Lesson Video: Related Time Rates | Nagwa Lesson Video: Related Time Rates | Nagwa

Lesson Video: Related Time Rates Mathematics

In this video, we will learn how to use derivatives to find the relation between the rates of two or more quantities in related rates problems.

17:20

Video Transcript

In this video, we’ll learn how to use derivatives to find the relation between the rates of two or more quantities in related rates problems.

In related rates problems, the idea is to compute the rate of change of one quantity in terms of the rate of change of another quantity, which can sometimes be more easily measured. This works because if two related quantities are changing over time, then the rates at which they change and, of course, their derivatives are also related.

Consider a spherical balloon being inflated with air for instance. Both the radius 𝑟 and the volume 𝑣 of the balloon are increasing over time. And since the volume of a sphere is in direct proportion to the cube of its radius, the rate of change of the volume and the rate of change of the radius are also related. Note also that the sign of the derivative can indicate to us whether the quantity is increasing or decreasing over time, depending on whether it’s positive or negative, respectively. These problems can be solved using implicit differentiation, which makes use of the chain rule, so make sure you have a sound understanding of how this works before watching this video.

Now for related rates problems, it’s usually best to jump straight into a problem and see how it all works.

The radius of a circle is increasing at a rate of three millimeters per second. Find the rate of change of the area of the circle when the radius of the circle is 15 millimeters.

Remember, when we think about the rate of change of a quantity, we’re thinking about its derivative with respect to time. So in this case, letting the area of the circle be 𝐴, the rate of change of the area will be d𝐴 by d𝑡, in other words, the derivative of the area with respect to time. The problem is we know the formula for the area of a circle in terms of its radius. Its area is equal to 𝜋 times the radius squared. This means we can differentiate 𝐴 with respect to 𝑟, but we’re going to struggle differentiating 𝐴 with respect to 𝑡.

Luckily, implicit differentiation, which is a special version of the chain rule, can help. This says that since 𝐴 is a function in 𝑟, the derivative of 𝐴 with respect to 𝑡 is equal to d𝐴 by d𝑟 times d𝑟 by d𝑡. And so we really need to find expressions or values for d𝐴 by d𝑟 and d𝑟 by d𝑡. Of course, we already identified that 𝐴 is equal to 𝜋𝑟 squared, so d𝐴 by d𝑟 is fairly straightforward. 𝜋 is, of course, a constant. And so we can use the general power rule for differentiation to find d𝐴 by d𝑟. It’s two 𝜋𝑟.

But what about d𝑟 by d𝑡? Well, in fact, we’re told that the radius of the circle increases at a rate of three millimeters per second. This means that d𝑟 by d𝑡, the rate of change of the radius, is positive, since it’s increasing, three. And so d𝐴 by d𝑡 is the product of these two expressions. It’s two 𝜋𝑟 times three, which is equal to six 𝜋𝑟. The rate of change of area of the circle with respect to time is now given as a function of its radius.

Our final job is to substitute 𝑟 equals 15 into this expression. And this will give us the instantaneous rate of change of the area when the radius is this value. d𝐴 by d𝑡 at 𝑟 equals 15 then is six times 𝜋 times 15, which is equal to 90𝜋. Since this is the rate of change of the area, which will be in square millimeters, with respect to time, the rate of change is 90𝜋 square millimeters per second.

So we’ve seen how the formula for the area of a circle can help us find the rate of change given information about the rate of change of the radius. In our next example, we’ll look at how we can link volume and surface area of a spherical balloon.

A spherical balloon leaks helium at a rate of 48 cubic centimeters per second. What is the rate of change of its surface area when its radius is 41 centimeters?

Ultimately, in this question, we’re looking to find the rate of change of a quantity. And we know that the rate of change is to do with its derivative with respect to time. Now, in fact, in this case, we want the rate of change of the surface area. So if we let the surface area be equal to 𝐴, its rate of change is d𝐴 by d𝑡.

We do have a slight problem, though. The surface area of a sphere whose radius is 𝑟 units is four 𝜋𝑟 squared. And so to find an expression for the rate of change of 𝐴, d𝐴 by d𝑡, we’re going to use implicit differentiation. This says that the derivative of 𝐴 with respect to 𝑡 is equal to the derivative of 𝐴 with respect to 𝑟 times d𝑟 by d𝑡. Now d𝐴 by d𝑟 can be calculated using the power rule for differentiation. Since 𝜋 is a constant, d𝐴 by d𝑟 is two times four 𝜋𝑟, which is eight 𝜋𝑟. And so d𝐴 by d𝑡 is the product of eight 𝜋𝑟 and d𝑟 by d𝑡.

Usually we’d look to find some value or expression for the rate of change of radius with respect to time. That’s d𝑟 by d𝑡. In this question, though, we’re given information about the rate at which helium leaks from the balloon. This is given in cubic centimeters per second. So actually, this is the rate of change of its volume. Specifically, if we let the volume of our sphere be 𝑣, we’re told information about d𝑣 by d𝑡. In fact, the balloon is leaking helium, so its derivative, the rate of change of 𝑣 with respect to time, is going to be negative. It’s negative 48. This doesn’t help us particularly at this point with finding d𝑟 by d𝑡, but we can perform a similar process using implicit differentiation but this time for d𝑣 by d𝑡.

We know that d𝑣 by d𝑡 could be calculated by finding the product of d𝑣 by d𝑟 and d𝑟 by d𝑡. We know the value of d𝑣 by d𝑡. It’s negative 48. So, in fact, if we can find an expression or value for d𝑣 by d𝑟, we can rearrange to find a formula or value for d𝑟 by d𝑡. And we know, in fact, that the volume of a sphere with radius 𝑟 is four-thirds 𝜋𝑟 cubed. And so its derivative with respect to 𝑟, d𝑣 by d𝑟, is three times four-thirds 𝜋𝑟 squared, which is four 𝜋𝑟 squared. And at this stage, we can also replace d𝑣 by d𝑡 with negative 48. And so now we see that we can find a formula for d𝑟 by d𝑡 by dividing both sides of this equation by four 𝜋𝑟 squared. That’s negative 48 over four 𝜋𝑟 squared, which becomes negative 12 over 𝜋𝑟 squared.

And now we see we can replace d𝑟 by d𝑡 in our earlier formula for d𝐴 by d𝑡. When we do, we get that d𝐴 by d𝑡 is eight 𝜋𝑟 times negative 12 over 𝜋𝑟 squared. We can simplify by cross canceling one value of 𝜋 and one 𝑟, since we know 𝑟 is not gonna be equal to zero. And so we have an expression for d𝐴 by d𝑡 in terms of 𝑟. It’s negative 96 over 𝑟. The rate of change of the surface area when 𝑟 is equal to 41 then is found by substituting 𝑟 equals 41 in. And it’s negative 96 over 41. The units for the rate of change will be square centimeters per second.

And actually this answer makes a lot of sense. A negative rate of change means that the surface area is decreasing over time. And we know that the balloon is leaking helium, so it must be getting smaller. d𝐴 by d𝑡 is negative 96 over 41 square centimeters per second.

In our next example, we’re going to demonstrate how to link two different rate of change values with a third rate of change value. This is within the context of finding the area of a rectangle.

The length of a rectangle is increasing at a rate of 15 centimeters per second and its width at a rate of 13 centimeters per second. Determine the rate at which the area of the rectangle increases when the length of the rectangle is 25 centimeters and its width is 12 centimeters.

We’re given quite a lot of information about the area and information about the length and width of a rectangle. Let’s begin by defining the length of our rectangle to be equal to 𝑙 centimeters and its width to be equal to 𝑤. And so it’s area 𝐴 will be equal to its length times its width 𝑙𝑤. That’s not quite enough, though. We’re looking to find information about the rate of change of the area, in other words, the derivative of area 𝐴 with respect to 𝑡. And of course, the information we have about the area of the rectangle is in terms of its length and its width.

Luckily, we can use a combination of implicit differentiation and the product rule to find an expression for d𝐴 by d𝑡 in terms of 𝑙, 𝑤, and their respective derivatives. Let’s begin with the product rule. 𝐴 itself is the product of what we assume are two differentiable functions 𝑙 and 𝑤. Now we can make that assumption because in the question we’re given the rate of change of 𝑙 and the rate of change of 𝑤. So the derivative with 𝐴 with respect to 𝑡 according to the product rule is 𝑙 times d𝑤 d𝑡 plus 𝑤 times d𝑙 by d𝑡.

And this is really useful because we’re told the length of the rectangle increases at a rate of 15 centimeters per second. And so the derivative of 𝑙 with respect to 𝑡 must be positive since it’s increasing and it must be equal to 15. Similarly, we can say that the derivative of 𝑤 with respect to 𝑡, the rate of change of its width, is 13. Substituting these values into our earlier expression for d𝐴 by d𝑡, and we get 13𝑙 plus 15𝑤. And actually that’s everything we need because we’re told that the length of the rectangle at a given point is 25 centimeters and its width is 12.

And so we’re going to substitute these values into our expression for d𝐴 by d𝑡. That gives us 13 times 25 plus 15 times 12, which is 505. This is the rate of change of its area. And area will be in square centimeters. And we’re also told that the time is given in seconds. And so we see the rate at which the area of the rectangle is increasing is 505 square centimeters per second. Note that even if we hadn’t been told that it was increasing, we could infer that from the sign of the derivative; it’s positive.

In our previous examples, we’ve looked at geometric applications of related rates problems. It’s important to realize that we can also apply similar processes when working with equations of curves. Let’s see what that might look like.

A particle is moving along the curve six 𝑦 squared plus two 𝑥 squared minus two 𝑥 plus five 𝑦 minus 13 equals zero. If the rate of change of its 𝑥-coordinate with respect to time as it passes through the point negative one, three is two, find the rate of change of its 𝑦-coordinate with respect to time at the same point.

So let’s begin by looking at what the question actually wants us to find. It wants us to find the rate of change of its 𝑦-coordinate. Now we recall that the rate of change of a quantity is its derivative with respect to time. So in this case, we’re looking to find d𝑦 by d𝑡. And in fact, we’re given information about the rate of change of the 𝑥-coordinate with respect to time. We’re given information about d𝑥 by d𝑡. At a given point, at the point negative one, three, the value of d𝑥 by d𝑡 is equal to two. And so what we’re going to do first is begin by differentiating the entire equation of our curve. We’ll need to use implicit differentiation to do so.

Remember, we’re differentiating with respect to 𝑡. And the derivative of zero is, of course, zero. We can then go through and differentiate term by term. And of course, we’re going to use implicit differentiation to do so. Let’s begin with six 𝑦 squared. We can differentiate six 𝑦 squared with respect to 𝑦. It will be two times six 𝑦. That simplifies to 12𝑦. And so that means the derivative of this expression with respect to 𝑡 will be 12𝑦 times d𝑦 by d𝑡. We perform a similar process for two 𝑥 squared. We differentiate it with respect to 𝑥. And that’s two times two 𝑥 or four 𝑥. And then we multiply that by d𝑥 by d𝑡.

Next, we differentiate negative two 𝑥 with respect to 𝑡. And to do so, we differentiate with respect to 𝑥 and then times that by d𝑥 by d𝑡. In a similar way, the derivative of five 𝑦 with respect to 𝑡 is five d𝑦 by d𝑡. Then the derivative of our constant, negative 13, is simply equal to zero. We’re now going to factor for d𝑦 by d𝑡 and d𝑥 by d𝑡 just to neat some things up a little bit. When we do, we find that d𝑦 by d𝑡 times 12𝑦 plus five plus d𝑥 by d𝑡 times four 𝑥 minus two is equal to zero.

Since we’re trying to find the value of d𝑦 by d𝑡 at a given point, we’re going to rearrange to make d𝑦 by d𝑡 the subject. We begin by subtracting the term containing d𝑥 by d𝑡 from both sides; then we divide through by 12𝑦 plus five. And so our expression for d𝑦 by d𝑡 is as shown. All that’s left is to perform a few substitutions. Firstly, we know the value of d𝑥 by d𝑡 is equal to two. And we’re interested in finding the rate of change of the 𝑦-coordinate with respect to time at the same point, when 𝑥 is equal to negative one and 𝑦 is equal to three.

And so we substitute all of these values into our expression for d𝑦 by d𝑡. We get negative two times four times negative one minus two over 12 times three plus five. And that gives us a value of 12 over 41. We aren’t given any units for the rate of change of the 𝑥-coordinate. Although, we might assume that it’s units per time. And so the rate of change of the 𝑦-coordinate with respect to time at the same point is 12 over 41 or 12 over 41 units per time.

Now that we’ve demonstrated a number of examples on how to answer related rates problems, let’s recap the key points from this lesson.

In this lesson, we learned that if two related quantities change over time, the rates at which they change, and such their derivatives, are also related. We saw that implicit differentiation, that special version of the chain rule, is hugely important in related rates problems. But we can also use other rules for differentiations such as the product rule. Finally, we saw how considering the sign of the derivative can give us information about whether that quantity itself is increasing or decreasing. If the final derivative, the rate of change, is positive, then the quantity is increasing. And if it’s negative, the quantity is decreasing.

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