Question

The top of a 45-foot ladder is sliding down a wall at a rate of 12 feet per second. How fast is the base of the ladder sliding away from the wall at the instant when the top of the ladder is 27 feet from the ground?

Answer 1

Do you think here you would have to assume that dy/dt is equal to dx/dt?

Answer 2

I think so

Answer 3

I can't necessarily figure out how to prove that to you, but it seems to be common sense, the ladder has to go somewhere, and the best I could do, is say for how much it moves down it has to move out. So this means that at any point in time before the ladder strikes the ground, the bottom of the ladder is moving 12 feet persecond.

Answer 4

how do I solve the problem?

Answer 5

oh, haha. Alright.

Answer 6

\[x^2+y^2=45^2\\
x^2=(45^2+27^3)\\
x=36
\]

Answer 7

Except a 1 foot displacement of the top of the ladder doesn't result in a 1 foot change in x. Consider y = 27, then x = 36 as just pointed out. Make y = 28 (1 foot earlier), and x now equals 45^2-28^2 = sqrt{1241) = 35.23 feet...

Answer 8

\[\large x^2 + y^2 = z^2\]\[\large x^2 +(27)^2 = (45)^2\]\[\large x = \sqrt{(45)^2 - (27)^2}= \sqrt{1296} = 36\]\[\large \frac{d}{dx}(x^2+y^2=z^2)\]\[\large 2xx' + 2yy' = 2zz'\]\[\large xx' +yy' = zz'\]

Answer 9

@DanielM_113 that should be \[x=\sqrt{45^2-y^2}\]

Answer 10

thanks

Answer 11

dy/dt=12
\[x=\sqrt{45-y^2}\\
\frac{\mathrm d x}{\mathrm d y} = ?\]

Answer 12

The Beatles had a revolution about the answer :-)

Answer 13

omgi jsut got it,.the ladder isn't changing speed soo z' would be 0!!!

Answer 14

\[\frac{dy}{dt}=-12\]\[x=\sqrt{45^2-y^2}=\sqrt{2025-y^2}\]\[\frac{dx}{dy} = -\frac{y}{\sqrt{2025-y^2}}\]\[\frac{dx}{dt} = \frac{dx}{dy}*\frac{dy}{dt} = -\frac{27}{\sqrt{2025-27^2}}*(-12) = \]

Answer 15

\[ \frac{\mathrm d x}{\mathrm d y} = \frac{\mathrm d (\sqrt{45^2-y^3})}{dy}\]
So you use substitution:
\[u=45^2-y^2\\
\mathrm d \sqrt{u}/\mathrm du= -\frac{1}{2\sqrt{u}}\]
And then use the chain rule.

Answer 16

@Jhannybean what do you mean z' would be 0?

Answer 17

\[\large xx' +yy' = zz'\]\[\large z'= 0\]\[\large (36)x' +(27)(12) = 0\]\[\large 36x' = -324 \implies x' = -9 \]

Answer 18

I'm not sure I understand @Jhannybean's method, but it did get the correct answer.

Answer 19

oh, now I see, z is just the length of the ladder. I was thinking she had somehow involved a third dimension :-)

Answer 20

You got it.

Answer 21

the ladder is moving away at 9 ft/s the negative just tells you the direction its moving.

Answer 22

Yes, that would be the most straightforward answer.

Answer 23

The Beatles song I alluded to is "Revolution 9", way before most of your time, I imagine :-)

Answer 24

Let me write it up as one post.

Answer 25

@whpalmer4 Not before mine. My Beatles music has gotten me 11k subscribers on YouTube.

Answer 26

\[\large x^2 + y^2 = z^2\]\[\large x^2 +(27)^2 = (45)^2\]\[\large x = \sqrt{(45)^2 - (27)^2}= \sqrt{1296} = 36\]\[\large \frac{d}{dx}(x^2+y^2=z^2)\]\[\large 2xx' + 2yy' = 2zz'\]\[\large xx' +yy' = zz'\]\[\large z'= 0\]\[\large (36)x' +(27)(12) = 0\]\[\large 36x' = -324 \implies x' = -9\]

Answer 27

well first of all remember the expression x^2+y^2=45......(1)
now let the ladder be at a distance y upwards and x leftwards and now it moves down a distance dy and leftwards by a distance dx..(dx and dy are very small displacements)
so now also...(x+dx)^2+(y+dy)^2=45.....(2)
now subtract....(1) from (2)...
we will get...(dx)^2+2x(dx) +(dy)^2-2y(dy)=0...
neglect dx^2 and dy^2...as they are very small..
so we get 2x(dx)=2y(dy)
or dy/dx=x/y
or (dy/dt)/(dx/dt)=x/y
or 12/(dx/dt)=36/27(by ...(1),,and it is given that y=27)
so dx/dt=9 leftwards...!!!!

Answer 28

@jhannybean To be picky, \(y' = -12\), and so \(\x' = 9\) The drawing ought to have the ladder base sliding to the right so that everything corresponds to our normal frame of reference :-)

Answer 29

Ohhhh my goodness. Yeah you're right -_-

Answer 30

Lots of interesting solutions. One of the few times I wish I could give out multiple medals!