# Related Rates Problem

cmajor47

## Homework Statement

A rocket is launched vertically and is tracked by a radar station located on the ground 5 mi from the launch pad. Suppose that the elevation angle θ of the line of sight to the rocket is increasing at 3° per second when θ=60°. What is the velocity of the rocket at this instant?

c2=x2+y2
tanθ=y/x

## The Attempt at a Solution

52=c2-y2
25=c2-y2
0=2c(dc/dt)-2y(dy/dt)
0=c(dc/dt) - y(dy/dt)

dc/dt=3°=π/60

θ=60°=π/3

tan π/3=y/5
5tan π/3=y
y=3

c2=52+32
c2=square root of 34

Can I just plug in 3 for y, square root of 34 for c, and π/60 for dc/dt

Homework Helper

## The Attempt at a Solution

52=c2-y2
25=c2-y2
0=2c(dc/dt)-2y(dy/dt)
0=c(dc/dt) - y(dy/dt)

This would be OK up to here,

dc/dt=3°=<pi>/60

But this is not the rate of change of c, but the rate of change of $$\theta$$.

For your definitions, x = 5 and y is the altitude of the rocket. So you are looking for dy/dt (since the rocket is going straight up) and you will need to differentiate

$$tan\theta = \frac{y}{x}$$ .

(The rate at which the hypotenuse, c, is changing is not needed in this problem.)

cmajor47
I realized my mistakes. I now have sec2θ dθ/dt = (dy/dt)/5.
Should I plug in dθ/dt in degrees or radians?

Homework Helper
I realized my mistakes. I now have sec2? d?/dt = (dy/dt)/5.
Should I plug in d?/dt in degrees or radians?

In mathematical expressions, angles are expressed in radians.