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March 31, 2015

March 31, 2015

Posted by **Nikki** on Sunday, September 12, 2010 at 1:59pm.

A man 6ft tall walks away from a lamp post (15ft) at 5ft/sec. How fast is his shadow lengthening?

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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 4:34pmThe 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 2:25pmlet 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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