# Thread: Help setting up rate of change of angle

1. ## Help setting up rate of change of angle

A camera mounted 132ft above the ground is videotaping a racecar moving at 264ft/sec. How fast will the camera angle be changing when the car is right in front of you? a half second later?

">" is the angle

tan>=y/x and dtan>/dt= x(dy/dt)-y(dx/dt)/x^2

y=132ft and dx/dt=264ft/sec

How do I find x. or where do I go from here

2. I'd do this problem a completely different way. I'm going to call the angle 'a.'

$tan(a) = \frac {x} {132}$ Notice that 132 is constant, don't need a y variable.

$a = Arctan(\frac{x}{132})$ I solve for 'a' so we can take a derivative here and find $\frac{da}{dx}$.

Then since we know $\frac{dx}{dt}$, the rest is straightforward.

3. what is the value of x?

4. x is the distance from directly underneath the camera that the car is. And therefore $\frac{dx}{dt}$ is just going to be equal to the speed of the car.

5. I am so lost. so does x=y and dx/dt=dy/dt? I am a 1st semester high school junior, so I need baby steps.

6. I think I understand the problem now.

t is the independant variable, time.

x is the position of the car.

a is the angle of the camera.

You can call them whatever you like, but those are the three quantities we're measuring. We're trying to find $\frac{da}{dt}$, the angle change per second. They tell us what $\frac{dx}{dt}$ is when they tell us the speed of the car. We then have to find $\frac{da}{dx}$ so we can plug it into the equation below.

$\frac{da}{dt}=\frac{da}{dx}*\frac{dx}{dt}$

That's why you need to find the derivative of the equation in my first reply.