Related Rate Problem (Involving Trig.)

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SUMMARY

The discussion focuses on solving a related rates problem involving trigonometry, specifically calculating the rate of change of the camera angle as a rocket ascends. The height function is defined as h(t) = 200t^2, with the camera positioned 150 m from the launch site. The differentiation process leads to the formula dθ/dt = (8t/3) * cos²(θ). At t = 4 seconds, the calculated rate of change of the angle is approximately 0.0234 rad/s. The solution is confirmed to be accurate, with a suggestion to utilize the identity sec²(θ) = 1 + tan²(θ) for simplification.

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Homework Statement



A rocket is moving into the air with a height function given by h(t) = 200t^2. A camera located 150 m away from the launch site is filming the launch. How fast must the angle of the camera be changing with respect to the horizontal 4 seconds after liftoff?

Homework Equations





The Attempt at a Solution



If we create a diagram, we will see that
tan(θ)=(200t^2)/150 or (4t^2)/3

Differentiating with respect to t,

sec^2(θ)dθ/dt=8t/3 which becomes dθ/dt=8t/3 * cos^2(θ)

At t=4s, tan(θ)=64/3, and then by sinθ=cosθtanθ, we know sinθ=(64/3)cosθ

Then by sin^2(θ)+cos^2(θ)=1, we know that cos^2(θ)=9/4105

Now evaluating the derivative at t=4s, we obtain dθ/dt=96/4105 rad/s≈0.0234 rad/s

I would just like to know if all my steps are accurate, and if my final answer is correct, or if I made an error along the way, leading to an incorrect result?
 
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Looks OK to me. You could have saved a couple of calculations by using ##\sec^2\theta=1+\tan^2\theta## instead of messing with the sines and cosines.
 
Okay thanks, and I completely forgot about that identity when doing this problem.
 

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