Question

A ladder 13 meters long rests on horizontal ground and leans against a vertical wall. The foot of the ladder is pulled away from the wall at the rate of 0.6 m/sec. How fast is the top sliding down the wall when the foot of the ladder is 5 m from the wall?

Answer 1

This is the one where I can't figure out the equation.

Answer 2

Let the horizontal distance from the wall be x and let the height of the top of the ladder be y. Note that the length of the ladder is the hypotenuse of a right triangle, so we have: \[ x^2 + y^2 = 13^2 = 169. \]The statement about the rate talks about a derivative (in specific, dx/dt - because it is the rate at which x is being change, wrt t). Do you see where it's heading or would you like some more guidance?

Answer 3

so, I just take the derivative and solve for y'?

Answer 4

2x*x'+2y*y'=0?

Answer 5

plug in x, y and x'

Answer 6

Wait, I'm confused again.

Answer 7

I don't have a y

Answer 8

or am I taking the derivative in respect to y?

Answer 9

even then, I'm confused.

Answer 10

Interpret the question. What is it asking you - as a derivative? I'm trying not to give away the solution so that you can think about the problem a little. Bear with me.

Answer 11

Ok, no worries. Um..

Answer 12

I'm assuming its in respect to time.

Answer 13

but when I get the derivative, I don't know what to plug in for y.

Answer 14

I know to plug in 0.6 to x' and 5 to x and I know I'm looking for y'

Answer 15

Is that wrong?

Answer 16

I figured it out I think.

Answer 17

I just get y by plugging in x to the original equation. I feel silly.

Answer 18

5, 12, 13

Answer 19

right.

Answer 20

You can solve for y when x = 5 use pythgorean like Tonks did.

Answer 21

Don't be surprised if dy/dt is negative.

Answer 22

Ok. Thanks!