Posted by Nikki on Sunday, September 12, 2010 at 1:59pm.
(It's a related rate question)
A man 6ft tall walks away from a lamp post (15ft) at 5ft/sec. How fast is his shadow lengthening?

I have the picture, and the constants, the man and the lamp post. I have 5ft/sec as dw/dt, and I know I'm looking for ds/dt, the rate of the shadow.
Problem is, I don't know the actual equation to figure this out. Help please?
Thanks!

Calculus  bobpursley, Sunday, September 12, 2010 at 4:34pm
The lamp post makes a right triangle.
let x be the distance from the lamppost to the man (dx/dt=5; given) Let L be the length of is shadow, measured from the man feet to the end of the shadow.
(dL/dt is the rate of the shadow lengthens.
Using similar triangle
6/15=L/(L+x) or
6L+6x=15L
6x=9 L
x=3/2 L
dx/dt= 3/2 dL/dt
well, then dL/dt=2/3 dx/dt= 2/3 5=10/3 ft/sec
Now if you were to ask at what rate the shadow tip was moving , that is dx/dt+dL/dt

Calculus  Heena Patel, Tuesday, October 12, 2010 at 2:25pm
let x be the rate of shadow.
x / (x + 5) = 6 / 15
15x = 6(x+5) = 6x + 30
9x = 18
x = 30/9 = 10/3
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