Question

A man 6 feet tall walks at a rate of 5 feet per second away from a light that is 15 feet above ground. When he is 10 feet from the base of the light,
a) at what rate is the tip of his shadow changing?
b) at what rate is the length of his shadow changing?

Answer 1

Some one please help! :/
I've been staring at this thing for 4 days now 0.o
and now I have a test on it tomorrow -.-"

I just only really need to know is what variables equal what and what to solve for.
I'll be able to do the rest from there, so someone please please please HELP!!!
Thanks :)

Answer 2

The message and the notifications are not working properly and I just got your msg.

Answer 3

yeah, thanks for coming! :D

Answer 4

okay...
so for part A is it solving for the rate of AE?

Answer 5

I should probably draw it a bit differently because as the man moves away the 10 feet will be changing. So we should probably keep that as the x variable.

Answer 6

Because of similar triangles: CD / AB = ED / EB

Answer 7

Let ED = y
CD / AB = ED / EB
6/15 = y / (y + x)
2/5 = y / (y + x)
2(y + x) = 5y
2y + 2x = 5y
2x = 3y
x = 3/2y

Answer 8

dx/dt = 3/2dy/dt
dx/dt = 5
5 = 3/2dy/dt
dy/dt = 2 * 5 /3 = 10/3 feet/sec is the answer to part b)

Answer 9

okay, thank you :) ...

Answer 10

and so for part a)
would I be looking for dx/dt + dy/dt ?
then would it be 5 + 10/3
25/3 feet/sec ?

Answer 11

Looks like it. Because the man is moving away from the light and so his shadow will be getting longer which we determined as dy/dt. To that we need to add the man's movement dx/dt.
What I am wondering is we never made use of the 10 feet in this problem.

Answer 12

Let me think about it and if I come across something I will leave a reply here.

Answer 13

okay, and thank you soo much! :)
and the answers match those that my teacher gave, so yeah, I just hadn't known how to do it :) Thank You!!!

Answer 14

oh, both answers are correct?

Answer 15

yep :)

Answer 16

cool. you are welcome.