# Thread: Word problems with differentiation (implicit, I think)

1. ## Word problems with differentiation (implicit, I think)

Doing homework/studying for test tomorrow.

"A street light is at the top of an 11 foot pole. A woman 6 feet tall walks away from the pole with a speed of 8 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 50 feet from the base of the pole?"

It includes a hint that says, "you should draw a picture of a right triangle with the vertical side representing the pole, and the other end of teh hypotenuse representing the tip of the woman's shadow. Where does the woman fit into this picture? Label her postion as a variable, and labelthe tip of her shadow as another variable. You might like to use similar triangles to find a relationship between these two variables."

Problem is, I have no idea HOW to state the relationship between the two triangles I am supposed to draw. Could anyone point me in the right direction? I'm fine with implicit differentiation and everything--just stuck on this one. Thanks.

2. corresponding sides of similar triangles are proportional ...

$\frac{11}{x+s} = \frac{6}{s}$

simplify the relationship between x and s, then take the time derivative to see how their rates of change are related.

3. scratch that Now I have seen Skeeters reply It is much simpler than I imagined.

4. Okay. I simplified, and got 6/5(x)=s. Then, I took the derivative of each side with respect to time t, and got 6/5(dx/dt)=5(ds/dt). I plug in the rate at which the woman walks (8 ft/sec [side note: how does she walk so freakin' fast?!] which I assumed to be the derivative of x in terms of t) and get 6/5(8)=ds/dt, so I thought 48/5 equaled the rate at which the tip of her shadow moves when she is 50 ft from the base of the pole. My homework software says that is wrong.

So, I figured I would plug in 50, her distance from the pole, at 6/5x=s. So, I get s=60. What do I do with this?!

5. How fast is the tip of her shadow moving when she is 50 feet from the base of the pole?"
tip of the shadow moves at the rate $\frac{dx}{dt} + \frac{ds}{dt}$ because it is the rate at which $(x+s)$ changes

distance from the base of the pole does not matter.