A street light is at the top of a 17 foot tall pole. A woman 6 ft tall walks away from the pole with a speed of 8 ft/s along a straight path. How fast is the tip of her shadow moving when she is 45 feet from the base of the pole?

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- May 28th 2008, 04:15 PMkeemarieerelated rates word problem
A street light is at the top of a 17 foot tall pole. A woman 6 ft tall walks away from the pole with a speed of 8 ft/s along a straight path. How fast is the tip of her shadow moving when she is 45 feet from the base of the pole?

- May 28th 2008, 04:57 PMJonboy
I found a very similar problem on the internet that I know will help.

It's here and it's Example 6. - May 28th 2008, 05:03 PMTheEmptySet
Always draw a diagram!! :)

Attachment 6543

Now we can use similar triangles to get

$\displaystyle \frac{x+y}{17}=\frac{y}{6}$ Now lets multiply th equation by 17 to get

$\displaystyle x+y=\frac{17}{6}y \iff x=\frac{11}{6}y \iff y=\frac{6}{11}x$

Now we know that the tip of the shawdow will be moving at the rate

$\displaystyle \frac{dx}{dt}+\frac{dy}{dt}$

Take the derivative of $\displaystyle y=\frac{6}{11}x$ with respect to time gives

$\displaystyle \frac{dy}{dt}=\frac{6}{11}\frac{dx}{dt}$

You should be able to finish from here. - May 28th 2008, 05:31 PMSoroban
Hello, keemariee!

With these "streetlight" problems, we must be very careful.

Sometimes they ask how fast the tip of the shadow is moving.

Sometimes they ask how fast how fast the shadow is growing.

Quote:

A street light is at the top of a 17 foot tall pole.

A woman 6 ft tall walks away from the pole with a speed of 8 ft/s along a straight path.

How fast is the tip of her shadow moving when she is 45 feet from the base of the pole?

Code:`A *`

| *

| *

| * C

17 | *

| | *

| 6| *

| | *

B *---------------*---------------* E

: y D x-y :

: - - - - - - x - - - - - - - - :

The streetlight is: $\displaystyle AB = 17$

The woman is: $\displaystyle CD = 6$

Let $\displaystyle y = BD.\quad\frac{dy}{dt} \,= \,8\text{ ft/sec}$

Let $\displaystyle x \:=\:BE\qquad\text{ then: }DE \:=\: x-y$

We have: .$\displaystyle \Delta CDE \sim \Delta ABE\quad\Rightarrow\quad \frac{x-y}{6} \:=\:\frac{x}{17}\quad\Rightarrow\quad x \:=\:\frac{17}{11}y$

Differentiate with respect to time: .$\displaystyle \frac{dx}{dt} \:=\:\frac{17}{11}\!\cdot\!\frac{dy}{dt}$

Since $\displaystyle \frac{dy}{dt} \,=\,8\!:\;\;\frac{dx}{dt}\:=\:\frac{17}{11}(8) \:=\:\frac{136}{11} \:=\:12\frac{4}{11}\text{ ft/sec}$

- Apr 10th 2013, 10:16 AMdokrbbRe: related rates word problem
Hi all, I know it's an old problem (since 2008...) but can someone show me how did you get this

$\displaystyle \Delta CDE \sim \Delta ABE\quad\Rightarrow\quad \frac{x-y}{6} \:=\:\frac{x}{17}\quad\Rightarrow\quad x \:=\:\frac{17}{11}y$

I know it should be something really simple but I can't see it, please :) - Apr 10th 2013, 10:25 AMtopsquarkRe: related rates word problem
You don't know how to prove they are similar triangles, or how to get the equation?

I'll assume the equation. The two similar triangles are CDE and ABE. So if we take the ratio of any two sides of CDE you can write a "similar" expression for ABE. For example, look at the sides x - y and 6 in CDE. These are the horizontal "leg" and vertical leg of CDE. So we know that the ratio we want from ABE will involve the horizontal (x) and vertical legs.

Thus

$\displaystyle \frac{\text{horizontal leg}}{\text{vertical leg}} = \frac{x - y}{6} = \frac{x}{17}$

-Dan - Apr 10th 2013, 10:39 AMdokrbbRe: related rates word problem
- Apr 10th 2013, 02:01 PMdokrbbRe: related rates word problem