Question

A street light is at the top of a 15 ft tall pole. A man 5.8 ft tall walks away from the pole with a speed of 5.5 ft/sec along a straight path. How fast is the tip of his shadow moving when he is 46 ft from the pole?

Answer 1

Similar triangles, so it should be proportional.

Answer 2

this is differential calculus btw

Answer 3

Yeah, it sounds like a related rates situation, but I don't think derivatives are necessary here.
I'll double check, though.

Answer 4

yeah we were just doing related rates so thats possible

Answer 5

Just to get a picture on the situation

Answer 6

yeah i get that

Answer 7

Hmm, might have to do it considering the rate of change of the angle, Θ.

Answer 8

Θ is getting smaller with time, and that can be expressed in terms of tan(Θ). I think you can use that relation to find the rate of change of the long base, x.

Answer 9

so how do i do it

Answer 10

Let's see. . . using the similar triangles, I am getting 75ft for x.
The angle, Θ, must then be arctan(15/75) at this moment.

Answer 11

I think how to do it is to see that \(Θ=arctan(15/x)\), so \(x=15/tan(Θ)\).
\(dΘ/dt = 1/(1+Θ^2)\), and \(dx/dt = dx/dΘ \cdot dΘ/dt\)

Answer 12

\[x'=5.5,y'=?\]
\[\frac{ x+y }{ 15 }=\frac{ y }{ 5.8 }\]divide by y and multiply by 15
\[\frac{ x }{ y }+1=\frac{ 15 }{ 5.8 }\]
\[x=(\frac{ 15-5.8 }{ 5.8 })y\]
\[x'=(\frac{ 9.2 }{ 5.8 })y'\]
shadow
\[x'+y'=9.2/8.5+5.5\]