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?

Homework Equations

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

 

[dynamicsolo]

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?

 

[dynamicsolo]

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.