# related rates

• Oct 29th 2006, 05:09 PM
turtle
related rates
i'm lost on the related rates chapter in calculus...so if anybody can please help me.

A man 6 ft tall walks at the rate of 5 ft/ sectoward a streetlight that is 16 ft above the ground. At what rate is the length of his shadow changing when he is 10 ft from the base of the light?
• Oct 29th 2006, 06:01 PM
splash
Did you leave out details about the light? And are the "particle" and the "man" the same thing?
• Oct 29th 2006, 06:13 PM
turtle
oops, sorry......i mixed two different problems........:confused:
• Oct 29th 2006, 07:04 PM
splash
Quote:

Originally Posted by turtle
i'm lost on the related rates chapter in calculus...so if anybody can please help me.

A man 6 ft tall walks at the rate of 5 ft/ sectoward a streetlight that is 16 ft above the ground. At what rate is the length of his shadow changing when he is 10 ft from the base of the light?

x'(t)=-5ft/s
Draw a right triangle base s+x, height 16 where s is the length of the shadow and x is the distance from the light.
http://i3.photobucket.com/albums/y79...elatedrate.jpg
You can see that (s+x)/16=s/6
Solve for x to get x=5s/3
Implicitly differentiate both sides x'=5s'/3
Solve for s'(t)=3x'(t)/5
Plug x'(t)=-5 to get s'(t)=-3ft/s

(I think this is right but I haven't done this in a few years so this is my disclaimer. Wait for someone to confirm if you want to be sure.)
• Oct 30th 2006, 02:52 AM
topsquark
Quote:

Originally Posted by splash
x'(t)=-5ft/s
Draw a right triangle base s+x, height 16 where s is the length of the shadow and x is the distance from the light.
http://i3.photobucket.com/albums/y79...elatedrate.jpg
You can see that (s+x)/16=s/6
Solve for x to get x=5s/3
Implicitly differentiate both sides x'=5s'/3
Solve for s'(t)=3x'(t)/5
Plug x'(t)=-5 to get s'(t)=-3ft/s

(I think this is right but I haven't done this in a few years so this is my disclaimer. Wait for someone to confirm if you want to be sure.)

Looks good to me! :)

-Dan