Lesson Explainer: Related Time Rates Mathematics • Higher Education

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 Cube’s Volume Using Related Rates

If 𝑉 is the volume of a cube with edge length π‘₯ and the cube expands as time passes, give an expression for dd𝑉𝑑.

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 π‘₯ is given by 𝑉=π‘₯. We begin by finding an expression for the derivative of 𝑉 with respect to the variable π‘₯ as dd𝑉π‘₯=3π‘₯=3π‘₯.

We then calculate the rate of change of the volume with respect to time, dd𝑉𝑑, by using implicit differentiation.

The expression of this rate of change is dddddddd𝑉𝑑=𝑉π‘₯π‘₯𝑑=3π‘₯π‘₯𝑑.

We can formalize this process in the following definition when two variables π‘₯ and 𝑦 are related by a given function.

Definition: Related Rate of Change for 𝑦 = 𝑓(π‘₯)

If we have two variables π‘₯ and 𝑦 that are related via the function 𝑦=𝑓(π‘₯) and are both changing with time, these variables can also be written as a function of 𝑑, π‘₯(𝑑) and 𝑦(𝑑), to emphasize their time dependence. We can express the rate of change of 𝑦 in terms of the rate of change of π‘₯ by using implicit differentiation and the chain rule as dddddddd𝑦𝑑=𝑦π‘₯π‘₯𝑑=𝑓′(π‘₯)π‘₯𝑑.

Let us consider an example in which an object is moving along a parabolic path described by 𝑦=π‘₯+2, as shown in the following diagram.

If we are given that, at a particular time 𝑑=5s, the π‘₯-coordinate is π‘₯=4cm and the rate at which the π‘₯-coordinate is changing is ddcmsπ‘₯𝑑=3/, then we are also able to find the rate at which the 𝑦-coordinate is changing, dd𝑦𝑑, at this specific time. From the definition of related rates of change, we know that dddddd𝑦𝑑=𝑦π‘₯π‘₯𝑑.

We can find the derivative of 𝑦 with respect to π‘₯ explicitly as dd𝑦π‘₯=2π‘₯; therefore, dddd𝑦𝑑=2π‘₯π‘₯𝑑.

We wish to evaluate the rate of change of dd𝑦𝑑 when 𝑑=5s, and at that time we are given that π‘₯=4cm and ddcmsπ‘₯𝑑=3/, so ddcms𝑦𝑑=2(4)(3)=24/.

Next, let us consider an example in which we have a circle with a radius, π‘Ÿ(𝑑), that is increasing with time 𝑑. 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 πœ†=π‘Ÿπ‘‘dd 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 𝐴(𝑑)=πœ‹(π‘Ÿ(𝑑)); hence, ddπ΄π‘Ÿ=2πœ‹π‘Ÿ.

The quantity we want to find is dd𝐴𝑑, but since the radius is a function of 𝑑 and the formula does not explicitly contain any of the 𝑑 variables, we have to differentiate this implicitly using the chain rule. In particular, dddddd𝐴𝑑=π΄π‘Ÿπ‘Ÿπ‘‘=2πœ‹π‘Ÿπœ†.

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, π‘Ÿ(𝑑), is increasing at the instant when π‘Ÿ=3cm. 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 𝑉(𝑑)=43πœ‹(π‘Ÿ(𝑑)); hence, ddπ‘‰π‘Ÿ=4πœ‹π‘Ÿ.

We can differentiate the volume 𝑉 implicitly with respect to time: dddddddd𝑉𝑑=π‘‰π‘Ÿπ‘Ÿπ‘‘=4πœ‹π‘Ÿπ‘Ÿπ‘‘.

But since we are given that the volume is increasing at a constant rate of 2 cm3/s, we have dd𝑉𝑑=2; hence, 2=4πœ‹π‘Ÿπ‘Ÿπ‘‘,dd which we can rearrange to find ddπ‘Ÿπ‘‘=12πœ‹π‘Ÿ.

Finally, we want to find the rate of change of the radius when π‘Ÿ=3cm, which we can substitute into the expression for the derivative as ddcmsπ‘Ÿπ‘‘=12πœ‹(3)=118πœ‹/.

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 𝑉 and surface area 𝐴 of a sphere of radius π‘Ÿ are given by 𝑉=43πœ‹π‘Ÿ,𝐴=4πœ‹π‘Ÿ.

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 dd𝑉𝑑 is negative and given by ddcms𝑉𝑑=βˆ’48/.

We can also differentiate the expression for the volume 𝑉 with respect to time implicitly using the explicit derivative of the volume with respect to π‘Ÿ: ddπ‘‰π‘Ÿ=4πœ‹π‘Ÿ; hence, dddddddd𝑉𝑑=π‘‰π‘Ÿπ‘Ÿπ‘‘=4πœ‹π‘Ÿπ‘Ÿπ‘‘.

Setting dd𝑉𝑑=βˆ’48, we have 4πœ‹π‘Ÿπ‘Ÿπ‘‘=βˆ’48,dd which we can solve for ddπ‘Ÿπ‘‘ to get ddπ‘Ÿπ‘‘=βˆ’12πœ‹π‘Ÿ.

The rate of change we want to find is that of the surface area dd𝐴𝑑, which we can compute implicitly using the explicit derivative for the surface area 𝐴 with respect to π‘Ÿ: ddπ΄π‘Ÿ=8πœ‹π‘Ÿ; hence, dddddddd𝐴𝑑=π΄π‘Ÿπ‘Ÿπ‘‘=8πœ‹π‘Ÿπ‘Ÿπ‘‘.

And substituting ddπ‘Ÿπ‘‘, we get dd𝐴𝑑=8πœ‹π‘Ÿο€Όβˆ’12πœ‹π‘Ÿοˆ=βˆ’96π‘Ÿ.

Finally, substituting the value π‘Ÿ=41cm, we find ddcms𝐴𝑑=βˆ’9641/.

Therefore, the rate of change of the surface area of the balloon when its radius π‘Ÿ=41cm is βˆ’9641 cm2/s.

In our next example, let’s 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 𝑀 cm and the length be 𝑙 cm. The area of the rectangle, 𝐴 cm2, is given by 𝐴=𝑀𝑙.

The rate of change of the area can be computed by finding the derivative of the product of two differentiable functions, 𝑀 and 𝑙. Recalling that the product rule states that (𝑀𝑙)β€²=𝑙𝑀′+𝑀𝑙′, we obtain dddddddd𝐴𝑑=(𝑀𝑙)𝑑=𝑙𝑀𝑑+𝑀𝑙𝑑.

Since the length and width are both increasing, the rate of change will be positive for both. Let ddcms𝑙𝑑=15/ and ddcms𝑀𝑑=13/ To determine the rate at which the area increases when 𝑙=25cm and 𝑀=12cm, we can substitute these values to find ddcms𝐴𝑑=(25)(13)+(12)(15)=505/.

Since the rate of change of the area, dd𝐴𝑑, is positive, we can say that the rate of change at which the area increases is 505 cm2/s.

Now, let’s 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 𝑦 is not purely given as a function of π‘₯.

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 𝑦-Coordinate

A particle is moving along the curve 6𝑦+2π‘₯βˆ’2π‘₯+5π‘¦βˆ’13=0. If the rate of change of its π‘₯-coordinate with respect to time as it passes through the point (βˆ’1,3) is 2, find the rate of change of its 𝑦-coordinate with respect to time at the same point.

Answer

In this example, we want to find the rate of change of the 𝑦-coordinate for a particle that moves along a curve, as shown in the diagram, at the instant of a particular point for (π‘₯,𝑦) and a given constant rate of change of the π‘₯-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 𝑦, dd𝑦𝑑, we begin by implicitly differentiating the equation of the curve with respect to 𝑑. Then, we substitute the rate of change of the π‘₯-coordinate, ddπ‘₯𝑑=2, at the point where π‘₯=βˆ’1 and 𝑦=3.

If we take the derivative of the equation of the curve with respect to 𝑑, we find 0=𝑑6𝑦+2π‘₯βˆ’2π‘₯+5π‘¦βˆ’13=12𝑦𝑦𝑑+4π‘₯π‘₯π‘‘βˆ’2π‘₯𝑑+5𝑦𝑑=(12𝑦+5)𝑦𝑑+(4π‘₯βˆ’2)π‘₯𝑑.dddddddddddddd

We can now substitute in the point π‘₯=βˆ’1, 𝑦=3 and the rate of change of the π‘₯-coordinate ddπ‘₯𝑑=2: 0=(12Γ—3+5)𝑦𝑑+(4Γ—βˆ’1βˆ’2)Γ—2=41π‘¦π‘‘βˆ’12.dddd

Then, rearranging for dd𝑦𝑑, we have dd𝑦𝑑=1241.

The rate of change of the 𝑦-coordinate is, therefore, 1241.

In the next example, let’s 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 man’s 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 man’s 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 π‘₯m be the distance away from the light pole on the ground and 𝑦m be the length of the shadow. In order to find the rate of change of the length of the man’s shadow, dd𝑦𝑑, we first find a relationship between this length, 𝑦, and the distance from the light pole, π‘₯. 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 𝑦 as a function of π‘₯: 5.5π‘₯+𝑦=1.8𝑦1.8(π‘₯+𝑦)=5.5𝑦𝑦=1837π‘₯.

We can take the derivative of this with respect to 𝑑 to find the relationship between their respective rates as dddd𝑦𝑑=1837π‘₯𝑑.

Now we can substitute the given rate, ddmsπ‘₯𝑑=1.5/, to find dd𝑦𝑑=1837Γ—1.5=2737.

Thus, the rate of change of the length of the man’s shadow to one decimal place is 0.7 m/s.

Now, let’s 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 𝑓(π‘₯)=√π‘₯+2. If its π‘₯-coordinate increases at a rate of 9√15 cm/s, find the rate of change of the distance between this point and the point (1,0) at π‘₯=3.

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 (π‘₯,𝑦)and (π‘Ž,𝑏) is given by 𝐷=(π‘₯βˆ’π‘Ž)+(π‘¦βˆ’π‘).

Therefore, if we take the function 𝑦=𝑓(π‘₯), the distance between the points ο€»π‘₯,√π‘₯+2ο‡οŠ¨ and (1,0) can be expressed as 𝐷=(π‘₯βˆ’1)+(π‘₯+2)=√2π‘₯βˆ’2π‘₯+3.

The derivative of this expression with respect to π‘₯ is dd𝐷π‘₯=4π‘₯βˆ’22√2π‘₯βˆ’2π‘₯+3.

The quantity we want to find is dd𝐷𝑑, but since π‘₯ is a function of 𝑑 and the formula does not explicitly contain any of the 𝑑 variables, we have to differentiate this implicitly using the chain rule. In particular, we have dddddddd𝐷𝑑=𝐷π‘₯π‘₯𝑑=ο€Ώ4π‘₯βˆ’22√2π‘₯βˆ’2π‘₯+3π‘₯𝑑.

We can now substitute the point π‘₯=3 and the rate of change of the π‘₯-coordinate, ddcmsπ‘₯𝑑=9√15/: dd𝐷𝑑=ο€Ώ4Γ—3βˆ’22√2Γ—3βˆ’2Γ—3+3×9√15=102√15Γ—9√15=45.

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 π‘Ž and 𝑏 and contained angle πœƒ has area 𝐴=12π‘Žπ‘πœƒsin. Suppose that π‘Ž=4, 𝑏=5, and the angle πœƒ is increasing at 0.6 rad/s. How fast is the area changing when πœƒ=πœ‹3?

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 πœƒ 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, π‘Ž=4 and 𝑏=5, can be expressed in terms of πœƒ as 𝐴=12Γ—4Γ—5πœƒ=10πœƒ.sinsin

The quantity we want to find is dd𝐴𝑑 at πœƒ=πœ‹3rad, given ddradsπœƒπ‘‘=0.6/. We can differentiate the expression for the area implicitly with respect to 𝑑 to find dddddd𝐴𝑑=π΄πœƒπœƒπ‘‘.

We can find the derivative of 𝐴 explicitly with respect to πœƒ as ddcosπ΄πœƒ=10πœƒ; therefore, ddcosdd𝐴𝑑=10πœƒπœƒπ‘‘.

We can now substitute the angle πœƒ=πœ‹3 and the rate of change of the angle ddπœƒπ‘‘=0.6 to find ddcos𝐴𝑑=10ο€»πœ‹3×0.6=3.

Therefore, the area is increasing at 3 cm2/s.

Suppose the rate of change of a particular quantity, 𝑋=𝑋(𝑑), is constant, which we denote by ̇𝑋, where ̇𝑋=𝑋𝑑=.ddconstant

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 𝑋 gains ̇𝑋>0 or loses ̇𝑋<0 by the amount ||̇𝑋|| after every unit of time. Equivalently, the net gain is ̇𝑋𝑑 at time 𝑑. If the initial value of the quantity is 𝑋=𝑋(0), then the quantity after time 𝑑 is given by 𝑋(𝑑)=𝑋+̇𝑋𝑑.

This clearly satisfies the rate of change, which can be verified upon differentiating this expression with respect to 𝑑.

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, 𝑀, of the rocket, at a particular time 𝑑, we will first determine the mass of the rocket at any given time 𝑑. 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 1β€Žβ€‰β€Ž000 kg, this is equivalent to 26β€Žβ€‰β€Ž000 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 ddkgs𝑀𝑑=βˆ’80/.

Since this is a constant rate of the quantity 𝑀, we can express that quantity after time 𝑑 with initial value 𝑀=𝑀(0): 𝑀(𝑑)=𝑀+𝑀𝑑𝑑.dd

On substituting the given rate and the initial value 𝑀=26000kg, we obtain 𝑀(𝑑)=26000βˆ’80𝑑.

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 𝑑. When 𝑑=25s, the mass of the rocket is 𝑀(25)=26000βˆ’80Γ—25=24000.kg

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 (<0) or increasing (>0) over time.

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