Calculating Angle of Elevation Rate of Change for an External Elevator

[squenshl]

Homework Statement

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.

Homework Equations

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?

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

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?

Why not just keep going and see what you get?

 

[squenshl]

Why not just keep going and see what you get?

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

 

[Dick]

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

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

 

[Ray Vickson]

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

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.

 

[squenshl]

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.

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

 

[Dick]

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

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