A 6 meter ladder is leaning on a vertical wall. The foot of the ladder is being pulled at a rate of 0.15m/s. Calculate the rate of which the height of the ladder is decreasing when the ladder is 3 meters off the ground.

I started by doing
dHdT=dHdR∗dRdT

Question

A 6 meter ladder is leaning on a vertical wall. The foot of the ladder is being pulled at a rate of 0.15m/s. Calculate the rate of which the height of the ladder is decreasing when the ladder is 3 meters off the ground.

I started by doing
dHdT=dHdR∗dRdT

anonymous · Answer

It didn't copy correctly.
$$\frac{ dH }{ dT }= \frac{ dH }{ dR }*\frac{ dR }{ dT }$$

anonymous · Answer

So first draw a picture of it ^_^

anonymous · Answer

You can start this problem a few ways, but I think the most intuitive is with labeling the ladder/wall/floor system as a right triangle.

anonymous · Answer

Also, before we get started, what're you defining as R?

anonymous · Answer

Okay Im back sorry.

anonymous · Answer

|dw:1426814548109:dw|

anonymous · Answer

Cool.  Now in your equation you were using H - which side is it?

anonymous · Answer

H is the wall. R is my 0.15m/s

anonymous · Answer

To use the rate like that doesn't actually do anything in this case.  If we label the picture with one more variable, B, - one that represents the base of the triangle
|dw:1426814779142:dw|
 - then the rate is actually
$$rate = \frac{dB}{dt} = .15m/s$$