Exploring related rates: Falling ladder

0:00So I've got a 10 foot ladder that's leaning against a wall. But it's on very slick ground, and it starts to slide outward. And right when it's-- and right at the moment that we're looking at this ladder, the base of the ladder is 8 feet away from the base of the wall.

0:14And it's sliding outward at 4 feet per second. And we'll assume that the top of the latter kind of glides along the side of the wall, it stays kind of in contact with the wall and moves straight down. And we see right over here, the arrow is moving straight down.

0:27And our question is, how fast is it moving straight down at that moment? So let's think about this a little bit. What do we know and what do we not know? So if we call the distance between-- let's call the distance between the base of the wall and the base the ladder, let's call that x.

0:44We know right now x is equal to 8 feet. We also know the rate at which x is changing with respect to time. The rate at which x is changing with respect to time is 4 feet per second. So we could call this dx dt. Now let's call the distance between the top of the ladder and the base of the latter h.

1:05Let's call that h. So what we're really trying to figure out is what dh dt is, given that we know all of this other information. So let's see if we can come up with the relationship between x and h and then take the derivative with respect to time, maybe using the chain rule.

1:25And see if we can solve for dh dt knowing all of this information. Well, we know the relationship between x and h at any time because of the Pythagorean theorem. We can assume this is a right angle. So we know that x squared plus h squared is going to be equal to the length of the ladder squared, is going to be equal to 100.

1:48And what we care about is the rate at which these things change with respect to time. So let's take the derivative with respect to time of both sides of this. We're doing a little bit of implicit differentiation. So what's the derivative with respect to time of x squared?

2:05Well the derivative of x squared with respect to x is 2x. And we're going to have to multiply that times the derivative of x with respect to t, dx dt. Just to be clear, this is the chain rule. This is the derivative of x squared with respect to x, which is 2x, times dx dt to get the derivative of x squared with respect to time.

2:28Just the chain rule. Now similarly, what's the derivative of h squared with respect to time? Well that's just going to be 2h, the derivative of h squared with respect to h is 2h times the derivative of h with respect to time. Once again, this right over here is the derivative of h squared with respect to h, times the derivative of h with respect to time, which gives us the derivative of h squared with respect to time.

2:55And what do we get on the right-hand side of our equation? Well the length of our ladder isn't changing. This 100 isn't going to change with respect to time. Derivative of a constant is just equal 0. So now we have it, a relationship between the rate of change of h with respect to time.

3:10The rate of change of x with respect to time. And then at a given point in time, when the length of x is x and h is h. But do we know what h is when x is equal to 8 feet? Well, we can figure it out. When x is equal to 8 feet, we can use the Pythagorean theorem again.

3:27We get 8 feet squared, plus h squared is going to be equal to 100. So 8 squared is 64. Subtract it from both sides, you get h squared is equal to 36. Take the positive square root, a negative square root doesn't make sense because then the ladder would be below the ground, it would be somehow underground.

3:48So we get h is equal to 6. So this is something that was essentially given by the problem. So now we know. We can look at this original thing right over here, we know what x is, that was given. Right now x is 8 feet. We know the rate of change of x with respect to time.

4:03It's 4 feet per second. We know what h is right now, it is 6. So then we can solve for the rate of h with respect to time. So let's do that. So we get 2 times 8 feet, times 4 feet per second, so times 4, plus 2h, is going to be plus 2 times, our height right now is 6, times the rate at which our height is changing with respect to t is equal to 0.

4:43And so we get 2 times 8 times 4 is 64. Plus 12 dh dt is equal to 0. We can subtract 64 from both sides, we get 12. 12 times the derivative of h with respect to time is equal to negative 64. And then we just have to divide both sides by 12. And so now we get a little bit of a drum roll.

5:07The derivative, the rate of change of h with respect to time is equal to negative 64 divided by 12. It's equal to negative 64 over 12, which is the same thing as negative 16 over 3, yeah that's right. Which is equal to-- let me scroll over to the right a little bit-- negative 5 and 1/3 feet per second.

5:30So we're done. But let's just do a reality check, does that make sense that we got a negative value right over here? Well our height is decreasing. So it completely makes sense that its rate of change is indeed negative. And we're done.