Related Rates: Calculating the Speed of a Moving Shadow

Click For Summary

Homework Help Overview

The problem involves related rates, specifically calculating the speed of the tip of a shadow cast by a man walking away from a street light mounted on a pole. The scenario includes a 15-foot tall pole and a 6-foot tall man walking at a speed of 6 ft/s, with the question focusing on the speed of the shadow when the man is 40 feet from the pole.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss drawing triangles to visualize the problem and consider the relationship between the man's distance from the pole and the speed of the shadow. There is uncertainty about whether the shadow's speed matches the man's speed and how to correctly apply similar triangles in the context of the problem.

Discussion Status

Some participants have attempted to set up equations based on similar triangles and derivatives, but there is a lack of consensus on the correct approach. One participant expresses confusion about incorporating the distance of 40 feet into their calculations, indicating that the discussion is ongoing without a clear resolution.

Contextual Notes

Participants are working within the constraints of the problem as stated, including the heights of the pole and the man, as well as the specified distance of 40 feet. There is an acknowledgment of potential misunderstandings regarding the application of geometric principles.

momogiri
Messages
52
Reaction score
0
Question
A street light is mounted at the top of a 15 foot tall pole. A man 6 ft tall walks away from the pole with a speed of 6 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

Attempt
Well actually, I've drawn it out and stuff, but I was wondering if the tip of shadow would be moving in the same speed as the man..
I'm a bit confused at how to tackle this problem.. :/
 
Physics news on Phys.org
Draw up some triangles, and remember to use Pythagoras.
 
Yeah, I drew up my triangle within a triangle, it's attached, so..

The change in speed of the shadow is not the same as the speed of the man, right?
And 40 ft refers to the distance between the pole and the man right? Not between the shadow tip and the pole?
 

Attachments

  • tri.JPG
    tri.JPG
    1.3 KB · Views: 470
Yes, the problem says "when he is 40 ft from the pole".

But you don't need Pythagoras since the length of the hypotenuse is not relevant. Use "similar triangles".
 
Last edited by a moderator:
HallsofIvy said:
But you don't need Pythagoras since the length of the hypotnuse is not relevant.

I was thinking of something else :( Sorry
 
Ok, so I tried to do it with similar triangles..

With my image that I attached earlier, I had the similar trigangles
y/6
and
(x+y)/15

(y being the distance of the shadow and the man
x being distance from man and pole)

So \frac{y}{6} = \frac{x+y}{15}
and I solved for x
15y = 6x+6y
9y = 6x
x = \frac{3}{2}y
and took derivatives of both sides wrt t
dx/dt = \frac{3}{2}dy/dt
And since the change in x is 6ft/sec
I just plugged in the numbers and found dy/dt to be 4
But it's wrong.

It's probably because I didn't incorporate 40ft as x somewhere, but I'm looking at what I did, and I just don't know what I should do :/
 

Similar threads

Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
12K
  • · Replies 3 ·
Replies
3
Views
10K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
5
Views
5K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
7
Views
7K