# Pg. 332 #29, 41, 53, 71, 73, 83, 87, 89, 93, 101, 102
---
Eric Nguyen  
20 Dec 2018


```
import numpy as np
import matplotlib.pyplot as plt
```

### *Find each logarithm*. *Round to six decimal places.*

#### 29. $\ln{5894}$


```
ans29 = round(np.log(5894), 6)
ans29
```

#### Answer 29:
> $8.68169$

### *Solve for $t$*.

#### 41. $e^{-0.02t} = 0.06$

$$
\begin{align}
e^{-0.02t} &= 0.06 \\
-0.02t &= \ln{0.06} \\
t &= \frac{\ln{0.06}}{-0.02}
\end{align}
$$


```
ans41 = np.log(0.06) / (-0.02)
ans41
```


```
# Verify
np.power(np.e, -0.02 * ans41)
```

#### Answer 41:
> $t = 140.670536$

### *Differentiate*.

#### 53. $y = \ln{\frac{x^2}{4}} \left(\text{Hint:} \ln{\frac{A}{B}} = \ln{A} - \ln{B}\right)$

$$
\begin{align}
y &= \ln{x^2} - \ln{4} \\
y' &= \frac{2x}{x^2} \\
&= \frac{2}{x}
\end{align}
$$


```
def pltfn(fn, xlim = (-10, 10), smoothness = 100, color = False):
    xp = np.linspace(xlim[0], xlim[1], smoothness)
    if (color):
      plt.plot(xp, fn(xp), color = color)
    else:
      plt.plot(xp, fn(xp))
qtn53 = lambda x: np.log(np.power(x, 2) / 4)
ans53 = lambda x: 2 / x
pltfn(qtn53)
pltfn(ans53)
```

#### Answer 53:
> $y' = \dfrac{2}{x}$

#### 59. $g(x) = e^x \ln{x^2}$

$$\begin{align}
g'(x) &= \frac{d}{dx} e^x \cdot \frac{d}{dx} \ln{x^2} \\
&= e^x \cdot \frac{2}{x} \\
&= \frac{2e^x}{x}
\end{align}$$


```
qtn59 = lambda x: np.power(np.e, x) * np.log(np.power(x, 2))
ans59 = lambda x: (2 * np.power(np.e, x)) / x
pltfn(qtn59, (5, 10))
pltfn(ans59, (5, 10))
```

#### Answer 59:
> $g'(x) = \dfrac{2e^x}{x}$

#### 71. Find the equation of the line tangent to the graph of $y = (\ln{x})^2$ at $x = 3$.                                                  

$$\begin{align}
y' &= 2 \cdot (\ln{x}) \cdot \frac{1}{x} \\
&= \frac{2\ln{x}}{x} \\
y &= f(x) \\
g(x) &= f'\left(a\right)\left(x-a\right)+f\left(a\right) \quad &\text{Tangent line equation.} \\
f'(3) &= \frac{2\ln{3}}{3} \quad &\text{Find the slope.} \\
&\approx 0.732408 \\
f(3) &\approx 1.206949 \\
g(x) &= 0.732408(x - 3) + 1.206949 \\
\end{align}$$


```
fn71 = lambda x: np.power(np.log(x), 2)
dfn71 = lambda x: (2 * np.log(x)) / x
```


```
slope71 = (2 * np.log(3)) / 3
slope71
```


```
fn71(3)
```


```
ans71 = lambda x: slope71 * (x - 3) + fn71(3)
pltfn(fn71, (0.5, 5))
pltfn(dfn71, (0.5, 5))
pltfn(ans71, (0.5, 5))
```

#### Answer 71:
> $g(x) = 0.732408(x - 3) + 1.206949$

#### 73. <span style="color: #0af">Advertising.</span>&emsp;A model for consumers' response to advertising is given by

$$ N(a) = 2000 + 500 \ln{a}, \quad a \geq 1,$$

#### where $N(a)$ is the number of units sold and $a$ is the amount spent on advertising, in thousands of dollars.

#### a) How many units were sold after spending $\$1000$ on advertising?


```
fn73 = lambda a: 2000 + 500 * np.log(a)
fn73(1000)
```

#### Answer 73a:
>  $N(1000) \approx 5453$

#### b) Find $N'(a)$ and $N'(10)$.

$$\begin{align}
N'(a) &= \frac{500}{a} \\
\end{align}$$


```
dfn73 = lambda a: 500 / a
dfn73(10)
```


```
pltfn(fn73, (0.25, 5))
pltfn(dfn73, (0.25, 5))
```


```
pltfn(dfn73, (8, 12), color="green")
plt.scatter(10, dfn73(10), color="green")
```

#### c) Find the maximum and minimum values, if they exist.

$$\begin{align}
N'(a) &= \frac{500}{a} \\
0 &= \frac{500}{a} & \text{Find the zeroes.} \\
0 &\neq 500
\end{align}$$

#### Answer 73c:
> There are no minimum or maximum values.

#### d) Find $\underset{a\to\infty}{\lim} N'(a)$. Discuss whether it makes sense to continue to spend more and more dollars on advertising.

$$\begin{align}
\underset{a\to\infty}{\lim} N'(a) &= \frac{500}{\infty} \\
\underset{a\to\infty}{\lim} N'(a) &= 0
\end{align}$$

#### Answer 73d:
> It does not make sense to spend more and more dollars on advertising, in thousands of dollars, as the effect of advertising rapidly diminishes for each thousand of dollars spent.

#### 83. <span style="color: #0af">Forgetting.</span>&emsp;Students in a zoology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After $t$ months, the average score $S(t)$, as a percentage was found to be given by

$$ S(t) = 78 - 15 \ln{\left(t + 1\right)}, \quad t \geq 0. $$

#### a) What was the average score when they initially took the test, $t = 0$?


```
fn83 = lambda t: 78 - 15 * np.log(t + 1)
pltfn(fn83, (0, 200))
ans83a = fn83(0)
ans83a
```

#### Answer 83a:
> $S(0) = 78.0$

#### b) What was the average score after 4 months?


```
ans83b = fn83(4)
ans83b
```

#### Answer 83b:
> $S(4) \approx 53.858$

#### c) What was the average score after 24 months?


```
ans83c = fn83(24)
ans83c
```

#### Answer 83c:
> $S(24) \approx 29.727$

#### d) What percentage of their original answers did the students retain after 2 years (24 months)?


```
ans83c / ans83a
```




    0.38098541829457677



#### Answer 83d:
> $38.1\%$

#### e) Find $S'(t)$.

$$\begin{align}
S(t) &= 78 - 15 \ln{\left(t + 1\right)}, & t \geq 0 \\
S'(t) &= -15 \cdot \frac{1}{t + 1}, & t \geq 0 \\
&= \frac{-15}{t + 1}, & t \geq 0
\end{align}$$


```
dfn83 = lambda t: -15 / (t + 1)
pltfn(fn83, (0, 200))
pltfn(dfn83, (0, 200))
```

#### f) Find the maximum and minimum values, if they exist.

$$\begin{align}
S'(t) &= \frac{-15}{t + 1} \\
0 &= \frac{-15}{t + 1} \\
0 &\neq -15
\end{align}$$

#### Answer 83f:
> There are no maximum or minimum values.

### *Solve for $t$*.
#### 87. $P = P_0e^{-kt}$

$$\begin{align}
\frac{P}{P_0} &= e^{-kt} \\
\ln{\left(\frac{P}{P_0}\right)} &= -kt \\
t &= \frac{\ln{\left(\frac{P}{P_0}\right)}}{-k}
\end{align}$$

### *Differentiate*.

#### 93. $f(t) = \ln{\dfrac{1 - t}{1 + t}}$

#### Answer 93:
$$\begin{align}
f(t) &= \ln{\left(1 - t \right)} - \ln{\left(1 + t\right)} \\
f'(t) &= \frac{-1}{1 - t} - \frac{1}{1 + t} \\
f'(t) &= \frac{-\left(1 + t\right)}{\left(1 - t\right)\left(1 + t\right)} - 
\frac{\left(1 - t\right)}{\left(1 + t\right)\left(1 - t\right)} \\
f'(t) &= \frac{-1 - t - 1 + t}{\left(1 - t\right)\left(1 + t\right)} \\
f'(t) &= \frac{-2}{\left(1 - t\right)\left(1 + t\right)} \\
\end{align}$$

#### 101. $f(x) = \ln{\dfrac{1 + \sqrt{x}}{1 - \sqrt{x}}}$

#### Answer 101:
$$\begin{align}
f(x) &= \ln{\left(1 + \sqrt{x}\right)} - \ln{\left(1 - \sqrt{x}\right)} \\
\frac{d}{dx} \sqrt{x} &= \frac{1}{2\sqrt{x}} \\
f'(x) &= \frac{\frac{1}{2\sqrt{x}}}{1 + \sqrt{x}} - \frac{-\frac{1}{2\sqrt{x}}}{1 - \sqrt{x}} \\
f'(x) &= \frac{1}{\left(2\sqrt{x}\right)\left(1 + \sqrt{x}\right)} + \frac{1}{\left(2\sqrt{x}\right)\left(1 - \sqrt{x}\right)} \\
f'(x) &= \frac{\left(1 - \sqrt{x}\right)}{\left(2\sqrt{x}\right)\left(1 + \sqrt{x}\right)\left(1 - \sqrt{x}\right)} + \frac{\left(1 + \sqrt{x}\right)}{\left(2\sqrt{x}\right)\left(1 - \sqrt{x}\right)\left(1 + \sqrt{x}\right)} \\
f'(x) &= \frac{2}{\left(2\sqrt{x}\right)\left(1 + \sqrt{x}\right)\left(1 - \sqrt{x}\right)} \\
f'(x) &= \frac{1}{\sqrt{x}\left(1 - \sqrt{x}\right)\left(1 + \sqrt{x}\right)} \\
f'(x) &= \frac{1}{\sqrt{x}\left(1 - x\right)} \\
\end{align}$$

#### 102. $f(x) = \ln{\left(\ln{x}\right)}^3$

#### Answer 102:
$$\begin{align}
f(x) &= 3 \ln{\left(\ln{x}\right)} \\
f'(x) &= \frac{\frac{3}{x}}{\ln{x}} \\
f'(x) &= \frac{3}{x\ln{x}}
\end{align}$$

a22cb2f3ff24a984f5196cd771571186513e8167
ipynb
Jupyter Notebook
2018-12/2018-12-20.ipynb
airicbear/calculus-homework
a765d3ba35b2b3794b9b2cce038152682eeb2cb8
MIT
2018-12/2018-12-20.ipynb
airicbear/calculus-homework
a765d3ba35b2b3794b9b2cce038152682eeb2cb8
MIT
2019-02-04T07:00:05.000Z
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airicbear/calculus-homework
a765d3ba35b2b3794b9b2cce038152682eeb2cb8
MIT
Qwen/Qwen-72B
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2. YES