2}\): Applying the Divergence Theorem
Calculate the surface integral
\[\iint_S \vecs F \cdot d\vecs S, \nonumber \]
where \(S\) is cylinder \(x^2 + y^2 = 1, \, 0 \leq z \leq 2\), including the circular top and bottom, and \(\vecs F = \left\langle \frac{x^3}{3} + yz, \, \frac{y^3}{3} - \sin (xz), \, z - x - y \right\rangle\).
Solution
We could calculate this integral without the divergence theorem, but the calculation is not straightforward because we would have to break the flux integral into three separate integrals: one for the top of the cylinder, one for the bottom, and one for the side. Furthermore, each integral would require parameterizing the corresponding surface, calculating tangent vectors and their cross product..
By contrast, the divergence theorem allows us to calculate the single triple integral
\[\iiint_E \text{div }\vecs F \, dV,\nonumber \]
where \(E\) is the solid enclosed by the cylinder. Using the divergence theorem (Equation \ref{divtheorem}) and converting to cylindrical coordinates, we have
\[ \begin{align*} \iint_S \vecs F \cdot d\vecs S &= \iiint_E \text{div }\vecs F \, dV, \\[4pt]
&= \iiint_E (x^2 + y^2 + 1) \, dV \\[4pt]
&= \int_0^{2\pi} \int_0^1 \int_0^2 (r^2 + 1) \, r \, dz \, dr \, d\theta \\[4pt]
&= \frac{3}{2} \int_0^{2\pi} d\theta \\[4pt]
&= 3\pi. \end{align*}\]


8.6: Acidity of Alpha Hydrogen Atoms- Enolate Ion Formation
After completing this section, you should be able to
- explain why the alpha hydrogens of carbonyl compounds are more acidic than the hydrogens in a typical hydrocarbon.
- list the properties that make lithium diisopropylamide a suitable reagent for converting a wide range of carbonyl compounds into their enolate anions.
- arrange a given list of carbonyl compounds in order of increasing or decreasing acidity.
- determine whether a given carbonyl‑containing compound is more or less acidic than selected other compounds, such as water, ammonia, alcohols, alkanes, alkenes, alkynes and amines.
- explain why dicarbonyl compounds, such as β ‑diketones, are more acidic than compounds that contain only a single carbonyl group.
Make certain that you can define, and use in context, the key terms below.
- β ‑diketone
- β ‑keto ester
α-hydrogens are weakly acidic because their conjugate base, called an enolate, is stabilized though conjugation with the π orbitals of the adjacent carbonyl. Removal of an α-hydrogen creates an sp 2 hybridized carbon anion. The negative charge and lone pair electrons of the carbon anion are contained in an unhybridized p orbital. Orbital overlap with p orbitals from the adjacent carbonyl pi bond allows for the lone pair electrons and negative charge to be shared with the electronegative Carbonyl oxygen, stabilizing the negative charge of the enolate ion. (Figure \(\PageIndex{1}\)).
Enolate conjugation can be represent by two major resonance forms represented by structures A and B which share the negative charge (Shown Below). In resonance form A, the negative charge is on a carbon while in resonance form B the negative change is on an oxygen.
Resonance form B is preferred because oxygen is highly electronegative and better able to stabilize the negative charge. Both resonance form contribute to an overall resonance hybrid but not equally. Due to its stability, resonance form A makes a greater contribution to the resonance hybrid structure. Form A’s increased contribution to the hybrid is seen when considering the electrostatic potential map of an acetone enolate. The red/yellow color shows there is a high electron density around the oxygen. The yellow/green color around the enolate carbon shows that there is a lower but still significant electron density present. Notice that the electron density on the enolate carbon is much larger than the non-enolate carbon (blue color).
The effect of an adjacent carbonyl on the acidity of α-hydrogens is seen when comparing the pK a of aldehydes (~16-18) and ketones (~19-21) to the pK a of an alkane (~50). The ability of the electronegative oxygen to stabilize a negative charge can be seen by considering the pK a of an allyl C-H bond (~42).
Having a C-H bond adjacent to a C=C double bond allows the conjugate base created by deprotonating an allyl C-H bond to be stabilized by conjugation. However, the stability gained only provides a 10 8 increase in acidity when compared to an alkane. In comparison, the presence of an electronegative oxygen allows the acidity of an α-hydrogen to be 10 20 times greater than that of an alkane.
Although α -hydrogens are weakly acidic, typical strong bases such as a hydroxide or alkoxides are only capable of forming the enolate ion in very low concentrations. This can leave a significant concentration of the electrophilic carbonyl carbon remaining, which can react with the bases or the enolate. To achieve the complete deprotonation of aldehyde or ketone reactants to their enolate conjugate bases, a very strong base such as LDA (lithium diisopropylamide) must be used. Complete deprotonation removes the carbonyl groups of the starting material from the reaction mixture and prevents their ability to form unwanted products. Also, the large alkyl substituents sterically hinder the ability of LDA and its corresponding amine (diisopropyl amine) to undergo nucleophilic addition to the carbonyl. In addition to aldehydes and ketones, LDA readily deprotonates a wide variety of carbonyl containing functional groups including esters, amides and nitriles. Ether solvents like tetrahydrofuran (THF) are commonly used for enolate formation reactions using LDA. Certain other strong bases, such as alkyl lithium and Grignard reagents, cannot be used to make enolate anions because they rapidly and irreversibly add to the carbonyl groups. Nevertheless, these very strong bases are useful in the preparation of LDA. By reacting n -butyllithium with diisopropylamine (pK a 36), LDA can be easily formed.
Example
Hydrogen atoms with two or more adjacent carbonyl groups are more acidic than typical α-hydrogens.
The enolate ions of compound such as β -diketones, β -keto-esters, and β -diesters are stabilized through additional resonance forms which share the negative charge with multiple carbonyl oxygens. The acidity of these compounds is increased to the point where weaker bases, such as sodium ethoxide (NaOCH 2 CH 3 ) can be used to form the enolate.
The pK a and acidic hydrogens of multiple functional groups are shown in the table below.
Table \(\PageIndex{1}\) Acidity Constants for Some Organic Compounds:
| Functional Group | Structure | pK a |
|---|---|---|
| carboxylic acid | H O–(C=O)R | 5 |
| nitro | RC H 2 –NO 2 | 9 |
| β -diketone * | R(O=C)–C H 2 –(C=O)R | 9 |
| β -ketoester * | R(O=C)–C H 2 –(C=O)OR | 11 |
| β -diester * | RO(O=C)–C H 2 –(C=O)OR | 13 |
| amide | RN H –(C=O)R | 15 |
| alcohol | RC H 2 –OH | 16 |
| aldehyde | RC H 2 –(C=O)H | 17 |
| ketone | RC H 2 –(C=O)R | 20 |
| thioester | RC H 2 –(C=O)SR | 21 |
| ester | RC H 2 –(C=O)OR | 25 |
| nitrile | RC H 2 –C≡N | 25 |
| sulfone | RC H 2 –SO 2 R | 25 |
| amide | RC H 2 –(C=O)N(CH 3 ) 2 | 30 |
| alkane | C H 3 –R | 50 |
* = a functional group with two ketones.
Identify the most acidic hydrogens in each of the following compounds, and rank the compounds in order of increasing acidity:
Strategy
Hydrogens on carbon next to a carbonyl group are acidic. In general, a β -dicarbonyl compound is most acidic, a ketone or aldehyde is next most acidic, and a carboxylic acid derivative is least acidic. Remember that alcohols, phenols, and carboxylic acids are also acidic because of their –OH hydrogens.
Solution
The acidity order is ( a ) > ( c ) > ( b ). Acidic hydrogens are shown in red.
Identify the most acidic hydrogens in each of the following molecules:
a. \(\ce{CH3CH2CHO}\) b. \(\ce{(CH3)3C–COCH3}\) c. \(\ce{CH3CO2H}\) d. \(\ce{Benzamide}\) e. \(\ce{CH3CH2CH2CN}\) f. \(\ce{CH3CON(CH3)2}\)
- Answer
-
a. \(\ce{CH3CH2CHO}\) b. \(\ce{(CH3)3CCOCH3}\) c. \(\ce{CH3CO2H}\) d. \(\ce{PhCONH2 }\) e. \(\ce{CH3CH2CH2CN}\) f. \(\ce{CH3CON(CH3)2}\)
Draw a resonance structure of the acetonitrile anion, – :CH 2 C≡N, and account for the acidity of nitriles.
- Answer
- – :CH 2 C≡N: <——> CH 2 =C=¨N: –


4.1: Interpreting the Partition Function
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When it is a good approximation to say that the energy of a molecule is the sum of translational, rotational, vibrational, and electronic components, we have
\[{\epsilon }_{i,j,k,m}={\epsilon }_{t,i}+{\epsilon }_{r,j}+{\epsilon }_{v,k}+{\epsilon }_{e,m} \nonumber \]
where the indices \(i\), \(j\), \(k\), and \(m\) run over all possible translational, rotational, vibrational, and electronic quantum states, respectively. Then the partition function for the molecule can be expressed as a product of the individual partition functions \(z_t\), \(z_r\), \(z_v\), and \(z_e\); that is,
\[\begin{align*} z_{\mathrm{molecule}} &=\sum_t{\sum_r{\sum_v{\sum_e{g_{t,i}}}}}g_{r,j}g_{v,k}g_{e,m}\mathrm{exp}\left(\frac{-{\epsilon }_{i,j,k,m}}{kT}\right) \\[4pt] &=\sum_t{g_{t,i}}exp\left(\frac{-{\epsilon }_{t,i}}{kT}\right)\sum_r{g_{r,j}}exp\left(\frac{-{\epsilon }_{r,j}}{kT}\right) \sum_v{g_{v,k}}exp\left(\frac{-{\epsilon }_{v,k}}{kT}\right)\sum_e{g_{e,m}}exp\left(\frac{-{\epsilon }_{e,m}}{kT}\right) \\[4pt] &=z_tz_rz_vz_e \end{align*} \nonumber \]
The magnitude of an individual partition function depends on the magnitudes of the energy levels associated with that kind of motion. Table 1 gives the contributions made to their partition functions by levels that have various energy values.
| \(\epsilon_{\boldsymbol{i}}\) | \(\frac{\boldsymbol{-}\epsilon_{\boldsymbol{i}}}{\boldsymbol{kT}}\) | \(\mathrm{exp}\left(\frac{\boldsymbol{-}\epsilon_{\boldsymbol{i}}}{\boldsymbol{kT}}\right)\) | Type of Motion |
|---|---|---|---|
| \({10}^{-2}\ kT\) | \(-{10}^{-2}\) | \(0.990\) | |
| \({10}^{-1}\ kT\) | \(-{10}^{-1}\) | \(0.905\) | |
| \(kT\) | \(-1\) | \(0.365\) | translational |
| \(5\ kT\) | \(-5\) | \(0.0067\) | rotational |
| \(10\ kT\) | \(-10\) | \(4.5\times {10}^{-5}\) | vibration |
| \(100\ kT\) | \(-100\) | \(3.7\times {10}^{-44}\) | electronic |
We see that only quantum states whose energy is less than \(kT\) can make substantial contributions to the magnitude of a partition function. Very approximately, we can say that the partition function is equal to the number of quantum states for which the energy is less than \(kT\). Each such quantum state will contribute approximately one to the sum that comprises the partition function; the contribution of the corresponding energy level will be approximately equal to its degeneracy. If the energy of a quantum state is large compared to \(kT\), the fraction of molecules occupying that quantum state will be small. This idea is often expressed by saying that such states are “unavailable” to the molecule. It is then said that the value of the partition function is approximately equal to the number of available quantum states. When most energy levels are non-degenerate, we can also say that the value of the partition function is approximately equal to the number of available energy levels.


16.17: Global Business Strategies for Responding to Cultural Differences
- Explain export strategies for global management.
- Explain standardization strategies for global management.
- Explain multidomestic strategies for global management.
- Explain transnational strategies for global management.
Global Business Strategies
A major concern for managers deciding on a global business strategy is the tradeoff between global integration and local responsiveness. Global integration is the degree to which the company is able to use the same products and methods in other countries. Local responsiveness is the degree to which the company must customize their products and methods to meet conditions in other countries. The two dimensions result in four basic global business strategies: export, standardization, multidomestic, and transnational. These are shown in the figure below.
Export Strategy
An export strategy is used when a company is primarily focused on its domestic operations. It does not intend to expand globally but does export some products to take advantage of international opportunities. It does not attempt to customize its products for international markets. It is not interested in either responding to unique conditions in other countries or in creating an integrated global strategy.
Standardization Strategy
A standardization strategy is used when a company treats the whole world as one market with little meaningful variation. The assumption is that one product can meet the needs of people everywhere. Many business-to-business companies can use a standardization strategy. Machines tools and equipment or information technologies are universal and need little customization for local conditions. CEMEX, the Mexico-based cement and building materials company, was able to expand globally using a standardization strategy. Apple uses a standardization strategy because its products do not have to be customized for local users. An iPod will look the same wherever you buy it. Domino’s Pizza also uses a standardization strategy. Although toppings may vary to meet local tastes, the basic recipes are the same and the store model of carryout or delivered pizza is the same everywhere. A standardization strategy produces efficiencies by centralizing many common activities, such as product design, gaining scale economies in manufacturing, simplifying the supply chain, and reducing marketing costs.
Multidomestic Strategy
A multidomestic strategy customizes products or processes to the specific conditions in each country. In the opening example, Lincoln Electric should have used a multidomestic strategy to customize its manufacturing methods to the conditions in each country where it built factories. Retailers often use multidomestic strategies because they must meet local customer tastes. 7-Eleven is an example of a company using a multidomestic strategy. It tailors the product selection, payment methods, and marketing to the values and regulations in each country where it operates. For example, in Japan, 7-Eleven allows customers to pay their utility bills at the store. In a company with a multidomestic strategy, overall management is centralized in the home country but country managers are given latitude to make adaptations. Companies sacrifice scale efficiencies for responsiveness to local conditions. Companies benefit from a multidomestic strategy because country managers understand local laws, customs, and tastes and can decide how to best meet them.
Transnational Strategy
A transnational strategy combines a standardization strategy and a multidomestic strategy. It is used when a company faces significant cost pressure from international competitors but must also offer products that meet local customer needs. A transnational strategy is very difficult to maintain because the company needs to achieve economies of scale through standardization but also be flexible to respond to local conditions. Ford Motor Company is adopting a transnational strategy. Ford is producing a “world car” that has many common platform elements that accommodate a range of add-ons. That way Ford benefits from the standardization of costly elements that the consumer does not see but can add custom elements to meet country laws, can customize marketing to local standards, and can provide unique products to meet local tastes.
Key Points
In today’s economy almost all companies must consider the opportunities presented by globalization, but global operations also present significant risks. Companies must research and plan thoroughly before engaging in international operations. And they must choose a strategy that matches their capabilities and objectives. The economies of standardization and the responsiveness of customization are competing pressures companies must resolve. The appropriate strategic choice is essential for a company to make the right choices.
Contributors and Attributions
- Global Business Strategies for Responding to Cultural Differences. Authored by : John/Lynn Bruton and Lumen Learning. License : CC BY: Attribution
- Image: International Business Strategies. Authored by : John/Lynn Bruton and Lumen Learning. License : CC BY: Attribution


8.1: Arithmetic Sequences
The following sequence of numbers has a pattern you are bound to recognize:
\(2, 4, 6, 8, 10, 12, 14, 16, 18, …\)
Likely, you would describe the sequence in words: the sequence of even numbers. Alternatively, can we describe the sequence mathematically? That is, can we describe the pattern of the sequence of even numbers using a formula? Absolutely! This section will explore arithmetic sequences, how to identify them, mathematically describe their terms, and the relationship between arithmetic sequences and linear functions. Let’s get started!
A sequence is a list of numbers: \(a_1 , a_2, a_3, a_4 , … , a_n, … \) A sequence can be a finite or infinite list. We call \(a_1\) the first term, \(a_2\) the second term, and \(a_n\) the “general term” or the \(n^{\text{th}}\) term. Sequences have a pattern. We describe the pattern in the general term \(a_n\).
For the sequence of even numbers: \(2, 4, 6, 8, 10, …\) the general term \(a_n = 2n\).
The general term \(a_n\) of a sequence is simply a function of \(n\), indicated above as \(f(n)\), where \(n\) is a natural number (or whole number if \(n\) starts with zero).
In the sequence of even numbers, what is the \(20^{\text{th}}\) term in the sequence?
Solution
The general term of the sequence of even numbers is \(a_n = 2n\). Since \(n =\) the term number, we are asked to find \(a_{20}\).
\(\begin{array}&& a_{20} = 2(20) = 40 &\text{Plug in the term-number \(n=20\) into the formula \(a_n=2n\)} \end{array}\)
Answer The \(20^{\text{th}}\) term of the sequence of even numbers is the number \(40\).
If the sequence: \(a_1, a_2, a_3, a_4 , … , a_{n−1}, a_n, …\) exhibits a pattern such that
\[a_n − a_{n−1} = d\]
For all \(n\), then the real number \(d\) is called the common difference , and the sequence is an arithmetic sequence .
A sequence is given. If the sequence is an arithmetic sequence, give the common difference. If the sequence is not an arithmetic sequence, explain how it fails to be arithmetic.
- \(25, 32, 39, 46, 53, 60, …\)
- \(2, 4, 8, 16, 32, …\)
- \(3^2 , 3^4, 3^6, 3^8, 3^{10}, …\)
- \(0, 1, 0, 1, 0, 1, …\)
Solution
- Is the sequence
\(25, 32, 39, 46, 53, 60, … \)
an arithmetic sequence?
\(\begin{array} &a_2 − a_1 &= 32 − 25 &= \textcolor{red}{7} \\ a_3 − a_2 &= 39 − 32 &= \textcolor{red}{7} \\a_4 − a_3 &= 46 − 39 &= \textcolor{red}{7} \\a_5 − a_4 &= 53 − 46 &= \textcolor{red}{7} \\a_6 − a_5 &= 60 − 53 &= \textcolor{red}{7} \end{array}\)
The sequence is arithmetic and the common difference is \(7\).
- Is the sequence
\(2, 4, 8, 16, 32, …\)
an arithmetic sequence?
\(\begin{array} &a_2 − a_1 &= 4 − 2 &= \textcolor{red}{2} \\ a_3 − a_2 &= 8 − 4 &= \textcolor{red}{4} \\a_4 − a_3 &= 16 − 8 &= \textcolor{red}{8} \\a_5 − a_4 &= 32 − 16 &= \textcolor{red}{16} \\ &&\textcolor{red}{2 \neq 4 \neq 8 \neq 16} \end{array}\)
The sequence is not arithmetic. \(a_n − a_{n-1}\) does not yield a common difference.
- Is the sequence
\(3^2 , 3^4, 3^6, 3^8, 3^{10}, …\)
an arithmetic sequence?
\(\begin{array} &3^4 − 3^2 &= 3^2 (3^2 − 1) &= 9 \cdot 8 &= \textcolor{red}{72} \\ 3^6 − 3^4 &= 3^4 (3^2 − 1) &= 81 \cdot 8 &= \textcolor{red}{648} \end{array}\)
Since \(a_3 − a_2 \neq a_2 − a_1\), we conclude the sequence is not arithmetic.
- Is the sequence
\(0, 1, 0, 1, 0, 1, …\)
an arithmetic sequence?
\(\begin{array} 1-0 &= \textcolor{red}{1} \\ 0-1 &= \textcolor{red}{-1} \end{array}\)
Since \(a_3 − a_2 \neq a_2 − a_1\), the sequence is not arithmetic.
Find the General Term of an Arithmetic Sequence
If a sequence is arithmetic, the general term \(a_n\) is determined using the common difference, \(d\), of the sequence. Functions of the form \(y = mx+b\), known as linear functions, have a strong relationship to arithmetic sequences. The slope \(m\) of a linear function is equivalent to the common difference \(d\) of an arithmetic sequence. Let’s compare arithmetic sequences to linear functions to build \(a_n\), the general term of an arithmetic sequence.
Find the general term \(a_n\) of each arithmetic sequence:
- \(4, 7, 10, 13, 16, …\)
- \(100, 80, 60, 40, 20, …\)
Solution
We will create a table of values for each sequence. The first column will be the term number, \(n\), starting with \(n = 1\). The second column will list the terms of the sequence. The common difference is shown on the side of the second column.
- The sequence \(4, 7, 10, 13, 16, …\) has the common difference \(d = 3\). But it’s also the slope \(m\) of the linear function \(f(x) = mx + b\).
\[m = \dfrac{\delta y}{\delta x} = \dfrac{a_n − a_{n-1}}{n − (n − 1))} = \dfrac{d}{1} = d\]
The above table essentially mimics any linear function, \(f(x) = mx+b\).
- Instead of \(x\), sequences use \(n\)−values.
- Instead of \(m =\) slope in linear functions, sequences use \(d =\) common difference.
- Instead of \(b\), a sequence notates the same value with \(a_0\).
If \(a_1\) denotes the first term of a sequence, then the general term of a sequence is:
\[a_n = f(n) = d \cdot n + a_0\]
To find the general term, \(a_n\), we will need to find the value \(a_0\). There are several ways to do this, but perhaps the simplest is to create an extra row where \(n = 0\), then use the common difference to find \(a_0\). The common difference pattern is maintained and \(a_0 + d = a_1\).
Find the value \(a_0\):
\(\begin{array} &&a_0 + 3 &= 4 \\&a_0 + 3 − 3 &= 4 − 3 \\&a_0 &= 1\end{array}\)
The general term of the sequence is:
\(a_n = 3n + 1\)
- Use the same strategy for Example \(8.1.3\)a to solve Example \(8.1.3\)b. Create a table, find the common difference, \(d\), and find the \(a_0\) term of the sequence \(100, 80, 60, 40, 20, …\)
The common difference \(d = −20\). Find the value \(a_0\).
\(\begin{array} &&a_0 − 20 &= 100 \\ &a_0 − 20 + 20 &= 100 + 20 \\ &a_0 &= 120 \end{array}\)
The general term of the sequence is:
\(a_n = −20n + 120\)
Try It! (Exercises)
For #1-5, the general term of a sequence is given.
- List the first \(5\) terms of the sequence: \(a_1, a_2, a_3, a_4, a_5\).
- Is the sequence arithmetic?
- \(a_n = n^2\)
- \(a_n = 4 − 5n\)
- \(a_n = 2^n\)
- \(a_n = \dfrac{1}{2}n\)
- \(a_n = 0.3n + 1\)
For #6-9, a table of values is given. State the general term, \(a_n\), of the arithmetic sequence.
-
\(n\) \(a_n\) \(1\) \(9\) \(2\) \(15\) \(3\) \(21\) \(4\) \(25\)
-
\(n\) \(a_n\) \(1\) \(42\) \(2\) \(38\) \(3\) \(34\) \(4\) \(30\)
-
\(n\) \(a_n\) \(1\) \(7\) \(2\) \(7.25\) \(3\) \(7.5\) \(4\) \(7.75\)
-
\(n\) \(a_n\) \(1\) \(65.4\) \(2\) \(52.2\) \(3\) \(39\) \(4\) \(25.8\)
For #10-15, find the general term of the arithmetic sequence. Assume the first term is \(a_1\).
- \(8, 15, 22, 29, …\)
- \(110, 85, 60, 35, …\)
- \(9, 7.4, 5.8, 4.2, …\)
- \(\dfrac{17}{2} , 8, \dfrac{15}{2} , 7, …\)
- \(−20, −8, 4, 16, 28, …\)
- \(4 \dfrac{1}{2} , 5 \dfrac{1}{4} , 6, 6 \dfrac{3}{4} , …\)
For #16-20, an arithmetic sequence is described. Find the general term \(a_n\).
- The arithmetic sequence has common difference \(d=8\). The first term \(a_1 = 28\).
- The arithmetic sequence has first term \(a_1 = 40\) and second term \(a_2 = 36\).
- The arithmetic sequence has first term \(a_1 = 6\) and third term \(a_3 = 24\).
- The arithmetic sequence has common difference \(d = −2\) and third term \(a_3 = 15\).
- The arithmetic sequence has common difference \(d = 3.6\) and fifth term \(a_5 = 10.2\).
- Explain how the formula for the general term given in this section: \(a_n = d \cdot n + a_0\) is equivalent to the following formula: \(a_n = a_1 + d(n − 1)\)
- Some sequences have a finite number of terms. Find the number of terms in the finite arithmetic sequence: \(3, 17, 31, … ,143\)
- Some sequences have a finite number of terms. Find the number of terms in the finite arithmetic sequence: \(80, 69, 58, … , −52\)
- Create a formula for finding the number of terms a finite arithmetic sequence when given the first and the last term of the sequence. Assume the first term is \(a_1\) and the last term is \(a_k\).


13.11: Natural Physiological Changes with Aging
Our bodies and our sexuality changes as we age. For more on this, see Chapter 11 on Sexuality Through the Lifespan. Here is a general list of some of the natural changes that occur with age related to sexual health.
- Decrease in hormones such as testosterone and estrogen
- Decrease in morning erections
- Decreased libido
- Decreased penile sensitivity
- Direct penile stimulation often required for erection
- Delayed ejaculation/orgasm, anorgasmia
- Less firm erection, some older men full rigidity seconds before ejaculation
- Ejaculatory urgency may be reduced
- Post orgasm/ejaculation, detumescence and testicular descent more rapid
- Less semen, less viable sperm
- Bladder control issues
- Longer refractory period
- Vaginal atrophy
- Thinning of vaginal tissue
- Lack of lubrication leading to vaginal dryness
- Pain with vaginal intercourse
- More time/stimulation needed to reach orgasm
Many of these physiological changes can present challenges, but there is also good news about aging and our sexual health. For some, the wisdom that comes with age, the knowing of one’s body, of a partner’s body, and less self-consciousness, can leave more room for pure sexual joy and freedom. In some cases, these physiological changes can present an opportunity for creativity that can lead to some of the best sex people have ever experienced. If erections are not always dependable for example or vaginas not always open to enveloping sex (aka penetrative) but people still want intimate connection, they become open to the many ways they can experience sensual and erotic pleasure in their bodies, ways that are less focused on one kind of sex, but open to many new ways of giving and receiving sexual pleasure.
It is empowering to be seen by a camera eye that is appreciative of older women’s beauty and experienced expression of sexuality. It is important, though, to understand that my feeling sensual/sexual does not depend on the male gaze. I must love and appreciate myself.
When I look in the mirror, or stretch and exercise, it is my challenge to get past notions of what constitutes a perfect body. The body that is sensitive, alert, alive, open, curious, willing, generous and relaxed is the body that has the most fun, whether hiking or playing pickleball or making love.
The woman who feels ageless is ageless. Jennie Orvino, born 1946


10.11: Variable Stars- One Key to Cosmic Distances
Learning Objectives
By the end of this section, you will be able to:
- Describe how some stars vary their light output and why such stars are important
- Explain the importance of pulsating variable stars, such as cepheids and RR Lyrae-type stars, to our study of the universe
Let’s briefly review the key reasons that measuring distances to the stars is such a struggle. As discussed in The Brightness of Stars, our problem is that stars come in a bewildering variety of intrinsic luminosities. (If stars were light bulbs, we’d say they come in a wide range of wattages.) Suppose, instead, that all stars had the same “wattage” or luminosity. In that case, the more distant ones would always look dimmer, and we could tell how far away a star is simply by how dim it appeared. In the real universe, however, when we look at a star in our sky (with eye or telescope) and measure its apparent brightness, we cannot know whether it looks dim because it’s a low-wattage bulb or because it is far away, or perhaps some of each.
Astronomers need to discover something else about the star that allows us to “read off” its intrinsic luminosity—in effect, to know what the star’s true wattage is. With this information, we can then attribute how dim it looks from Earth to its distance. Recall that the apparent brightness of an object decreases with the square of the distance to that object. If two objects have the same luminosity but one is three times farther than the other, the more distant one will look nine times fainter. Therefore, if we know the luminosity of a star and its apparent brightness, we can calculate how far away it is. Astronomers have long searched for techniques that would somehow allow us to determine the luminosity of a star—and it is to these techniques that we turn next.
Variable Stars
The breakthrough in measuring distances to remote parts of our Galaxy, and to other galaxies as well, came from the study of variable stars. Most stars are constant in their luminosity, at least to within a percent or two. Like the Sun, they generate a steady flow of energy from their interiors. However, some stars are seen to vary in brightness and, for this reason, are called variable stars . Many such stars vary on a regular cycle, like the flashing bulbs that decorate stores and homes during the winter holidays.
Let’s define some tools to help us keep track of how a star varies. A graph that shows how the brightness of a variable star changes with time is called a light curve (Figure \(\PageIndex{1}\)). The maximum is the point of the light curve where the star has its greatest brightness; the minimum is the point where it is faintest. If the light variations repeat themselves periodically, the interval between the two maxima is called the period of the star. (If this kind of graph looks familiar, it is because we introduced it in Diameters of Stars.)
Pulsating Variables
There are two special types of variable stars for which—as we will see—measurements of the light curve give us accurate distances. These are called cepheid and RR Lyrae variables, both of which are pulsating variable stars . Such a star actually changes its diameter with time—periodically expanding and contracting, as your chest does when you breathe. We now understand that these stars are going through a brief unstable stage late in their lives.
The expansion and contraction of pulsating variables can be measured by using the Doppler effect. The lines in the spectrum shift toward the blue as the surface of the star moves toward us and then shift to the red as the surface shrinks back. As the star pulsates, it also changes its overall color, indicating that its temperature is also varying. And, most important for our purposes, the luminosity of the pulsating variable also changes in a regular way as it expands and contracts.
Cepheid Variables
Cepheids are large, yellow, pulsating stars named for the first-known star of the group, Delta Cephei. This, by the way, is another example of how confusing naming conventions get in astronomy; here, a whole class of stars is named after the constellation in which the first one happened to be found. (We textbook authors can only apologize to our readers for the whole mess!)
The variability of Delta Cephei was discovered in 1784 by the young English astronomer John Goodricke. The star rises rather rapidly to maximum light and then falls more slowly to minimum light, taking a total of 5.4 days for one cycle. The curve in Figure \(\PageIndex{1}\)represents a simplified version of the light curve of Delta Cephei.
Several hundred cepheid variables are known in our Galaxy. Most cepheids have periods in the range of 3 to 50 days and luminosities that are about 1000 to 10,000 times greater than that of the Sun. Their variations in luminosity range from a few percent to a factor of 10.
Polaris, the North Star, is a cepheid variable that, for a long time, varied by one tenth of a magnitude, or by about 10% in visual luminosity, in a period of just under 4 days. Recent measurements indicate that the amount by which the brightness of Polaris changes is decreasing and that, sometime in the future, this star will no longer be a pulsating variable. This is just one more piece of evidence that stars really do evolve and change in fundamental ways as they age, and that being a cepheid variable represents a stage in the life of the star.
The Period-Luminosity Relation
The importance of cepheid variables lies in the fact that their periods and average luminosities turn out to be directly related. The longer the period (the longer the star takes to vary), the greater the luminosity. This period-luminosity relation was a remarkable discovery, one for which astronomers still (pardon the expression) thank their lucky stars. The period of such a star is easy to measure: a good telescope and a good clock are all you need. Once you have the period, the relationship (which can be put into precise mathematical terms) will give you the luminosity of the star.
Let’s be clear on what that means. The relation allows you to essentially “read off” how bright the star really is (how much energy it puts out). Astronomers can then compare this intrinsic brightness with the apparent brightness of the star. As we saw, the difference between the two allows them to calculate the distance.
The relation between period and luminosity was discovered in 1908 by Henrietta Leavitt (Figure \(\PageIndex{2}\)), a staff member at the Harvard College Observatory (and one of a number of women working for low wages assisting Edward Pickering, the observatory’s director; see Annie Cannon: Classifier of the Stars). Leavitt discovered hundreds of variable stars in the Large Magellanic Cloudand Small Magellanic Cloud, two great star systems that are actually neighboring galaxies (although they were not known to be galaxies then). A small fraction of these variables were cepheids (Figure \(\PageIndex{3}\)).
These systems presented a wonderful opportunity to study the behavior of variable stars independent of their distance. For all practical purposes, the Magellanic Clouds are so far away that astronomers can assume that all the stars in them are at roughly the same distance from us. (In the same way, all the suburbs of Los Angeles are roughly the same distance from New York City. Of course, if you are in Los Angeles, you will notice annoying distances between the suburbs, but compared to how far away New York City is, the differences seem small.) If all the variable stars in the Magellanic Clouds are at roughly the same distance, then any difference in their apparent brightnesses must be caused by differences in their intrinsic luminosities.
Leavitt found that the brighter-appearing cepheids always have the longer periods of light variation. Thus, she reasoned, the period must be related to the luminosity of the stars. When Leavitt did this work, the distance to the Magellanic Clouds was not known, so she was only able to show that luminosity was related to period. She could not determine exactly what the relationship is.
To define the period-luminosity relation with actual numbers (to calibrate it), astronomers first had to measure the actual distances to a few nearby cepheids in another way. (This was accomplished by finding cepheids associated in clusters with other stars whose distances could be estimated from their spectra, as discussed in the next section of this chapter.) But once the relation was thus defined, it could give us the distance to any cepheid, wherever it might be located (Figure \(\PageIndex{4}\)).
Here at last was the technique astronomers had been searching for to break the confines of distance that parallax imposed on them. Cepheids can be observed and monitored, it turns out, in many parts of our own Galaxy and in other nearby galaxies as well. Astronomers, including Ejnar Hertzsprung and Harvard’s Harlow Shapley, immediately saw the potential of the new technique; they and many others set to work exploring more distant reaches of space using cepheids as signposts. In the 1920s, Edwin Hubble made one of the most significant astronomical discoveries of all time using cepheids, when he observed them in nearby galaxies and discovered the expansion of the universe. As we will see, this work still continues, as the Hubble Space Telescope and other modern instruments try to identify and measure individual cepheids in galaxies farther and farther away. The most distant known variable stars are all cepheids, with some about 60 million light-years away.
John Goodricke
The brief life of John Goodricke (Figure \(\PageIndex{5}\)) is a testament to the human spirit under adversity. Born deaf and unable to speak, Goodricke nevertheless made a number of pioneering discoveries in astronomy through patient and careful observations of the heavens.
Born in Holland, where his father was on a diplomatic mission, Goodricke was sent back to England at age eight to study at a special school for the deaf. He did sufficiently well to enter Warrington Academy, a secondary school that offered no special assistance for students with handicaps. His mathematics teacher there inspired an interest in astronomy, and in 1781, at age 17, Goodricke began observing the sky at his family home in York, England. Within a year, he had discovered the brightness variations of the star Algol and suggested that an unseen companion star was causing the changes, a theory that waited over 100 years for proof. His paper on the subject was read before the Royal Society (the main British group of scientists) in 1783 and won him a medal from that distinguished group.
In the meantime, Goodricke had discovered two other stars that varied regularly, Beta Lyrae and Delta Cephei, both of which continued to interest astronomers for years to come. Goodricke shared his interest in observing with his older cousin, Edward Pigott, who went on to discover other variable stars during his much longer life. But Goodricke’s time was quickly drawing to a close; at age 21, only 2 weeks after he was elected to the Royal Society, he caught a cold while making astronomical observations and never recovered.
Today, the University of York has a building named Goodricke Hall and a plaque that honors his contributions to science. Yet if you go to the churchyard cemetery where he is buried, an overgrown tombstone has only the initials “J. G.” to show where he lies. Astronomer Zdenek Kopal, who looked carefully into Goodricke’s life, speculated on why the marker is so modest: perhaps the rather staid Goodricke relatives were ashamed of having a “deaf-mute” in the family and could not sufficiently appreciate how much a man who could not hear could nevertheless see.
RR Lyrae Stars
A related group of stars, whose nature was understood somewhat later than that of the cepheids, are called RR Lyrae variables, named for the star RR Lyrae, the best-known member of the group. More common than the cepheids, but less luminous, thousands of these pulsating variables are known in our Galaxy. The periods of RR Lyrae stars are always less than 1 day, and their changes in brightness are typically less than about a factor of two.
Astronomers have observed that the RR Lyrae stars occurring in any particular cluster all have about the same apparent brightness. Since stars in a cluster are all at approximately the same distance, it follows that RR Lyrae variables must all have nearly the same intrinsic luminosity, which turns out to be about 50 L Sun. In this sense, RR Lyrae stars are a little bit like standard light bulbs and can also be used to obtain distances, particularly within our Galaxy. Figure \(\PageIndex{6}\) displays the ranges of periods and luminosities for both the cepheids and the RR Lyrae stars.
Key Concepts and Summary
Cepheids and RR Lyrae stars are two types of pulsating variable stars. Light curves of these stars show that their luminosities vary with a regularly repeating period. RR Lyrae stars can be used as standard bulbs, and cepheid variables obey a period-luminosity relation, so measuring their periods can tell us their luminosities. Then, we can calculate their distances by comparing their luminosities with their apparent brightnesses, and this can allow us to measure distances to these stars out to over 60 million light-years.
Glossary
- cepheid
- a star that belongs to a class of yellow supergiant pulsating stars; these stars vary periodically in brightness, and the relationship between their periods and luminosities is useful in deriving distances to them
- light curve
- a graph that displays the time variation of the light from a variable or eclipsing binary star or, more generally, from any other object whose radiation output changes with time
- period-luminosity relation
- an empirical relation between the periods and luminosities of certain variable stars
- pulsating variable star
- a variable star that pulsates in size and luminosity
- RR Lyrae
- one of a class of giant pulsating stars with periods shorter than 1 day, useful for finding distances
Contributors and Attributions
-
Andrew Fraknoi (Foothill College), David Morrison (NASA Ames Research Center), Sidney C. Wolff (National Optical Astronomy Observatory) with many contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/astronomy ).


24.1: Introduction (2016)
Henry David Thoreau’s “Where I Lived and What I Lived For” is an example of how one person can improve the way they interact with their environment, their society, and improve their own well-being. It describes one man’s way of existing or being alive and his attempt to persuade his readers to live as he did to create a better world. Thoreau includes themes in Walden that are timeless and relevant today and that exist throughout many cultures, and different societies.
Thoreau made the decision to live in a cabin near Walden Pond because he “wished to live deliberately” (Thoreau 1028). Walden is not about Thoreau trying to isolate himself from society or becoming a hermit. He wanted to create an “experiment” to see if it was possible to live the way one wished rather than the way society expected one to live (1025). Thoreau did not live in the woods to be self-sufficient, but to be self-reliant. He wanted to learn more about himself rather than focus on the material gains of his crops. While looking at some local farms he was interested in the landowner told Thoreau, “its bounding on the river, which… protected it by its fogs from frosts in the spring, though that was nothing to me” (1024). Thoreau was clearly not interested in making a fortune in agriculture or isolating himself from people. Walden Pond was only a couple miles from Concord, Massachusetts and he often visited friends and family. The land the cabin sat on belonged to Thoreau’s friend Ralph Waldo Emerson (1026).
Thoreau made multiple references to the Revolutionary War in Walden which adds the theme of civil disobedience to his writing. He moved into the cabin on Walden Pond on July 4, 1845 and it was only a short distance from the Concord Battle Ground (1026). While he did not wish to isolate himself completely from society Thoreau did feel that there was something wrong in the way that people in society lived their everyday lives. He wanted to “wake” Americans up and he decided to do this by leading by example (1028). And he even went so far as to compare the cultural norm to a disease, “We have the Saint Vitus’ dance and cannot possibly keep our heads still,” saying that people were moving uncontrollably and without reason (1029-1030). He wanted to put a stop to senseless action and endeavors reveal a path towards conscious living. It did not necessarily have to be separate from society, actually Thoreau wanted to see this change in society.
While Thoreau preferred to write about America and his homeland he did incorporate some global values into Walden which are shown in his references to Hinduism Greek mythology. He quotes the Hindu triad when he says, “from the circumstance in which it is placed, mistakes its own character, until the truth is revealed to it by some holy teacher, and then it knows itself to be Brahme”(1031). Thoreau was using this text from another culture to show how he wanted to be a spiritual leader to his readers and show them the way to learning about themselves.
We are reading Henry David Thoreau’s Walden because it encapsulates one man’s perception of a world that existed hundreds of years ago. While the lifestyle has changed, from the day Thoreau wrote his famous text to today, the themes and ideas still remain relatively similar. Studying this text is gaining insight from a man who views the world with a sense of purpose and that embracing the moment of time that you’re in is all we can do and all we should do. Thoreau intends his readers to come from all fragments of time and apply his outtake to their own lives in order to keep society afloat.
Work Cited
Thoreau, Henry David. “Walden: Where I Lived and What I Lived For.” Comp. Abby Goode. Currents In American Literature. N.p.: LAD Custom, 2016. 95-158. Print.


2.3: Critical Thinking
Writing and Critical Thinking
We have seen how writing in college requires you to use higher order thinking skills such as analyzing, evaluating, and creating. These higher level abilities require critical thinking, which is a core skill emphasized by college professors, not just for college academic life, but also for navigating the choices you make as a citizen and member of your family and different communities. This is because critical thinking skills are perhaps the most fundamental skills involved in making judgments and solving problems. You use them every day, and you can continue improving them.
The ability to think critically about a matter—to analyze a question, situation, or problem down to its most basic parts—is what helps us evaluate the accuracy and truthfulness of statements, claims, and information we read and hear. It is the sharp knife that, when honed, separates fact from fiction, honesty from lies, and the accurate from the misleading. We all use this skill to one degree or another almost every day. For example, we use critical thinking every day as we consider the latest consumer products and why one particular product is the best among its peers. Is it a quality product because a celebrity endorses it? Because a lot of other people may have used it? Because it is made by one company versus another? Or perhaps because it is made in one country or another? These are questions representative of critical thinking.
The academic setting demands more of us in terms of critical thinking than everyday life. It demands that we evaluate information and analyze myriad issues. It is the environment where our critical thinking skills can be the difference between success and failure. In this environment, we must consider information in an analytical, critical manner. We must ask questions—What is the source of this information? Is this source an expert one, and what makes that so? Are there multiple perspectives to consider about an issue? Do multiple sources agree or disagree on an issue? Does quality research substantiate information or opinion? Do I have any personal biases that may affect my consideration of this information?
It is only through purposeful, frequent, intentional questioning such as this that we can sharpen our critical thinking skills and improve as students, learners and researchers.
—Dr. Andrew Robert Baker, Foundations of Academic Success: Words of Wisdom
Defining Critical Thinking
Thinking comes naturally. You don’t have to make it happen—it just does. But you can make it happen in different ways. For example, you can think positively or negatively. You can think with “heart,” and you can think with rational judgment. You can also think strategically and analytically, as well as mathematically and scientifically. These are a few of multiple ways in which the mind can process thought.
What are some forms of thinking you use? When do you use them, and why?
As a college student, you are tasked with engaging and expanding your thinking skills. One of the most important of these skills is critical thinking. Critical thinking is important because it relates to nearly all tasks, situations, topics, careers, environments, challenges, and opportunities. It’s not restricted to a particular subject area.
Critical thinking is clear, reasonable, reflective thinking focused on deciding what to believe or do. It means asking probing questions like, “How do we know?” or “Is this true in every case or just in this instance?” It involves being skeptical and challenging assumptions, rather than simply memorizing facts or blindly accepting what you hear or read.
Imagine, for example, that you’re reading a history textbook. You wonder who wrote it and why because you detect certain assumptions in the writing. You find that the author has a limited scope of research focused only on a particular group within a population. In this case, your critical thinking reveals that there are “other sides to the story.”
Who are critical thinkers, and what characteristics do they have in common? Critical thinkers are usually curious and reflective people. They like to explore and probe new areas and seek knowledge, clarification, and new solutions. They ask pertinent questions, evaluate statements and arguments, and they distinguish between facts and opinion. They are also willing to examine their own beliefs, possessing a manner of humility that allows them to admit lack of knowledge or understanding when needed. They are open to changing their mind. Perhaps most of all, they actively enjoy learning, and seeking new knowledge is a lifelong pursuit.
This may well be you!
No matter where you are on the road to being a critical thinker, you can always more fully develop your skills. Doing so will help you develop more balanced arguments, express yourself clearly, read critically, and absorb important information efficiently. Critical thinking skills will help you in any profession or any circumstance of life, from science to art to business to teaching.
| Critical Thinking IS | Critical Thinking is NOT |
|---|---|
| Skepticism | Memorizing |
| Examining assumptions | Group thinking |
| Challenging reasoning | Blind acceptance of authority |
| Uncovering biases | Biased viewpoints |
Critical Thinking in Action
The following video, from Lawrence Bland, presents the major concepts and benefits of critical thinking.
Critical Thinking and Logic
Critical thinking is fundamentally a process of questioning information and data. You may question the information you read in a textbook, or you may question what a politician or a professor or a classmate says. You can also question a commonly held belief or a new idea. With critical thinking, anything and everything is subject to question and examination.
Logic’s Relationship to Critical Thinking
The word logic comes from the Ancient Greek logike , referring to the science or art of reasoning. Using logic, a person evaluates arguments and strives to distinguish between good and bad reasoning, or between truth and falsehood. Using logic, you can evaluate ideas or claims people make, make good decisions, and form sound beliefs about the world. [1]
Questions of Logic in Critical Thinking
Let’s use a simple example of applying logic to a critical-thinking situation. In this hypothetical scenario, a man has a PhD in political science, and he works as a professor at a local college. His wife works at the college, too. They have three young children in the local school system, and their family is well known in the community.
The man is now running for political office. Are his credentials and experience sufficient for entering public office? Will he be effective in the political office? Some voters might believe that his personal life and current job, on the surface, suggest he will do well in the position, and they will vote for him.
In truth, the characteristics described don’t guarantee that the man will do a good job. The information is somewhat irrelevant. What else might you want to know? How about whether the man had already held a political office and done a good job? In this case, we want to ask, "How much information is adequate in order to make a decision based on logic instead of assumptions?"
The following questions, also presented in Figure 1, below, are ones you may apply to formulate a logical, reasoned perspective in the above scenario or any other situation:
- What’s happening? Gather the basic information and begin to think of questions.
- Why is it important? Ask yourself why it’s significant and whether or not you agree.
- What don’t I see? Is there anything important missing?
- How do I know? Ask yourself where the information came from and how it was constructed.
- Who is saying it? What’s the position of the speaker and what is influencing them?
- What else? What if? What other ideas exist and are there other possibilities?
Problem-Solving with Critical Thinking
For most people, a typical day is filled with critical thinking and problem-solving challenges. In fact, critical thinking and problem-solving go hand-in-hand. They both refer to using knowledge, facts, and data to solve problems effectively. But with problem-solving, you are specifically identifying, selecting, and defending your solution. Below are some examples of using critical thinking to problem solve:
- Your roommate was upset and said some unkind words to you, which put a crimp in your relationship. You try to see through the angry behaviors to determine how you might best support your roommate and help bring your relationship back to a comfortable spot.
- Your campus club has been languishing because of lack of participation and funds. The new club president, though, is a marketing major and has identified some strategies to interest students in joining and supporting the club. Implementation is forthcoming.
- Your final art class project challenges you to conceptualize form in new ways. On the last day of class when students present their projects, you describe the techniques you used to fulfill the assignment. You explain why and how you selected that approach.
- Your math teacher sees that the class is not quite grasping a concept. She uses clever questioning to dispel anxiety and guide you to new understanding of the concept.
- You have a job interview for a position that you feel you are only partially qualified for although you really want the job and you are excited about the prospects. You analyze how you will explain your skills and experiences in a way to show that you are a good match for the prospective employer.
 But there are some gaps between what you want and what you feel you can afford. You analyze your income, savings, and budget to better calculate what you will need to stay in college and maintain your desired level of spending.
Problem-Solving Action Checklist
Problem-solving can be an efficient and rewarding process, especially if you are organized and mindful of critical steps and strategies. Remember, too, to assume the attributes of a good critical thinker. If you are curious, reflective, knowledge-seeking, open to change, probing, organized, and ethical, your challenge or problem will be less of a hurdle, and you’ll be in a good position to find intelligent solutions.
| STRATEGIES | ACTION CHECKLIST [2] | |
|---|---|---|
| 1 | Define the problem |
|
| 2 | Identify available solutions |
|
| 3 | Select your solution |
|
The following video demonstrates problem solving in our personal lives.
It's Not About the Nail by Jason Headley. All Rights Reserved. Standard YouTube license.
While this video is a lighthearted take on relationship problems, it is true that problem solving in any relationship can be difficult, whether that relationship is between two people in an intimate relationship, or between two roommates, or between two friends.
Example of Relationship Problem Solving using Critical Thinking
Anika’s roommate, Eli, was upset and he said some unkind words to her, which put a crimp in the relationship. Anika wants to try to see through the angry behaviors to determine how she might best support Eli and help bring the relationship back to a comfortable spot. How should Anika proceed?
First, let’s help Anika identify the problem. How would you describe the problem? Here are some options:
- Eli behaved inappropriately
- Anika is being too sensitive
- Anika and Eli are probably not a great fit as roommates
The best problem description is the first one – Eli behaved inappropriately. All of us at some point get upset and say things at times that hurt others. That doesn’t mean it’s acceptable, but it also doesn’t mean that Anika and Eli are unable to solve the problem and continue their relationship.
Now that Anika has identified the problem – Eli behaved inappropriately – what are some possible solutions? Here are some options:
- Anika should ignore Eli – eventually he’ll figure out that he did something wrong and apologize.
- Anika should ask Eli to sit down and discuss the problem right away.
- Anika should give Eli a little space and then talk to him.
Ignoring Eli is probably not a good solution. While it’s perfectly natural for Anika to be angry about Eli’s unkind words and want to ignore him, that doesn’t address the problem. The other two options are better solutions.
Think through the pros and cons of each of the two possible solutions. Which one do you think is best?
It’s a great idea for Anika to sit down and discuss the problem with Eli – this is a much healthier option than ignoring the problem. Whether or not she talks to Eli right away is likely a judgment call depending on the situation. If Eli is really upset, it might be better to wait a little while until he calms down and can have a more productive discussion.
Anika observed that Eli seemed pretty upset, and so decided to wait a few hours before talking with him. When she did talk to him later that afternoon, Anika told Eli that she valued his friendship, but that his unkind words were hurtful to her.
Eli explained that he has been getting continually more upset since he got a poor grade on his French mid-term. He needs to pass the class for his major, and has been stressed about the situation. He apologized for his earlier words, and Anika forgave him.
By using problem-solving skills, Anika was able to handle this situation in an effective way that resulted in a relatively quick resolution rather than letting it escalate into a more difficult situation that would be harder to resolve and would likely result in more pain on both sides of the relationship.
Exercise
Exercise 1
Use the "Problem Solving Action Checklist" to find a way solve the following problem.
Problem: You are overwhelmed with both work and school and don't know what to do.  How can you solve this problem and decrease your stress levels?
Define the problem:
Identify available solutions:
Select your solution:
Explain why you selected this solution:
Contributors
- Adapted from Critical Thinking in Basic Reading and Writing. Provided by: Lumen Learning. License: CC BY: Attribution.


8: Carl Rogers and Abraham Maslow
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- 8.2: Carl Rogers and Humanistic Psychology
- Carl Rogers is the psychologist many people associate first with humanistic psychology, but he did not establish the field in the way that Freud established psychoanalysis. A few years older than Abraham Maslow, and having moved into clinical practice more directly, Rogers felt a need to develop a new theoretical perspective that fit with his clinical observations and personal beliefs. Thus, he was proposing a humanistic approach to psychology and, more specifically, psychotherapy before Maslow.
-
- 8.3: Abraham Maslow and Holistic-Dynamic Psychology
- Maslow stands alongside Rogers as one of the founders of humanistic psychology. Although he began his career working with two of the most famous experimental psychologists in America, he was profoundly influenced by the events that led into World War II. He became devoted to studying the more virtuous aspects of personality, and he may be viewed as one of the founders of positive psychology.