Help with Related Rates: 12-ft ladders rests against wall...

[jml14]

Problem: A ladder 12ft. long rests against a vertical wall. If the bottom of the ladder slips away from the wall at a rate of 1ft/sec , how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6ft away from the wall?

Now my teacher likes me to do this in a four step process were I have to show a diagram, write out what is given, write out what i have to find, and how the given and find steps link to get the answer.

So far I have a diagram and the given.

The Given is dr/dt = 1 ft/sec 'r' stands for rate and 'A' stands for area

I think I am suppose to be finding dA/dt, but this is were im lost and I do not understand how to carry on from here. If someone could help me that would be greatly appreciated!

 

[Deleted member 4993]

Re: Help with Related Rates Please

Problem: A ladder 12ft. long rests against a vertical wall. If the bottom of the ladder slips away from the wall at a rate of 1ft/sec , how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6ft away from the wall?

Now my teacher likes me to do this in a four step process were I have to show a diagram, write out what is given, write out what i have to find, and how the given and find steps link to get the answer.

So far I have a diagram and the given.

The Given is dr/dt = 1 ft/sec <<< What is the variable 'r' here - related to your picture?

I think I am suppose to be finding dA/dt, <<< What is the variable 'A' here - related to your picture?

 

[BigGlenntheHeavy]

Re: Help with Related Rates Please

(12)^2 = x^2+y^2, 0 =2x(dx/dt)+2y(dy/dt), x(dx/dt) =-y(dy/dt)

When x = 6, dx/dt =1 and y = 6sqrt3, hence dy/dt = -sqrt(3)/3 ft/sec.

 

[daon]

Re: Help with Related Rates Please

Problem: A ladder 12ft. long rests against a vertical wall. If the bottom of the ladder slips away from the wall at a rate of 1ft/sec , how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6ft away from the wall?

Now my teacher likes me to do this in a four step process were I have to show a diagram, write out what is given, write out what i have to find, and how the given and find steps link to get the answer.

So far I have a diagram and the given.

The Given is dr/dt = 1 ft/sec 'r' stands for rate and 'A' stands for area

I think I am suppose to be finding dA/dt, but this is were im lost and I do not understand how to carry on from here. If someone could help me that would be greatly appreciated!

Lets use 'x' and 'y' instead of r and whatever else. Let x be the distance the bottom of the ladder is from the wall, and y the distance the top of the ladder is from the ground.

The bottom of the ladder is moving. \(\displaystyle \frac{dx}{dt} = 1 \frac{ft}{s}\). You're asked to find \(\displaystyle \frac{dy}{dt}\) i.e. the rate at which the top of the latter is falling to the ground as it slips.

You're also given that the ladder is 12 feet long. We can set up a right triangle with this information having legs 'x' and 'y' and hypottenuse 12.

\(\displaystyle x^2+y^2=12^2\)

If we take the derivavtive impicitly, we get:

\(\displaystyle 2x\frac{dx}{dt}+2y\frac{dy}{dt} = 0\)

We know dx/dt, and we're given that x=6ft for what we want to find. Plug in the 1 for dx/dt and 6 for x, and solve for dy/dt.

\(\displaystyle 2(6)(1)+2y\frac{dy}{dt} = 0\)
\(\displaystyle 12+2y\frac{dy}{dt} = 0\)
\(\displaystyle 2y\frac{dy}{dt} = -12\)
\(\displaystyle \frac{dy}{dt} = \frac{-6}{y}\)

Okay?

Now, we don't have a number yet. We need to replace 'y' with something. Use your triangle again with the fact that x=6.

\(\displaystyle x^2+y^2=12^2\)
\(\displaystyle 6^2+y^2=12^2\)
\(\displaystyle 36+y^2=144\)
\(\displaystyle y^2=108\)
\(\displaystyle y=\sqrt{108} =6\sqrt{3} ft\)

Now, plug this y into your dy/dt above:

\(\displaystyle \frac{dy}{dt} = \frac{-6}{y} = \frac{-6}{6\sqrt{3}} = \frac{-1}{\sqrt{3}} \frac{ft}{s}\)

Which is equivilant to BigGlenntheHeavy's answer above, only his denominator is rationalized.

 

[galactus]

Re: Help with Related Rates Please

You mentioned area, but the problem is not related to that. But, if the problem did say,

"how fast is the area the of the triangle (formed by the ladder, floor, and wall) changing at that instant", then you could use the formula for the area of a right triangle.

\(\displaystyle A=\frac{1}{2}xy\)

Differentiate implicitly via the product rule (the product rule is often overlooked in these cases, be careful):

\(\displaystyle \frac{1}{2}\left(x\cdot\frac{dy}{dt}+y\cdot\frac{dx}{dt}\right)\)

Plug in the givens and the dy/dt from the previous problem:

\(\displaystyle \frac{1}{2}\left(\frac{-6}{\sqrt{3}}+6\sqrt{3}\right)=2\sqrt{3}\)

 

[Deleted member 4993]

Re: Help with Related Rates Please

Problem: A ladder 12ft. long rests against a vertical wall. If the bottom of the ladder slips away from the wall at a rate of 1ft/sec , how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6ft away from the wall?

Now my teacher likes me to do this in a four step process were I have to show a diagram, write out what is given, write out what i have to find, and how the given and find steps link to get the answer.

So far I have a diagram and the given.

The Given is dr/dt = 1 ft/sec 'r' stands for rate <<< That cannot be correct - 'r' is rate of what?

 

[galactus]

Re: Help with Related Rates Please

I took it as dx/dt=1. I assume that's a typo.