Related Rates Ladder Problem with Angles The problem is as follows: A 13-foot ladder leans against the side of a building, forming an angle θ with the ground. Given that the foot of the ladder is being pulled away from the building at the rate of 0.1 feet per second, what is the rate of change of θ when the top of the ladder is 12 feet above the ground?
Starting off I found it difficult to visualize the problem in my head. I guess they meant theta to be on the 'bottom' of the triangle. I drew a quick diagram in Gimp.
So I start off by identifying what I have and what I need.
$$\frac{dx}{dt} = .1ft/s$$
$$\frac{d\theta}{dt} = ?$$
I use trigonometry (shrug) to find a relationship between $\theta$ and $x$.
$$\frac{x}{13} = cos(\theta)$$
And this is where I get stuck. I know I can use implicit differentiation but that only renders $\frac{d\theta}{dt}$, how should I go about getting $\theta{d\theta}{dy}$? Thanks!
 A: There are some problems of notation. For example, we are in effect  told that $\frac{dx}{dt}=0.1$.  (It is not $\frac{dt}{dx}$.)
And we are asked to find how fast the angle is changing, so we want $\frac{d\theta}{dt}$.  
The relationship $\frac{x}{13}=\cos\theta$ is right. Differentiate both sides with respect to $t$.  We get
$$\frac{1}{13}\cdot \frac{dx}{dt}=-\sin\theta\frac{d\theta}{dt}\tag{1}$$
(Chain Rule).
Now freeze the situation at the instant when the height is $12$. At that instant, we have $x=5$, and $\frac{dx}{dt}=0.1$, and $\sin\theta=\frac{12}{13}$. 
Finally, solve for $\frac{d\theta}{dt}$. We get that at that instant $\frac{d\theta}{dt}=-\frac{0.1}{12}$ radians per second. 
A: So, let's look at the values that we have:
Given a right triangle with legs $x$ and $y$, and hypotenuse $z$:
$$\frac{dx}{dt} = 0.1$$
$$z = 13$$
When $y = 12$
$$x^2 +12^2 = 13^2$$
$$x = 5$$
Using our SOHCAHTOA, we can say that: 
$$\cos(\theta) = \frac{x}{13}$$
Taking the derivative:
$$-\sin(\theta)*\frac{d\theta}{dt} = \frac{1}{13}*\frac{dx}{dt}$$
Note that:
$$\cos(\theta) = \frac{5}{13}$$
Then we can say that at that certain instant
$$\sin(\theta) = \frac{12}{13}$$
So,
$$-\frac{12}{13}*\frac{d\theta}{dt} = \frac{1}{13}*\frac{1}{10}$$
Therefore,
$$\frac{d\theta}{dt} =  -\frac{1}{120}$$
A: Actually, $\frac{dx}{dt}=0.1 ft/s=\frac{1}{10}ft/s$. You other observation is true: $\frac{x}{13ft}=\cos\theta$. Now you can use implicit differentiation. Apply $\frac{d}{dt}$ to both sides to get:
$$\frac{1}{13ft}\frac{dx}{dt}=\frac{d\cos\theta}{dt} \Rightarrow \frac{1}{130}s^{-1}=\frac{d\cos\theta}{dt}$$
You can finish by using the chain rule to compute $\frac{d\cos\theta}{dt}$:
$$\frac{d\cos\theta}{dt}= \frac{d\cos\theta}{d\theta}\frac{d\theta}{dt}=-\sin\theta\frac{d\theta}{dt}$$
So that $\displaystyle\frac{d\theta}{dt}=\frac{1}{-130\sin\theta}s^{-1}$. Now we want to evaluate this at $y=12ft$. Notice that the length of the ladder doesn't change, so $\displaystyle\sin\theta=\frac{12ft}{13ft}=\frac{12}{13}$.
Therefore:
$$\displaystyle\frac{d\theta}{dt}=\frac{1}{-130\sin\theta}s^{-1}=-\frac{1}{120}\mathrm{rad} \cdot s^{-1}$$
And we're done. (By the definition of a radian, $\mathrm{rad} \cdot s^{-1}$ is the same as $s^{-1}$ dimensionally speaking, so the units are correct.)
