Word problem involving rates and derivatives..

[tarynt1]

A rocket is rising vertically according to the position equation s = 60t^2. If a camera 4000 feet away is filming the launch, find the rate of change of the angle of elevation after 3 seconds. Use radians.

I know that I need to set up a right triangle, with x being 4000ft, y being 60t^2, and the angle of elevation being theta.

tan(theta) = (60t^2)/4000, and theta = arctan ((60t^2)/4000), and after 3 is substituted for t, theta = arctan ((540)/4000). I realize I need to take the derivative (d/dt) of tan(theta) = (60t^2)/4000, but beyond that I am having some trouble putting the pieces together. I need to know what needs to be done in order to solve for d(theta)/dt, and most importantly, why.

Any help would be greatly appreciated. Thanks!

 

[soroban]

Hello, tarynt1!

A rocket is rising vertically according to the position equation \(\displaystyle s \,= \,60t^2.\)
If a camera 4000 feet away is filming the launch,
find the rate of change of the angle of elevation after 3 seconds. Use radians.

I would use: \(\displaystyle \L\,\tan\theta\:=\:\frac{s}{4000}\;\;\Rightarrow\;\;4000\cdot\tan\theta \:=\:s\)

Differentiate with respect to time: \(\displaystyle \L\:4000\cdot\sec^2\theta\cdot\frac{d\theta}{dt}\:=\:\frac{ds}{dt}\)

. . and we have: \(\displaystyle \L\:\frac{d\theta}{dt} \:=\:\frac{1}{4000}\cdot\cos^2\theta\cdot\frac{ds}{dt}\;\;\) [1]


We have: \(\displaystyle \L\,s\,=\,60t^2\;\;\Rightarrow\;\;\frac{ds}{dt}\,=\,120t\)
. . When \(\displaystyle \L t\,=\,3,\;\frac{ds}{dt}\,=\,360\;\;\) [2]


When \(\displaystyle \L t\,=\,3:\;s\,=\,60\cdot3^2\,=\,540\;\;\Rightarrow\;\;\tan\theta\,=\,\frac{540}{4000}\,=\,\frac{27}{200}\)

. . \(\displaystyle \L\sec^2\theta\:=\:\tan^2\theta\,+\,1\:=\:\left(\frac{27}{200}\right)^2\,+\,1 \:=\:\frac{40,729}{40,000}\)

. . Hence: \(\displaystyle \L\,\cos^2\theta\:=\:\frac{40,000}{40,729}\;\;\) [3]


Substitute [2] and [3] into [1]:

. . \(\displaystyle \L\:\frac{d\theta}{dt}\:=\:\frac{1}{4000}\,\cdot\,\frac{40,000}{40,729}\,\cdot\,360 \:=\:0.088389108\)


Therefore, the rate of change of the angle of elevation at \(\displaystyle t\,=\,3\)
. . is approximately 0.088 radians/sec.

 

[tarynt1]

Thank you for your help; I really understand the problem now!