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Mathematics LibreTexts

3.9: Related Rates

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Learning Objectives
  • Express changing quantities in terms of derivatives.
  • Find relationships among the derivatives in a given problem.
  • Use the Chain Rule to find the rate of change of one quantity that depends on the rate of change of other quantities.
  • Apply the modified version of Polya's Problem-Solving Strategy to solve related rate problems.

We have seen that for quantities that change over time, the rates at which these quantities change are given by derivatives. If two related quantities are changing over time, the rates at which the quantities change are related. For example, suppose a balloon is being filled with air. In that case, both the balloon's radius and the balloon's volume are increasing. In this section, we consider several problems in which two or more related quantities are changing, and we study how to determine the relationship between the rates of change of these quantities.

In many real-world applications, related quantities are changing with respect to time. For example, let's consider the balloon example again. The rate of change in the volume, V, is related to the rate of change in the radius, r. In this case, we say that \frac{dV}{dt} and \frac{dr}{dt} are related rates because V is related to r. Here, we study several examples of related quantities that are changing with respect to time, and we look at how to calculate one rate of change given another rate of change.

Example \PageIndex{1}: Inflating a Balloon

A spherical balloon is being filled with air at the constant rate of 2\,\text{cm}^3\text{/sec} (Figure \PageIndex{1}). How fast is the radius increasing when the radius is 3 cm?

Three balloons are shown at Times 1, 2, and 3. These balloons increase in volume and radius as time increases.
Figure \PageIndex{1}: As the balloon is filled with air, both the radius and the volume are increasing with respect to time.

Solution

The volume of a sphere of radius r centimeters isV=\dfrac{4}{3} \pi r^3\,\text{cm}^3.\nonumberSince the balloon is being filled with air, both the volume and the radius are functions of time. Therefore, t seconds after beginning to fill the balloon with air, the volume of air in the balloon isV(t)=\dfrac{4}{3} \pi \big[r(t)\big]^3\text{cm}^3.\nonumberDifferentiating both sides of this equation with respect to time and applying the Chain Rule, we see that the rate of change in the volume is related to the rate of change in the radius by the equationV^{\prime}(t)=4 \pi \big[r(t)\big]^2 r^{\prime}(t).\nonumberThe balloon is being filled with air at the constant rate of 2 \,\text{cm}^3\text{/sec}, so V^{\prime}(t)=2\,\text{cm}^3\text{/sec}. Therefore,2\,\text{cm}^3\text{/sec}= \left( 4 \pi \big[r(t)\big]^2\;\text{cm}^2 \right) \cdot \left( r^{\prime}(t)\;\text{cm/s} \right) ,\nonumber which impliesr^{\prime}(t)=\dfrac{1}{2 \pi \big[r(t)\big]^2}\;\text{cm/sec}.\nonumberWhen the radius r=3 cm,r^{\prime}(t)=\dfrac{1}{18 \pi }\;\text{cm/sec}.\nonumber

Checkpoint \PageIndex{1}

What is the instantaneous rate of change of the radius when r=6 cm?

Answer

\frac{1}{72 \pi } cm/sec, or approximately 0.0044 cm/sec

While memorizing procedures is the absolute worst way to enjoy Mathematics, applications (also known as "word problems") tend to confound students so much that it is beneficial to develop a procedure for solving them. Let's outline a modified problem-solving strategy developed by Polya.

In 1945 George Polya published the book How To Solve It which quickly became his most prized publication. It sold over one million copies and has been translated into 17 languages. In this book, he identifies basic principles of problem-solving. I have modified these problem-solving strategies to fit Calculus best (we will see these again in a section titled Optimization Problems).

Polya's Modified Problem-Solving Process
  1. Read the given problem.
  2. Understand the given problem. This might require you to rephrase the problem in terms that you can understand or, more commonly, to draw a picture of the situation.
  3. Label unknowns. All problems presented in mathematics have unknown quantities. In this step, you take the time to label these unknowns using variables. This step is often combined with the previous step (especially when drawing pictures).
  4. List the Givens and Wants. This step is dedicated to making a table with one column dedicated to the given information and another to the wanted or desired information. Normally, we only list the given and desired rates in this table.
  5. Create a Master Equation involving the variables within the given and wanted rate information. This is a critical step in all problem-solving processes involving Polya's method.
    Warning: The Master Equation should be a relation between two variables (those listed in the Givens and Wants table). If your Master Equation has more than two variables, you need to find a relation between the excess variables to reduce the number of final variables in your Master Equation to two.
  6. Find the Rate Equation from the Master Equation. This almost always requires implicit differentiation.
  7. Substitute in any constants given within the problem.
  8. Solve the resulting equation for the desired "wanted" piece of information.

Note that when solving a related-rates problem, it is crucial not to substitute known values too soon. For example, suppose the value for a changing quantity is substituted into an equation before both sides are differentiated. In that case, the quantity will behave as a constant, and its derivative will not appear in the new equation found during the rate equation step. We examine this potential error in the following example.

Examples of Polya's Problem-Solving Process

Let's implement the described strategy to solve several related-rates problems. The first example involves a plane flying overhead. The relationship we are studying is between the speed of the aircraft and the rate at which the distance between the plane and a person on the ground is changing.

Example \PageIndex{2}: An Airplane Flying at a Constant Elevation

An airplane is flying overhead at a constant elevation of 4000 ft. A man is viewing the plane from a position 3000 ft from the base of a radio tower. The airplane is flying horizontally away from the man. If the plane is flying at the rate of 600 ft/sec, at what rate does the distance between the man and the plane increase when the plane passes over the radio tower?

Solution

Read. This step is only listed so you can have a "win." All of us can read a problem, but understanding requires a bit more effort.

Understand and Label. Draw a picture, introducing variables to represent the quantities involved.

A right triangle is made with a person on the ground, an airplane in the air, and a radio tower at the right angle on the ground. The hypotenuse is s, the distance on the ground between the person and the radio tower is x, and the side opposite the person (that is, the height from the ground to the airplane) is 4000 ft.
Figure \PageIndex{2}: An airplane is flying at a constant height of 4000 ft. The distances between the person and the airplane and the person and the place on the ground directly below the airplane are changing. We denote those quantities with the variables s and x, respectively.

As shown, x denotes the distance between the man and the position on the ground directly below the airplane. The variable s denotes the distance between the man and the plane. Note that both x and s are functions of time. We do not introduce a variable for the plane's height because it remains at a constant elevation of 4000 ft. Since an object's height above the ground is measured as the shortest distance between the object and the ground, the line segment of length 4000 ft is perpendicular to the line segment of length x feet, creating a right triangle.

Givens and Wants.

Given Want
\dfrac{dx}{dt} = 600 feet per second \dfrac{ds}{dt}\left|_{x = 3000}\right.

Since x denotes the horizontal distance between the man and the point on the ground below the plane, dx/dt represents the speed of the plane. We are told the plane's speed is 600 ft/sec. Therefore, \frac{dx}{dt}=600 ft/sec. Since we are asked to find the rate of change in the distance between the man and the plane when the plane is directly above the radio tower, we need to find ds/dt when x=3000 ft.

As was mentioned in Polya's modified problem-solving process, we only list the given and desired rate information. That is, don't bother listing the facts that the height of the plane is always 4000 feet, and the distance from the tower to our position is always 3000 feet.

Master Equation. The variables within our table above are x and s. From the figure, we can use the Pythagorean Theorem to write an equation relating these two variables:[x(t)]^2+4000^2=[s(t)]^2.

Rate Equation. Differentiating this equation with respect to time and using the fact that the derivative of a constant is zero, we arrive at the equationx\dfrac{dx}{dt}=s\dfrac{ds}{dt}.\nonumber

Substitute. When we "stop time" to do the problem, we know that \frac{dx}{dt}=600 ft/sec and x=3000 ft. Thus, (3000)(600) = s \dfrac{ds}{dt}. \nonumber

Solve. You can see that the only thing we are missing is s. However, we can use the Pythagorean Theorem to determine the distance s when x=3000 ft and the height is 4000 ft. Solving the equation3000^2+4000^2=s^2 \nonumber for s, we have s=5000 ft at the time of interest. Using these values, we conclude that ds/dt is a solution of the equation(3000)(600)=(5000) \cdot \dfrac{ds}{dt}. \nonumber Therefore,\dfrac{ds}{dt}=\dfrac{3000 \cdot 600}{5000}=360\,\text{ft/sec}.\nonumber Note: When solving related-rates problems, it is important not to substitute values for the variables too soon. For example, in the Master Equation step, we related the variable quantities x(t) and s(t) by the equation[x(t)]^2+4000^2=[s(t)]^2.\nonumber Since the plane remains at a constant height, it is not necessary to introduce a variable for the height, and we are allowed to use the constant 4000 to denote that quantity. However, the other two quantities are changing. If we mistakenly substituted x(t)=3000 into the equation before differentiating, our equation would have been3000^2+4000^2=[s(t)]^2.\nonumber After differentiating, our equation would become0=s(t)\dfrac{ds}{dt}.\nonumber As a result, we would incorrectly conclude that \frac{ds}{dt}=0.

Checkpoint \PageIndex{2}

What is the speed of the plane if the distance between the person and the plane is increasing at the rate of 300 ft/sec?

Answer

500 ft/sec

Example \PageIndex{3}: A Rocket Launch

A rocket is launched so that it rises vertically. A camera is positioned 5000 ft from the launch pad. When the rocket is 1000 ft above the launch pad, its velocity is 600 ft/sec.

A photo of a rocket lifting off.
Figure \PageIndex{3}: (credit: modification of work by Steve Jurvetson, Wikimedia Commons)

Find the necessary rate of change of the camera's angle as a function of time so that it stays focused on the rocket.

Solution

Read. Again, this is a freebie step. Make sure you read the problem thoroughly!

Understand and Label. Draw a picture introducing the variables.

A right triangle is formed with a camera at one of the nonright angles and a rocket at the other nonright angle. The angle with the camera measures θ. The distance from the rocket to the ground is h; note that this is the side opposite the angle with measure θ. The side adjacent to the angle with measure θ is 5000 ft.
Figure \PageIndex{4}: A camera is positioned 5000 ft from the launch pad of the rocket. The rocket's height and the camera's angle are changing with respect to time. We denote those quantities with the variables h and \theta ,respectively.

Let h denote the height of the rocket above the launch pad and \theta be the angle between the camera lens and the ground.

Givens and Wants. We are trying to find the rate of change in the camera angle with respect to time when the rocket is 1000 ft off the ground. That is, we need to find \frac{d \theta }{dt} when h=1000 ft. At that time, we know the velocity of the rocket is \frac{dh}{dt}=600 ft/sec.

Given Want
\dfrac{dh}{dt}\left|_{h = 1000}\right. = 600 feet per second \dfrac{d \theta}{dt}\left|_{h = 1000} \right.

Master Equation. Using the fact that we have drawn a right triangle, it is natural to think about trigonometric functions. Recall that \tan{(\theta)} is the ratio of the length of the opposite side of the triangle to the length of the adjacent side. Thus, we have\tan{(\theta)} = \dfrac{h}{5000}.\nonumber This gives us the equationh=5000\tan{(\theta)}.\nonumber

Rate Equation. Differentiating this equation with respect to time t, we obtain\dfrac{dh}{dt} = 5000\sec^2{(\theta)} \dfrac{d\theta}{dt}.\nonumber

Substitute. Letting h = 1000 and \frac{dh}{dt} = 600, we get600=5000\sec^2{(\theta)} \dfrac{d\theta}{dt}. \nonumber

Solve. We want to find \frac{d\theta}{dt} when h=1000 ft. We need to determine \sec^2{(\theta)}. Recall that \sec{(\theta)} is the ratio of the length of the hypotenuse to the length of the adjacent side. We know the length of the adjacent side is 5000 ft. To determine the length of the hypotenuse, we use the Pythagorean Theorem, where the length of one leg is 5000 ft, the length of the other leg is h=1000 ft, and the length of the hypotenuse is c feet as shown in the following figure.

A right triangle has one angle with measure θ. The hypotenuse is c, the side length opposite the angle with measure θ is 1000, and the side adjacent to the angle with measure θ is 5000.

We see that1000^2+5000^2=c^2\nonumber and we conclude that the hypotenuse isc=1000\sqrt{26}\,\text{ft}.\nonumber Therefore, when h=1000, we have\sec^2{(\theta)} = \left(\dfrac{1000\sqrt{26}}{5000}\right)^2=\dfrac{26}{25}.\nonumber Doing a final substitution into our formula from the Substitute step, we arrive at600=5000\left(\dfrac{26}{25}\right)\dfrac{d\theta}{dt}.\nonumber Therefore, \frac{d \theta }{dt}=\frac{3}{26} rad/sec.

Checkpoint \PageIndex{3}

What rate of change is necessary for the elevation angle of the camera if the camera is placed on the ground at a distance of 4000 ft from the launch pad and the velocity of the rocket is 500 ft/sec when the rocket is 2000 ft off the ground?

Answer

\frac{1}{10} rad/sec

In the following example, we consider water draining from a cone-shaped funnel. We compare the rate at which the level of water in the cone is decreasing with the rate at which the volume of water is decreasing.

Example \PageIndex{4}: Water Draining from a Funnel

Water is draining from the bottom of a cone-shaped funnel at the rate of 0.03\,\text{ft}^3\text{/sec}. The height of the funnel is 2 ft, and the radius at the top of the funnel is 1 ft. At what rate is the height of the water in the funnel changing when the height of the water is \frac{1}{2} ft?

Solution

This time, we will shorthand our steps to mirror the pacing you should get used to.

Read. Done.

Understand and Label. See the picture below.

A funnel is shown with height 2 and radius 1 at its top. The funnel has water to height h, at which point the radius is r.
Figure \PageIndex{5}: Water is draining from a funnel of height 2 ft and radius 1 ft. The height of the water and the radius of the top layer of water change over time. We denote these quantities with the variables h and r.

Let h denote the height of the water in the funnel, r denote the radius of the water at its surface, and V denote the volume of the water.

Givens and Wants.

Given Want
\dfrac{dV}{dt} = -0.03 \text{ ft}^3\text{/sec} \dfrac{dh}{dt}\left|_{h = 1/2} \right.

Master Equation.V=\dfrac{1}{3} \pi r^2h.\nonumber

Caution: Substitute Out a Variable

Since the Master Equation involves more than two variables, we want to find a relation to rewrite r in terms of either h or V.

From the figure, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the same. Therefore, \frac{r}{h}=\frac{1}{2} or r=\frac{h}{2}. Using this fact, the equation for volume can be simplified toV=\dfrac{1}{3} \pi \left(\dfrac{h}{2}\right)^2h=\dfrac{ \pi }{12}h^3.\nonumber

Rate Equation.\dfrac{dV}{dt}=\dfrac{\pi}{4}h^2\dfrac{dh}{dt}.\nonumber

Substitute.−0.03 = \dfrac{\pi}{4}\cdot \left(\dfrac{1}{2}\right)^2 \cdot \dfrac{dh}{dt},\nonumber

Solve. The previous step implies−0.03=\dfrac{\pi}{16}\dfrac{dh}{dt}.\nonumber It follows that\dfrac{dh}{dt}=−\dfrac{0.48}{\pi} \approx −0.153\,\text{ft/sec}.\nonumber

Checkpoint \PageIndex{4}

At what rate is the height of the water changing when the height of the water is \frac{1}{4} ft?

Answer

−0.61 ft/sec


Key Concepts

  • To solve a related rates problem, first draw a picture that illustrates the relationship between the two or more related quantities that are changing with respect to time.
  • Regarding the quantities, state the information given and the rate to be found.
  • Find an equation relating the quantities.
  • Use differentiation, applying the Chain Rule as necessary, to find an equation that relates the rates.
  • Be sure not to substitute a variable quantity for one of the variables until after finding an equation relating the rates.

Glossary

related rates
are rates of change associated with two or more related quantities that are changing over time


This page titled 3.9: Related Rates is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Roy Simpson.

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