In this explainer, we will learn how to use derivatives to find the relation between the rates of two or more quantities in related rates problems.
The rate of change of quantities that change over time, such as displacement or velocity, can be calculated using derivatives. If two related quantities are changing over time, then the rates at which they change, and hence their derivatives, are also related.
Consider a spherical balloon being inflated with air; both the radius and the volume of the balloon are increasing over time. The volume of a sphere is in direct proportion to the cube of its radius; hence, the rate of change of the volume and the rate of change of the radius are also related.
The sign of the derivative can also indicate whether the rate of a change of a particular quantity is increasing or decreasing over time, depending on whether its positive or negative respectively.
These problems can be solved by an application of implicit differentiation and are useful for when you want to find the unknown rate of some quantity, by relating it to rates of other quantities that are known.
Implicit differentiation makes use of the chain rule and allows you to find the derivative of some quantity with respect to time using implicitly defined functions, as usually for related rates we do not know the explicit relationship between that quantity and time.
For these problems, it can be useful to begin by looking at how to apply the process in a specific example with the volume of a cube.
Example 1: Finding an Expression That Represents the Rate of Change of a Cubeβs Volume Using Related Rates
If is the volume of a cube with edge length and the cube expands as time passes, give an expression for .
Answer
In this example, we want to find an algebraic expression for the rate of change of the volume expressed in terms of the rate of change of the edge length of the cube. Let us visualize the problem with a simple diagram that shows we have a cube that is expanding.
The volume, in cubic units, of a cube with edge length is given by . We begin by finding an expression for the derivative of with respect to the variable as
We then calculate the rate of change of the volume with respect to time, , by using implicit differentiation.
The expression of this rate of change is
We can formalize this process in the following definition when two variables and are related by a given function.
Definition: Related Rate of Change for π¦ = π(π₯)
If we have two variables and that are related via the function and are both changing with time, these variables can also be written as a function of , and , to emphasize their time dependence. We can express the rate of change of in terms of the rate of change of by using implicit differentiation and the chain rule as
Let us consider an example in which an object is moving along a parabolic path described by , as shown in the following diagram.
If we are given that, at a particular time , the -coordinate is and the rate at which the -coordinate is changing is then we are also able to find the rate at which the -coordinate is changing, , at this specific time. From the definition of related rates of change, we know that
We can find the derivative of with respect to explicitly as therefore,
We wish to evaluate the rate of change of when , and at that time we are given that and , so
Next, let us consider an example in which we have a circle with a radius, , that is increasing with time . We can visualize this problem with a simple diagram that shows we have a circle that is expanding.
If we are given that the rate of change is known and constant, then we can use this to find the rate of change of the area of the circle. The area at any given time is given by hence,
The quantity we want to find is , but since the radius is a function of and the formula does not explicitly contain any of the variables, we have to differentiate this implicitly using the chain rule. In particular,
Now let us return to our example of a spherical balloon being inflated with air, as shown in the following diagram.
Suppose the balloon is inflated at a constant rate of 2 cm3/s and we wish to find out how fast the radius, , is increasing at the instant when . As the balloon inflates, both the radius and the volume are increasing with respect to time; therefore, their rates of change are positive. Since the balloon is spherical, we know that the volume is given by hence,
We can differentiate the volume implicitly with respect to time:
But since we are given that the volume is increasing at a constant rate of 2 cm3/s, we have hence, which we can rearrange to find
Finally, we want to find the rate of change of the radius when , which we can substitute into the expression for the derivative as
It is important to fully define all of the expressions; a common misconception is to confuse variables and constants. Related rate problems usually involve multiple expressions; some represent quantities, while others represent rates of change; some are variable and will change over time, while others are constants.
We will now look at a few examples to practice and deepen our understanding of the types of problems involving related rates of change. These examples have an explicit expression for the relationship between the quantity you want to find the rate of and other quantities.
Example 2: Finding the Rate of Change of the Surface Area of a Shrinking Sphere given the Rate of Change of Its Volume Using Related Rates
A spherical balloon leaks helium at the rate of 48 cm3/s. What is the rate of change of its surface area when its radius is 41 cm?
Answer
In this example, we want to find the rate of change of the surface area of the balloon, at the instant of a particular radius and a given constant rate of change for the volume. While the solution will be algebraic at the start, the final answer will be numerical after we substitute the given values.
The volume and surface area of a sphere of radius are given by
Since the balloon leaks helium, its volume is decreasing over time, which we can visualize with a diagram.
This means the rate of change of the volume is negative and given by
We can also differentiate the expression for the volume with respect to time implicitly using the explicit derivative of the volume with respect to : hence,
Setting , we have which we can solve for to get
The rate of change we want to find is that of the surface area , which we can compute implicitly using the explicit derivative for the surface area with respect to : hence,
And substituting , we get
Finally, substituting the value , we find
Therefore, the rate of change of the surface area of the balloon when its radius is cm2/s.
In our next example, letβs find the rate of change in the area of an expanding rectangle using related rates.
Example 3: Finding the Rate of Change in the Area of an Expanding Rectangle Using Related Rates
The length of a rectangle is increasing at a rate of 15 cm/s and its width at a rate of 13 cm/s. Determine the rate at which the area of the rectangle increases when the length of the rectangle is 25 cm and its width is 12 cm.
Answer
In this example, we want to find the rate of change of the area of an expanding rectangle, at the instant of a length and width and a given constant rate of change for the length and width, as shown in the diagram.
While the solution will be algebraic at the start, the final answer will be numerical after we substitute in the given values.
Let the width of the rectangle be cm and the length be cm. The area of the rectangle, cm2, is given by
The rate of change of the area can be computed by finding the derivative of the product of two differentiable functions, and . Recalling that the product rule states that , we obtain
Since the length and width are both increasing, the rate of change will be positive for both. Let and To determine the rate at which the area increases when and , we can substitute these values to find
Since the rate of change of the area, , is positive, we can say that the rate of change at which the area increases is 505 cm2/s.
Now, letβs consider a slightly different example; here, the equation of the curve is an implicit relationship between the variables; that is, the equation for the curve is not purely given as a function of .
Example 4: A Related Rates Problem on a Particle Moving along a Given Curve Where It Is Required to Find the Rate of Change in Its π¦-Coordinate
A particle is moving along the curve . If the rate of change of its -coordinate with respect to time as it passes through the point is 2, find the rate of change of its -coordinate with respect to time at the same point.
Answer
In this example, we want to find the rate of change of the -coordinate for a particle that moves along a curve, as shown in the diagram, at the instant of a particular point for and a given constant rate of change of the -coordinate.
While the solution will be algebraic at the start, the final answer will be numerical after we substitute the given values.
In order to find the rate of change of , , we begin by implicitly differentiating the equation of the curve with respect to . Then, we substitute the rate of change of the -coordinate, , at the point where and .
If we take the derivative of the equation of the curve with respect to , we find
We can now substitute in the point , and the rate of change of the -coordinate :
Then, rearranging for , we have
The rate of change of the -coordinate is, therefore, .
In the next example, letβs find the rate of change of the length of the shadow of a man moving away from a light pole.
Example 5: Finding the Rate of Change of the Shadow of a Man Moving Away from a Light Pole Using Related Rates
A 1.8-metre-tall man is moving away from a streetlight at 1.5 m/s. Given that the lamp on the pole is 5.5 m above the ground, what is the rate of change of the length of the manβs shadow? Give your answer to one decimal place.
Answer
In this example, we want to find the rate of change of the length of a manβs shadow as he moves away from a streetlight at a particular speed, as shown in the diagram.
While the solution will be algebraic at the start, the final answer will be numerical after we substitute the given values.
Let be the distance away from the light pole on the ground and be the length of the shadow. In order to find the rate of change of the length of the manβs shadow, , we first find a relationship between this length, , and the distance from the light pole, . We can do this using similar triangles, as shown in the figure.
Since the angles in both triangles are identical, the ratios of the lengths of the sides of the triangle are equal too. We can use this to express as a function of :
We can take the derivative of this with respect to to find the relationship between their respective rates as
Now we can substitute the given rate, , to find
Thus, the rate of change of the length of the manβs shadow to one decimal place is 0.7 m/s.
Now, letβs consider an example where we have to find the rate of change of the distance between a fixed point and a point moving on the curve, at a particular rate.
Example 6: Finding the Rate of Change of the Distance between a Fixed Point and a Point Moving on the Curve of a Root Function Using Related Rates
A point is moving on the curve of the function . If its -coordinate increases at a rate of cm/s, find the rate of change of the distance between this point and the point at .
Answer
In this example, we want to find the rate of change of the distance between a given point and a point moving on a curve, as shown in the diagram, at a particular rate and position.
We note that the distance between two points and is given by
Therefore, if we take the function , the distance between the points and can be expressed as
The derivative of this expression with respect to is
The quantity we want to find is , but since is a function of and the formula does not explicitly contain any of the variables, we have to differentiate this implicitly using the chain rule. In particular, we have
We can now substitute the point and the rate of change of the -coordinate, :
Therefore, the rate of change of the distance is given by 45 cm/s.
In the next example, we will determine how fast the area is changing in a triangle with an increasing angle and two sides of constant length.
Example 7: A Related Rates Problem of a Triangle with Two Sides of Constant Length While the Included Angle Increases at a Given Rate
A triangle with sides and and contained angle has area . Suppose that , , and the angle is increasing at 0.6 rad/s. How fast is the area changing when ?
Answer
In this example, we want to find the rate of change of the area of a triangle at the instant when the angle is a particular value. The contained angle is increasing for a triangle where two of the sides are fixed, as shown in the diagram.
The area of the triangle with the given sides, and , can be expressed in terms of as
The quantity we want to find is at , given . We can differentiate the expression for the area implicitly with respect to to find
We can find the derivative of explicitly with respect to as therefore,
We can now substitute the angle and the rate of change of the angle to find
Therefore, the area is increasing at 3 cm2/s.
Suppose the rate of change of a particular quantity, , is constant, which we denote by , where
Since we know that the sign of the derivative can indicate whether the rate of change of a particular quantity is increasing or decreasing over time, depending on whether it is positive or negative, respectively, this means that quantity gains or loses by the amount after every unit of time. Equivalently, the net gain is at time . If the initial value of the quantity is , then the quantity after time is given by
This clearly satisfies the rate of change, which can be verified upon differentiating this expression with respect to .
Finally, we will look at an example where we can use this and the knowledge of a particular rate of change of a quantity to solve a differential equation and determine how much a particular quantity has changed after a given period of time.
Example 8: Finding the Final Value of a Quantity given Its Initial Value and Constant Rate of Change
Given that a rocket of mass 26 tonnes is burning fuel at a constant rate of 80 kg/s, find the mass of the rocket 25 seconds after take-off.
Answer
In this example, a rocket is expelling fuel at a constant rate as shown in the diagram.
In order to find the final value of the mass, , of the rocket, at a particular time , we will first determine the mass of the rocket at any given time . While the solution will be algebraic at the start, the final answer will be a number after we substitute in the given values.
The initial mass of the rocket is 26 tonnes. Since 1 tonne is 1βββ000 kg, this is equivalent to 26βββ000 kg.
Since the mass of the rocket decreases over time as the rocket burns fuel, at a constant rate of 80 kg/s the rate of change of the mass is negative and given by
Since this is a constant rate of the quantity , we can express that quantity after time with initial value :
On substituting the given rate and the initial value , we obtain
This is intuitive since the rocket loses 80 kg after every second and it clearly satisfies the rate of change, which can be verified upon differentiating this expression with respect to . When , the mass of the rocket is
Therefore, the mass of the rocket 25 seconds after take-off is 24 tonnes.
Key Points
- If two or more quantities are related, then their rates of change are also related. This allows us to find unknown rates of change by using knowledge about related rates.
- Related rate problems are an application of implicit differentiation via the chain and product rules.
- The sign of a particular rate of a quantity tells you that quantity is decreasing or increasing over time.
