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?
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        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.



<pre>  
h = the height of the rocket
&theta; = the angle of elevation of the camera.

{{{drawing(150,100,-.5,2.5,-.5,1.5,triangle(0,0,2,0,2,1),locate(.5,.31,theta),
locate(1,0,25),locate(2,0,A), locate(2.1,.6,h))}}}

{{{matrix(2,3,

tan(theta),""="",h/25,
25tan(theta),""="",h)}}}

Take derivatives of both sides with respect to time t:

{{{matrix(1,3,

25sec^2(theta)*expr(matrix(1,2,d,theta)/matrix(1,2,d,t)),""="",matrix(1,2,d,h)/matrix(1,2,d,t))}}}

{{{matrix(1,2,d,h)/matrix(1,2,d,t)}}} the upward speed of the rocket, which is given as 550 miles/hour

{{{matrix(1,3,

25sec^2(theta)*expr(matrix(1,2,d,theta)/matrix(1,2,d,t)),""="",550)}}}

Divide both sides by 25

{{{matrix(1,3,

sec^2(theta)*expr(matrix(1,2,d,theta)/matrix(1,2,d,t)),""="",22)}}}

Change {{{sec^2(theta)}}} to {{{1^""/cos^2(theta)}}}

{{{matrix(1,3,

(1/cos^2(theta))*expr(matrix(1,2,d,theta)/matrix(1,2,d,t)),""="",22)}}}

Multiply through by {{{cos^2(theta)}}}:

{{{matrix(1,3,

matrix(1,2,d,theta)/matrix(1,2,d,t),""="",22cos^2(theta))}}}

Next we calculate the value of {{{matrix(1,2,d,theta)/matrix(1,2,d,t)}}} when
the rocket is 25 miles high.

The figure we are "freezing" the rocket at looks like this, so that's when
{{{theta=45^o=pi/4}}}:

{{{drawing(150,200,-.5,2.5,-.5,1.5,triangle(0,0,2,0,2,1),locate(.4,.3,theta=45^o=pi/4), locate(2,-.2,A),
locate(1,0,25), locate(2.1,.5,25))}}}

Since {{{cos(pi/4)=sqrt(2)/2}}}, we substitute:

{{{matrix(1,3,

matrix(1,2,d,theta)/matrix(1,2,d,t),""="",22*(sqrt(2)/2)^2)}}} = {{{22/2}}} = 11 radians/hour

That's  11 radians per hour.

That's about 10.5 degrees per minute.

@ikleyn
</pre>

Solved correctly.