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!
Best Answer
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.