Question

A 25-ft ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.18 ft/sec, how fast, in ft/sec, is the top of the ladder sliding down the wall, at the instant when the bottom of the ladder is 20 ft from the wall?

Answer 1

x^2 + y^2 = 625
2x dy/dx + 2y dy/dt= 0
x=20
y=15
dy/dt=0.18
What do I do next?

Answer 2

@AihberKhan

Answer 3

@Compassionate

Answer 4

@triciaal

Answer 5

You're using y for the vertical component leading up to the ladder, yes?
It looks like we have x' = 0.18, it's the base of the ladder sliding away.
And we're trying to solve for y', the rate at which the top of the latter is sliding down the wall.

Answer 6

Your differentiation looks good!
2xx'+2yy'=0
Plug in all of your information, solve for y'
:)

Answer 7

What do I plug it into? Im lost there. .-.

Answer 8

You differentiated, and a bunch of new variables popped up.
These "instantaneous rates of change".

They gave us information about how the ladder is moving in the x direction,
and you solved for the x and y components,
x' = 0.18
x = 20
y = 15

Plug those in... solve for y'

2xx'+2yy'=0

Answer 9

You took a derivative and have an equation involving 4 unknowns.
See how you have three of the pieces of information to plug into it? :O

Answer 10

Oh okay. Thank you so much for your help. I have an appointment and I hve to go, so Ill have to finish this later, but still appreciate your assistance. cx

Answer 11

np

Answer 12

Im still lost on what to plug in and is 0.18 x' or y' because i have it as dy/dt?

Answer 13

@AihberKhan

Answer 14

@hartnn

Answer 15

Sorry... I am not sure about this question

@KaleidoscopicsInk

Answer 16

It's okay.

Answer 17

y' = dy/dt.
I was just using other notation.