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.

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### Video Transcript

In this video, weβ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 π and the volume π£ 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 itβ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, itβ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, weβre thinking about its derivative with respect to time. So in this case, letting the area of the circle be π΄, the rate of change of the area will be dπ΄ by dπ‘, 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 π΄ with respect to π, but weβre going to struggle differentiating π΄ with respect to π‘.

Luckily, implicit differentiation, which is a special version of the chain rule, can help. This says that since π΄ is a function in π, the derivative of π΄ with respect to π‘ is equal to dπ΄ by dπ times dπ by dπ‘. And so we really need to find expressions or values for dπ΄ by dπ and dπ by dπ‘. Of course, we already identified that π΄ is equal to ππ squared, so dπ΄ by dπ is fairly straightforward. π is, of course, a constant. And so we can use the general power rule for differentiation to find dπ΄ by dπ. Itβs two ππ.

But what about dπ by dπ‘? Well, in fact, weβre told that the radius of the circle increases at a rate of three millimeters per second. This means that dπ by dπ‘, the rate of change of the radius, is positive, since itβs increasing, three. And so dπ΄ by dπ‘ is the product of these two expressions. Itβs two ππ times three, which is equal to six ππ. 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 π equals 15 into this expression. And this will give us the instantaneous rate of change of the area when the radius is this value. dπ΄ by dπ‘ at π 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 weβ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, weβ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, weβ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 π΄, its rate of change is dπ΄ by dπ‘.

We do have a slight problem, though. The surface area of a sphere whose radius is π units is four ππ squared. And so to find an expression for the rate of change of π΄, dπ΄ by dπ‘, weβre going to use implicit differentiation. This says that the derivative of π΄ with respect to π‘ is equal to the derivative of π΄ with respect to π times dπ by dπ‘. Now dπ΄ by dπ can be calculated using the power rule for differentiation. Since π is a constant, dπ΄ by dπ is two times four ππ, which is eight ππ. And so dπ΄ by dπ‘ is the product of eight ππ and dπ by dπ‘.

Usually weβd look to find some value or expression for the rate of change of radius with respect to time. Thatβs dπ by dπ‘. In this question, though, weβ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 π£, weβre told information about dπ£ by dπ‘. In fact, the balloon is leaking helium, so its derivative, the rate of change of π£ with respect to time, is going to be negative. Itβs negative 48. This doesnβt help us particularly at this point with finding dπ by dπ‘, but we can perform a similar process using implicit differentiation but this time for dπ£ by dπ‘.

We know that dπ£ by dπ‘ could be calculated by finding the product of dπ£ by dπ and dπ by dπ‘. We know the value of dπ£ by dπ‘. Itβs negative 48. So, in fact, if we can find an expression or value for dπ£ by dπ, we can rearrange to find a formula or value for dπ by dπ‘. And we know, in fact, that the volume of a sphere with radius π is four-thirds ππ cubed. And so its derivative with respect to π, dπ£ by dπ, is three times four-thirds ππ squared, which is four ππ squared. And at this stage, we can also replace dπ£ by dπ‘ with negative 48. And so now we see that we can find a formula for dπ by dπ‘ by dividing both sides of this equation by four ππ squared. Thatβs negative 48 over four ππ squared, which becomes negative 12 over ππ squared.

And now we see we can replace dπ by dπ‘ in our earlier formula for dπ΄ by dπ‘. When we do, we get that dπ΄ by dπ‘ is eight ππ times negative 12 over ππ squared. We can simplify by cross canceling one value of π and one π, since we know π is not gonna be equal to zero. And so we have an expression for dπ΄ by dπ‘ in terms of π. Itβs negative 96 over π. The rate of change of the surface area when π is equal to 41 then is found by substituting π equals 41 in. And itβ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. dπ΄ by dπ‘ is negative 96 over 41 square centimeters per second.

In our next example, weβ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.

Weβre given quite a lot of information about the area and information about the length and width of a rectangle. Letβs begin by defining the length of our rectangle to be equal to π centimeters and its width to be equal to π€. And so itβs area π΄ will be equal to its length times its width ππ€. Thatβs not quite enough, though. Weβre looking to find information about the rate of change of the area, in other words, the derivative of area π΄ with respect to π‘. 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 dπ΄ by dπ‘ in terms of π, π€, and their respective derivatives. Letβs begin with the product rule. π΄ itself is the product of what we assume are two differentiable functions π and π€. Now we can make that assumption because in the question weβre given the rate of change of π and the rate of change of π€. So the derivative with π΄ with respect to π‘ according to the product rule is π times dπ€ dπ‘ plus π€ times dπ by dπ‘.

And this is really useful because weβre told the length of the rectangle increases at a rate of 15 centimeters per second. And so the derivative of π with respect to π‘ must be positive since itβs increasing and it must be equal to 15. Similarly, we can say that the derivative of π€ with respect to π‘, the rate of change of its width, is 13. Substituting these values into our earlier expression for dπ΄ by dπ‘, and we get 13π plus 15π€. And actually thatβs everything we need because weβre told that the length of the rectangle at a given point is 25 centimeters and its width is 12.

And so weβre going to substitute these values into our expression for dπ΄ by dπ‘. 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 weβ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 hadnβt been told that it was increasing, we could infer that from the sign of the derivative; itβs positive.

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

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

So letβs begin by looking at what the question actually wants us to find. It wants us to find the rate of change of its π¦-coordinate. Now we recall that the rate of change of a quantity is its derivative with respect to time. So in this case, weβre looking to find dπ¦ by dπ‘. And in fact, weβre given information about the rate of change of the π₯-coordinate with respect to time. Weβre given information about dπ₯ by dπ‘. At a given point, at the point negative one, three, the value of dπ₯ by dπ‘ is equal to two. And so what weβre going to do first is begin by differentiating the entire equation of our curve. Weβll need to use implicit differentiation to do so.

Remember, weβre differentiating with respect to π‘. And the derivative of zero is, of course, zero. We can then go through and differentiate term by term. And of course, weβre going to use implicit differentiation to do so. Letβs begin with six π¦ squared. We can differentiate six π¦ squared with respect to π¦. It will be two times six π¦. That simplifies to 12π¦. And so that means the derivative of this expression with respect to π‘ will be 12π¦ times dπ¦ by dπ‘. We perform a similar process for two π₯ squared. We differentiate it with respect to π₯. And thatβs two times two π₯ or four π₯. And then we multiply that by dπ₯ by dπ‘.

Next, we differentiate negative two π₯ with respect to π‘. And to do so, we differentiate with respect to π₯ and then times that by dπ₯ by dπ‘. In a similar way, the derivative of five π¦ with respect to π‘ is five dπ¦ by dπ‘. Then the derivative of our constant, negative 13, is simply equal to zero. Weβre now going to factor for dπ¦ by dπ‘ and dπ₯ by dπ‘ just to neat some things up a little bit. When we do, we find that dπ¦ by dπ‘ times 12π¦ plus five plus dπ₯ by dπ‘ times four π₯ minus two is equal to zero.

Since weβre trying to find the value of dπ¦ by dπ‘ at a given point, weβre going to rearrange to make dπ¦ by dπ‘ the subject. We begin by subtracting the term containing dπ₯ by dπ‘ from both sides; then we divide through by 12π¦ plus five. And so our expression for dπ¦ by dπ‘ is as shown. All thatβs left is to perform a few substitutions. Firstly, we know the value of dπ₯ by dπ‘ is equal to two. And weβre interested in finding the rate of change of the π¦-coordinate with respect to time at the same point, when π₯ is equal to negative one and π¦ is equal to three.

And so we substitute all of these values into our expression for dπ¦ by dπ‘. 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 arenβt given any units for the rate of change of the π₯-coordinate. Although, we might assume that itβs units per time. And so the rate of change of the π¦-coordinate with respect to time at the same point is 12 over 41 or 12 over 41 units per time.

Now that weβve demonstrated a number of examples on how to answer related rates problems, letβ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 itβs negative, the quantity is decreasing.