Question

Can someone help me and walk me through a related rates problem? I have a hard time understanding how it works and how to set up equations.

A man walks along a straight path at a constant speed of 4 feet per second. A searchlight is on the ground 20 feet from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 15 feet from the point on the path which is closest to the searchlight?

Answer 1

doesn't say 15 feet before or after, so not exactly clear, but we can solve both

Answer 2

if we call the distance between the man and the point on the path closet to the light \(x\) then we get a relation between \(x\) and \(\theta\) via \(\tan(\theta)=\frac{x}{20}\)

Answer 3

then your are told the speed of the man is \(4\) so \(x'=4\) and you want \(\theta '\)

Answer 4

in my picture actually \(x'=-4\) is the man is walking from left to right, because in that case \(x\) is decreasing, but nvm lets leave \(x'=4\)

Answer 5

then with the relation \[\frac{x}{20}=\tan(\theta)\] take the derivative wrt time and get
\[\frac{x}{20}=\sec^2(\theta)\theta'\]

Answer 6

damn i meant
\[\frac{x'}{20}=\sec^2(\theta)\theta '\]

Answer 7

since \(x'=4\) you get
\[\frac{1}{5}=\sec^2(\theta)\theta '\]

Answer 8

now solve for \(\theta '\) using the fact that you are given \(x=15\)

Answer 9

pythagoras gives the hypotenuse as \(25\) or really just recognizing this as a 3 - 4 - 5 right triangle only it is 15-20-25

Answer 10

this makes \(\sec(\theta)=\frac{25}{20}=\frac{5}{4}\) and so \(\sec^2(\theta)=\frac{25}{16}\)

Answer 11

now \[\frac{1}{5}=\sec^2(\theta)\theta '\] is
\[\frac{1}{5}=\frac{25}{16}\theta '\] and you can easily solve for \(\theta '\)

Answer 12

Thank you so much for the walkthrough! It helped a lot!
However, when I was working with a study group, they approached it differently and got a different answer.
We had the same picture and setup, but instead of taking the derivative of tan, they set up their equation like this \[\arctan(\frac{ x }{ 20 })=\theta \] and then took the derivative of both sides, making it \[\frac{ x }{ 20(1+\frac{ x ^{2} }{ 400 }) }= \frac{ d \theta }{ dt }\] and plugged in 4 for x' which equals .147 rads/second or something like that. Can this be another way of approaching this problem and we just screwed up somewhere, or is it wrong to do it like this?

Answer 13

the work should be identical just another way of looking at it