Question 1181712
.
A rocket is being launched vertically over a point 𝐴 on the ground with
a velocity of 550 𝑚𝑖/ℎ𝑟. 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 in the post by @CPhill is incorrect and his answer is incorrect,  too.


        The error is in the last step,  where @CPhill converts the value of  11 mi/hr  to  radians/hr.


        @CPhill mistakenly treats this value of  11 mi/hr as miles per hour.


        Actually,  it is  11 radians per hour,  and conversion is just done on the way  and  IS  NOT  NEEDED  anymore.


        Below is my correct solution.



<pre>
Here's how to solve this related rates problem:


**1. Diagram and Variables:**

*   Draw a right triangle.
*   Point A is one vertex (where the rocket launches).
*   The photographer is at another vertex, 25 miles away from A.
*   The rocket's altitude is the vertical leg of the triangle (let's call it *y*).
*   The distance from A to the photographer is the horizontal leg (25 miles).
*   The angle of elevation from the photographer to the rocket is θ.


**2. Given Information:**

*   dy/dt = 550 mi/hr (rocket's velocity)
*   We want to find dθ/dt when y = 25 miles.


**3. Relate Variables:**

We can relate θ and y using the tangent function:

tan(θ) = y / 25


**4. Implicit Differentiation:**

Differentiate both sides of the equation with respect to time (t):

sec²(θ) * (dθ/dt) = (1/25) * (dy/dt)


**5. Solve for dθ/dt:**

dθ/dt = (1/25) * (dy/dt) / sec²(θ)
dθ/dt = (cos²(θ)/25) * (dy/dt)


**6. Find cos(θ) when y = 25 miles:**

When y = 25 miles, the triangle is a right isosceles triangle, so θ = 45 degrees or π/4 radians.  Therefore, cos(θ) = cos(45°) = 1/√2.


**7. Substitute and Calculate:**

dθ/dt = ((1/√2)² / 25) * 550 mi/hr
dθ/dt = (1/50) * 550 mi/hr = 11 radians/hr.


    +-----------------------------------------------------------------------------+
    |    Conversion from mi/hr of the left side to radians/hr in the right side   |
    |               is just made inside the previous formula.                     |
    |         This conversion is already built into the coefficients.             |
    +-----------------------------------------------------------------------------+


**Answer:** The angle of elevation is changing at a rate of 11 radians per hour when the rocket reaches an altitude of 25 miles.
</pre>

Solved and answered correctly.


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<H3>Post-solution note</H3>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;11 &nbsp;radiance per hour is &nbsp;{{{(11*57.2958)/60}}} = 10.50423 &nbsp;degrees per minute,


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;if you want to have the answer in units that are adequate to the problem.



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This problem was posted to this forum several years ago (about 5 years ago).


It was solved by Edwin under this link


<A HREF=https://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.1181710.html>https://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.1181710.html</A>


https://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.1181710.html



Edwin' solution was conceptually correct, &nbsp;but had an error in his implementation, &nbsp;leading to incorrect answer.


I found that solution via &nbsp;Google search, &nbsp;and placed there my corrected solution.
My solution follows to the &nbsp;Edwin's idea/design, &nbsp;but fixes/repairs that error.


Now both my solutions to this problem produce the same answer, &nbsp;bringing peace in your mind.