Question

A 25 ft ladder is leaning against a vertical wall. The bottom of the ladder is pulled horizontally away from the wall at .2 in/sec. Determine how fast the top of the ladder is sliding when the top of the ladder is 20 ft above the floor? I really need this... so please help!...Thanks

Answer 1

are you in calculus?

Answer 2

yes and i am not sure if i have the correct answer, and it it getting graded, so i want it to be correct.

Answer 3

ok well I can show you how I would calculate it and then we can compare



Is this your drawing?

You have the rate in inches so you will have to make a conversion.

Answer 4

yes it is

Answer 5

Cool, so you have to find the rate at which the y distance is shrinking -> therefore I need an equation for y

Using Pythagorean Theorem \[A^2 + B^2 = C^2 -> x^2 + y^2 = (246)^2 -> y = \sqrt{(246)^2 - x^2}\]

Do you have this equation?

Answer 6

When i convert from the 25 feet to inches, yes i do, i put my work into feet, so that may have confused me. But go on.

Answer 7

Then you need to solve for x in terms of time:

\[x = 0.2t\]

Plus this into the previous to get y in terms of time

\[y(t) = \sqrt{(246)^2 - (0.2t)^2}\]

Solve for t when y = 20 ft = 240 in

Answer 8

I am staying in inches so converting feet to inches

Answer 9

Ok i will, thanks.

Answer 10

Solve for time when y = 240 in

\[240 = \sqrt{(246)^2 - (0.2t)^2} -> t = 270 seconds\] I dont have a handy calculator so I am using an online one, I think that should be the number

Answer 11

Then you have the constraints solved for and need to find the rate at which y shrinks:

\[\frac{dy}{dt} = \frac{1}{2}[(246)^2 - (0.2t)^2]^\frac{-1}{2}(-0.8t) \]

when t = 270 seconds

Answer 12

I get \[\frac{dy}{dt} = -0.45 in/sec\] rate going down the wall

Answer 13

same, thank you

Answer 14

i have to say that seems like loads of extra work
not that it is wrong, just extra work

Answer 15

from your picture you have \(x^2+y^2=25^2\) taking derivatives gives
\[2xx'+2yy'=0\] solve to get
\[y'=-\frac{xx'}{y}\]