# Related Rates Problem

1. Jun 30, 2008

### cmajor47

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
c2=x2+y2
tanθ=y/x

3. 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

2. Jun 30, 2008

### dynamicsolo

This would be OK up to here,

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.)

3. Jul 2, 2008

### cmajor47

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

4. Jul 2, 2008

### dynamicsolo

In mathematical expressions, angles are expressed in radians.