# Video: Using Implicit Differentiation to Solve a Rates of Change Problem

The top of a 25-foot ladder is leaning against a vertical wall in a museum, and the bottom is on the floor. The bottom of the ladder is being pulled horizontally along the floor, away from the wall at a rate of −9 ft/s. The top of the ladder remains in contact with the wall as it slides. When the bottom of the ladder is 20 ft from the wall, what is the rate of change of the distance from the top of the ladder to the floor?

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### Video Transcript

The top of a 25-foot ladder is leaning against a vertical wall in a museum, and the bottom is on the floor. The bottom of the ladder is being pulled horizontally along the floor away from the wall at a rate of negative nine feet per second. The top of the ladder remains in contact with the wall as it slides. When the bottom of the ladder is 20 foot from the wall, what is the rate of change of the distance from the top of the ladder to the floor?

To visualize a problem, let’s draw a quick sketch of the ladder resting against the wall. There’s quite a lot of information given to us in this question. So let’s go through the numbers that we’ve been given. The ladder is 25 foot and we’re told the ladder is being pulled away from the wall at a rate of negative nine foot per second. But what does it mean by negative nine foot per second. A negative rate of change indicates a decrease, so the distance from the wall to the bottom of the ladder is decreasing. So it’s actually moving in this direction. And as this happens, the top of the ladder will raise up against the wall.

The rate of this raising is what we’ve been asked to find. Let’s call the distance from the wall to the bottom of the ladder 𝑥 and the distance, from the ground to the top of the ladder 𝑦. So we want to calculate the rate of change of which the top of the ladder is moving. And when we want to calculate the rate of change, we use derivatives. But we need a starting point. We need some kind of equation to work with. But if we look here, we’ve actually ended up with a right-angle triangle and we know a very useful formula for dealing with right-angled triangles. It’s the Pythagorean theorem. That is, 𝑎 squared plus 𝑏 squared equals 𝑐 squared, where 𝑐 is the hypotenuse.

So for our question, we can say that 𝑥 squared plus 𝑦 squared equals 25 squared. And 25 squared is 625. So this gives us an equation to work with. And remember, we want to find the rate of change of the distance 𝑦, with respect to time. So we want to find d𝑦 by d𝑡. So let’s differentiate the equation that we’ve got with respect to time. And because this is a function of 𝑥 and 𝑦 and we’re differentiating with respect to time 𝑡, we need to do implicit differentiation, which is actually just an extension of the chain rule and gives us the formula d𝑓 by d𝑡 equals d𝑓 by d𝑥 multiplied by d𝑥 by d𝑡.

So we’ll differentiate implicitly term by term starting with 𝑥 squared. So we have that the derivative of 𝑥 squared with respect to 𝑡 is the derivative of 𝑥 squared with respect to 𝑥 multiplied by d𝑥 by d𝑡. But we know the derivative of 𝑥 squared with respect to 𝑥. It’s just two 𝑥. So this gives us two 𝑥 multiplied by d𝑥 by d𝑡. And now, we differentiate 𝑦 squared implicitly. So the derivative over 𝑦 squared with respect to 𝑡 is the derivative of 𝑦 squared with respect 𝑦 multiplied by d𝑦 by d𝑡. And we know the derivative of 𝑦 squared with respect to 𝑦 is just two 𝑦. So this is two 𝑦 multiply by d𝑦 by d𝑡. And so differentiating our equation with respect to 𝑡 gives us two 𝑥 d𝑥 by d𝑡 add two 𝑦 d𝑦 by d𝑡 equals zero. Notice that here 625 differentiated to zero because constant differentiate to zero.

Remember that we’re trying to find d𝑦 by d𝑡. So let’s rearrange this for d𝑦 by d𝑡. We start by subtracting two 𝑥 d𝑥 by d𝑡 from both sides and then divide both sides by two. We can then cancel the twos to give us d𝑦 by d𝑡 equals negative 𝑥 over 𝑦 d𝑥 by d𝑡. So now, we need to substitute values for 𝑥, 𝑦, and d𝑥 by d𝑡. Well, remember, 𝑥 is the horizontal distance from the wall to the bottom of the ladder. And actually, we were given that information in the question because we want the rate when the ladder is 20 foot from the wall. d𝑥 by d𝑡 is the rate of change of 𝑥 with respect to time. That’s the rate of change of the horizontal distance with respect to time which we were given in the question. It’s negative nine feet per second.

And finally, do we know what the value of 𝑦 is? Well, actually, we can use the Pythagorean theorem on our triangle, knowing that the hypotenuse is 25 foot and one length is 20 foot. The Pythagorean theorem gives us that 20 squared add 𝑦 squared equals 25 squared. And so 𝑦 squared equals 25 squared minus 20 squared, which is 625 minus 400 which is 225. And so 𝑦 equals the square root of 225 which is 15. So now we can substitute all of these values to find d𝑦 by d𝑡. This is going to be negative 20 over 15 multiplied by negative nine, which gives us the answer of 12 feet per second, which seems like a sensible answer because it’s positive, which indicates an increasing rate of change. Which is true because the distance from the ground to the top of the ladder is increasing as the ladder is being raised up the wall.