<s>[INST] Implicit differentiation question with a ladder and a house. A $13$ ft ladder is leaning against a house when its base starts to slide away from the house, at the instant when the base is $12$ ft from the house, the base is moving at a rate of $5$ ft/sec.
Question:
Let $ \theta $ be the angle between the ladder and the ground. at what rate is $\theta$ changing when the base is $12$ ft form the house?
I said $ \tan (\theta) = y x^{-1} $ then used implicit differentiation to take the derivative with respect to time.
\begin{align*}
\frac{d \theta}{dt} \sec^{2} (\theta) &= \frac {dy}{dt}\left( \frac {dx}{dt} -x^{-2} \right)\\
\implies\frac{d \theta}{dt} &= \cos^{2}(\theta) \left(\frac {dy}{dt} \left( \frac {dx}{dt} -x^{-2}\right) \right)
\end{align*}
Does this make sense or did I mess something up?
 [/INST] With $\theta = \arctan \frac{y}{x}$ we get 
$$d \theta =d \arctan \frac{y}{x} = \frac{ d (\frac{y}{x})}{1+ (\frac{y}{x})^2}= \frac{ x \cdot d y - y \cdot d x}{x^2 + y^2}$$
Now use $x^2 + y^2 = l^2= $ const. and get 
$$ x dx + y d y = 0$$
Substituting from this $d y = -\frac{x}{y} dx$ into the first equality we get
$$ d \theta = \frac{ - \frac{x^2}{y} dx  - y dx}{x^2 + y^2} = -\frac{d x}{y}$$
Note that this is the result we would get if we looked at the angle formed with the wall, approximated the angle with the tangent, and assumed that only the bottom part is moving. 
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