# Homework Help: Rate of Change Question

1. Dec 4, 2017

### squenshl

1. The problem statement, all variables and given/known data
A building has an external elevator. The elevator is rising at a constant rate of $2 \; \text{ms}^{-1}$.
Sarah is stationary, watching the elevator from a point 30m away from the base of the elevator shaft.
Let the angle of elevation of the elevator floor from Sarah's eye level be $\theta$.

Find the rate at which the angle of elevation is increasing when the elevator floor is 20m above Sarah’s eye level.

2. Relevant equations

3. The attempt at a solution
Am I trying to find $\frac{d\theta}{dt} = \frac{d\theta}{dh}\times \frac{dh}{dt}$. We know $\frac{dh}{dt} = 2$. Am I on the right track???

2. Dec 4, 2017

### Ray Vickson

Why not just keep going and see what you get?

3. Dec 4, 2017

### squenshl

Ok then. $h = 30 \tan{(\theta)}$ so $\frac{dh}{d\theta} = 30\sec^2{(\theta)}$. Hence, $\frac{d\theta}{dt} = \frac{\cos^2{(\theta)}}{15}$.

4. Dec 4, 2017

### Dick

Fine so far. But you should be able to express $\frac{d\theta}{dt}$ as a numerical value. What is $\cos(\theta)$ when $h=20$?

5. Dec 4, 2017

### Ray Vickson

Good. That is exactly what I was hoping you would do.

I was basically encouraging you to do more on your own, by taking a chance and trying out something---win or lose. That is how we all learned.

6. Dec 5, 2017

### squenshl

I found the length of the hypotenuse then calculated $\theta = \cos^{-1}\left(\frac{30}{\sqrt{1300}}\right)$ then threw that solution into $\frac{d\theta}{dt}$.

7. Dec 6, 2017

### Dick

Right. Though you actually didn't need to find $\theta$, right? All you need is $\cos(\theta)=\frac{30}{\sqrt{1300}}$.