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

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)   (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)   (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: the upward speed of the rocket, which is given as 550 miles/hour Divide both sides by 25 Change to Multiply through by : Next we calculate the value of when the rocket is 25 miles high. The figure we are "freezing" the rocket at looks like this, so that's when : Since , we substitute: = = 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)   (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: the upward speed of the rocket, which is given as 550 miles/hour Divide both sides by 25 Change to Multiply through by : Next we calculate the value of when the rocket is 25 miles high. The figure we are "freezing" the rocket at looks like this, so that's when : Since , we substitute: radians/hour That's about 15.6 radians per hr. That's about 14.8 degrees per minute. Edwin

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