# Related Rate

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I don't know how to solve this one. Can anyone help me?

A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks from the pole with a speed of 5ft/s along the straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

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$\frac{15*5}{6} = 12.5\frac{ft}{s}$?

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You've got some similar triangles. Therefore:

Since both the man and the pole have constant heights with a constant ratio of 6:15, so too will the ratio of man's distance to shadow's distance be constant, with the same ratio. So if the man is walking at 5ft/s, the shadow is moving at (15/6)(5ft/s)=12.5ft/s.

EDIT: Since this is in the calculus forum, I guess you want a calculus solution. You can write the total length (y) as a function of the man's distance from the pole (x) with y=f(x)=15x/6=5x/2. Also, you can write the man's distance as a function of time: x=g(t)=5t. Combine for f(g(t)) to get total distance as a function of time, y=(15/6)(5t)=75t/6=12.5t. Then you just take its derivative at t=8 to find the ft/s there, which in this case will always just be 12.5.

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I'm having doubts about 12.5 ft/s...

Post height = $H$

Object height = $h$

Shadow speed = $V$

Object speed = $v$

$\frac{H}{h}=\frac{V}{V-v} =>V=\frac{Hv}{H-h}$

Then it should be $8.(3)\frac{ft}{s}$

Is this correct?

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Er, yes... I was a tad mixed up there. The ratio is NOT

total distance:man's distance::pole height:man's height

but RATHER

total distance:total distance - man's distance::pole height:man's height

So:

The ratio of the shadow to the whole is 2 to 5 (simplified from 6 to 15). Therefore the ratio of the man's distance to the whole is 3 to 5.

The speed is constant, so the ratio of the man's speed to the shadow's speed is also 3 to 5.

The man's speed is 5ft/s.

Therefore, the shadow's speed is 25/3.

I put the blame entirely on you.

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