Related Rates Question: Outdoor sensor, rate of shadow

[grapz]

An outdoor sensor light is at the top of a 3-m gate. Faye is 1.5 m tall, and walks away from the gate in a straight path with a speed of 0.5 m/s. How fast is the tip of her shadow moving when she is 10 m from the gate?

Is this how u do it?

Use similar triangles. 1.5/ l = 3/x + l , x = l

so know u have to find dP/dT (the rate of her shadow)

which is dP/dT = 2 (dX/dT) = 1m/s? <--- i'm not sure about this part, or why it works. please explain to me thx

 

[soroban]

Hello, grapz!

These "shadow" problems have a built-in source of confusion.

An outdoor sensor light is at the top of a 3-m gate.
Faye is 1.5 m tall, and walks away from the gate in a straight path with a speed of 0.5 m/s.
How fast is the tip of her shadow moving when she is 10 m from the gate?


Is this how u do it?

Use similar triangles: \(\displaystyle \:\frac{1.5}{L}\:=\:\frac{3}{x\,+\,L}\;\;\Rightarrow\;\;x\:=\:L\)

so know u have to find \(\displaystyle \frac{dP}{dT}\), (the rate of her shadow)

. . which is: \(\displaystyle \,\frac{dP}{dT}\:=\:2\left(\frac{dx}{dt}\right)\:=\:1\text{ m/s}\)?

I'm not sure about this part or why it works.

I have to guess what your diagram looks like.
I assume it looks like this . . .

        *
        |   *
        |       *
       3|           *
        |           |   *
        |        1.5|       *
        |           |           *
      - * - - - - - * - - - - - - - * -
        :     x     :       L       :
        : - - - - - - P - - - - - - :

You work absolutely correct!

If you still don't understand your solution,
. . label the diagram like this:

        *
        |   *
        |       *
       3|           *
        |           |   *
        |        1.5|       *
        |           |           *
      - * - - - - - * - - - - - - - * -
        :     x     :      P-x      :
        : - - - - - - P - - - - - - :

So we have: \(\displaystyle \:\L\frac{1.5}{P\,-\,x} \:=\:\frac{3}{P}\;\;\Rightarrow\;\;P \:=\:2x\)

Then: \(\displaystyle \L\:\frac{dP}{dt}\:=\:2\left(\frac{dx}{dt}\right)\)


Hope this helps . . .

 

[grapz]

ahhhhh yes, thank you, that was what i was thinking