# Similar Triangles, Light and Shadow.

#### Hypnos_16

1. Homework Statement

A man 6 feet tall walks at a rate of 5 feet per second away from a light that is attached to a pole 15 feet above the ground. At what rate is the length of his shadow changing when he is 30 feet from the base of the pole?

I get that this is really like a two similar triangle problem, the way the light is shining and the shadow cast by the man

so i called the larger triangle ∆abc
and the smaller ∆xyz

The larger is the one that is caused by the light
a (x value) = 30 feet (Distance from Light Pole) + Shadow Length
b (y value) = 15 feet (Light Pole)
c (z value) = the Pythagorean value

The smaller triangle
x (x value) = Shadow Length
y (y value) = 6 feet (Man's Height)
z (z value) = the Pythagorean Value

Now i have figured out some values.
a = 30 feet
b = 15 feet
y = 6 feet
c2 = (302 + (b - y)2)
c = √981

2. Homework Equations

I assume a2 + b2 = c2
since it's a right triangle
but i don't have enough values

3. The Attempt at a Solution

i don't really have one. Sorry if that was confusing. I can try and clarify if need be.

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#### HallsofIvy

Homework Helper
You don't need to know the hypotenuse of either triangle so the Pythagorean theorem is not needed. Use "similar triangles" instead.

Let "x" be the man's distance from the lamp post, s the length of his shadow. Then one triangle, the one with the lamp post as a side, has a leg of length x+ s and a side of length 15. The other triangle, the one with the man as a side, has corresponding sides of length 6 and s. Since those are similar triangles, we have
[tex]\frac{x+s}{15}= \frac{x}{6}[/itex]

Differentiate that with respect to time, t, getting an equation relating dx/dt and dy/dt. You are told what dx/dt is and can use that to find dy/dt.

#### Hypnos_16

You don't need to know the hypotenuse of either triangle so the Pythagorean theorem is not needed. Use "similar triangles" instead.

Let "x" be the man's distance from the lamp post, s the length of his shadow. Then one triangle, the one with the lamp post as a side, has a leg of length x+ s and a side of length 15. The other triangle, the one with the man as a side, has corresponding sides of length 6 and s. Since those are similar triangles, we have
[tex]\frac{x+s}{15}= \frac{x}{6}[/itex]

Differentiate that with respect to time, t, getting an equation relating dx/dt and dy/dt. You are told what dx/dt is and can use that to find dy/dt.
should it be
[tex]\frac{x+s}{15}= \frac{s}{6}[/itex]
since in this case the s is with the man who's 6 feet?
Also then do i solve for a variable like
x = (15s / 6) - s
What equation can i use here to solve this?

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