A 12-ft ladder is leaning against a vertical wall when Jack be...

Question

A 12-ft ladder is leaning against a vertical wall when Jack begins pulling the foot of the ladder away from the wall at a rate of 0.2 ft/s. What is the configuration of the ladder at the instant when the vertical speed of the top of the ladder equals the horizontal speed of the foot of the ladder?

Accepted Answer

Welcome back, everyone. In this problem, a 10-meter ladder is positioned against a hose. If the base of the ladder is being moved away from the hose at a speed of 0.4 m per second, determine the position of the ladder when the speed of the top of the ladder descending the wall equals the speed of the base moving away. Now, what are we trying to figure out here? Well, let's try to sketch our problem to get a better understanding, OK? So, we have a ladder that's positioned against a wall, OK. This looks like a ladder, right? And we know that the side of our wall will make a 90 degree angle with the base with the ground, where our ladder is 10 m. So let's call the length of the ladder L. Let's also call the distance. Off the top of the ladder to the base of the wall to be Y and the distance from the bottom of the ladder to the to the base to be X. Now we're trying to determine the position of the ladder, OK? In other words, the values of Y and X when the speed of the top of the ladder descending the wall equals the speed of the base moving away. In other words, if we have our speed here that it's moving downwards, the derivative of Y with respect to T. and the speed here that it's moving. Outwards at the base, the derivative of X with respect to T. Then basically we're trying to find X and Y. OK. When the absolute value of DYDT is equal to the absolute value of DXDT. Now how are we going to do that? Well, first we need to find a way to relate X, Y, and L. And we can use that by the Pythagorean theorem because notice here that a right triangle is formed. Thus by the Pythagorean theorem. That tells us then that X2 + Y2 will be equal to L2102, which is 100. Now to incorporate derivative of Y and X with respect to T, we can differentiate this expression or this equation with respect to T, OK? So differentiating both sides. With respect to tea. OK. Then we're going to get 2 XDXDT. Plus 2DYDT. To be equal to 0 because the derivative of a constant equals 0. Now, in this case, we can cancel 2 throughout our entire equation, divide our entire equation by 2. And then if we do some transposition here, OK, then we should get YDUYBT OK, to be equal to negative XDXDT. No, what do we know? Well, remember our condition from our problem statement. We want to find it when the speed of the top of the ladder descending equals the speed of the base moving away, where DYDT equals DXDT. But we already know that the ladder is being moved away from the hose at a speed of 0.4 m per second. So that means this equals 0.4 m per second, DYDT and DXDT, OK. And in that case, if we substitute it into our problem here, OK. Then that means that why, well, let me keep that in red here. Why multiplied by 0.4, OK. Why I multiplied by 0.4 here. Is going to be equal to -0.4 multiplied by X. And now remember that this is our absolute value, OK? So we have an absolute value here. And if we do some transposition, OK, uh, if we cancel out for the absolute value of 0.4, then we're gonna have 0.4 Y being equal to 0.4 X. And if we divide through by 0.4, that means X equals Y. So we know that X will be equal to Y when the speeds on both the change in vertical distance and the change in horizontal distance are equal. So now we can substitute that into our equation, OK? So, substituting. X equalsy into the relationship for those variables. That means X2 + X2 will be equal to 100. So 2 X2 equals 100, which means X2 equals 50, and thus X is going to be equal to the square root of 50, which we can simplify to be 5 root 2 m. And if X equals 5 2 m, then that also means Y will be equal to 5 root 2 m. Thus this tells us that the base of the ladder is 5 root 2 m away from the wall and the top of the ladder is 5 root 2 m above the ground when the speed of the top of the ladder descending equals descending the wall equals the speed of the base moving away. Thanks a lot for watching everyone. I hope this video helped.

A 12-ft ladder is leaning against a vertical wall when Jack begins pulling the foot of the ladder away from the wall at a rate of 0.2 ft/s. What is the configuration of the ladder at the instant when the vertical speed of the top of the ladder equals the horizontal speed of the foot of the ladder?

Key Concepts

Related Rates

Pythagorean Theorem

Instantaneous Rate of Change

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