Related Rates Problem

  • Thread starter cmajor47
  • Start date
  • #1
57
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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
 

Answers and Replies

  • #2
dynamicsolo
Homework Helper
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4

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 [tex]\theta[/tex].

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

[tex]tan\theta = \frac{y}{x}[/tex] .

(The rate at which the hypotenuse, c, is changing is not needed in this problem.)
 
  • #3
57
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I realized my mistakes. I now have sec2θ dθ/dt = (dy/dt)/5.
Should I plug in dθ/dt in degrees or radians?
 
  • #4
dynamicsolo
Homework Helper
1,648
4
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.
 

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