SOLUTION: A rocket is being launched vertically over a point 𝐴 on the ground with a velocity of 550 𝑚𝑖Τℎ𝑟. Twenty five miles away from point 𝐴 on the ground, there is a p

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A rocket is being launched vertically over a point 𝐴 on the ground with a velocity of 550 𝑚𝑖Τℎ𝑟. Twenty five miles away from point 𝐴 on the ground, there is a p      Log On


   

Question 1181710: A rocket is being launched vertically over a point 𝐴 on the ground with
a velocity of 550 𝑚𝑖Τℎ𝑟. Twenty five miles away from point 𝐴 on the
ground, there is a photographer video-taping the launch. At what rate
is the angle of elevation of the camera changing when the rocket
achieves an altitude of 25 miles?
Found 3 solutions by Alan3354, ikleyn, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A rocket is being launched vertically over a point 𝐴 on the ground with
a velocity of 550 𝑚𝑖Τℎ𝑟
----------------------------
IDK what that means.

Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
A rocket is being launched vertically over a point 𝐴 on the ground with
a velocity of 550 𝑚𝑖Τℎ𝑟. Twenty five miles away from point 𝐴 on the
ground, there is a photographer video-taping the launch. At what rate
is the angle of elevation of the camera changing when the rocket
achieves an altitude of 25 miles?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        The solution by  Edwin is good conceptually,  but his implementation has an error,  which leads to wrong answer.

        In this my post,  I fix this error and bring a correct solution and answer to you. 


  
h = the height of the rocket
θ = the angle of elevation of the camera.





Take derivatives of both sides with respect to time t:



matrix%281%2C2%2Cd%2Ch%29%2Fmatrix%281%2C2%2Cd%2Ct%29 the upward speed of the rocket, which is given as 550 miles/hour



Divide both sides by 25



Change sec%5E2%28theta%29 to 1%5E%22%22%2Fcos%5E2%28theta%29



Multiply through by cos%5E2%28theta%29:



Next we calculate the value of matrix%281%2C2%2Cd%2Ctheta%29%2Fmatrix%281%2C2%2Cd%2Ct%29 when
the rocket is 25 miles high.

The figure we are "freezing" the rocket at looks like this, so that's when
theta=45%5Eo=pi%2F4:



Since cos%28pi%2F4%29=sqrt%282%29%2F2, we substitute:

 = 22%2F2 = 11 radians/hour

That's  11 radians per hour.

That's about 10.5 degrees per minute.

@ikleyn

Solved correctly.



Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
 
You don't need to clarify anything. This is an ordinary related rates problem
from calculus.

A rocket is being launched vertically over a point 𝐴 on the ground with
a velocity of 550 𝑚𝑖Τℎ𝑟. Twenty five miles away from point 𝐴 on the
ground, there is a photographer video-taping the launch. At what rate
is the angle of elevation of the camera changing when the rocket
achieves an altitude of 25 miles?
  
h = the height of the rocket
θ = the angle of elevation of the camera.





Take derivatives of both sides with respect to time t:



matrix%281%2C2%2Cd%2Ch%29%2Fmatrix%281%2C2%2Cd%2Ct%29 the upward speed of the rocket, which is given as 550 miles/hour



Divide both sides by 25



Change sec%5E2%28theta%29 to 1%5E%22%22%2Fcos%5E2%28theta%29



Multiply through by cos%5E2%28theta%29:



Next we calculate the value of matrix%281%2C2%2Cd%2Ctheta%29%2Fmatrix%281%2C2%2Cd%2Ct%29 when
the rocket is 25 miles high.

The figure we are "freezing" the rocket at looks like this, so that's when
theta=45%5Eo=pi%2F4:



Since cos%28pi%2F4%29=sqrt%282%29%2F2, we substitute:



 radians/hour

That's about 15.6 radians per hr.

That's about 14.8 degrees per minute.

Edwin