problem solving rates

A rate is a type of ratio where something of one unit is compared with something else of another unit. Rates are applied in many real-world scenarios like unit conversions, speed/velocity, and per-item price.

• Unit Conversions (Prealgebra)
• Speed and Other Rates (Prealgebra)

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Chapter 9: Radicals

9.10 Rate Word Problems: Work and Time

If it takes Felicia 4 hours to paint a room and her daughter Katy 12 hours to paint the same room, then working together, they could paint the room in 3 hours. The equation used to solve problems of this type is one of reciprocals. It is derived as follows:

[latex]\text{rate}\times \text{time}=\text{work done}[/latex]

For this problem:

[latex]\begin{array}{rrrl} \text{Felicia's rate: }&F_{\text{rate}}\times 4 \text{ h}&=&1\text{ room} \\ \\ \text{Katy's rate: }&K_{\text{rate}}\times 12 \text{ h}&=&1\text{ room} \\ \\ \text{Isolating for their rates: }&F&=&\dfrac{1}{4}\text{ h and }K = \dfrac{1}{12}\text{ h} \end{array}[/latex]

To make this into a solvable equation, find the total time [latex](T)[/latex] needed for Felicia and Katy to paint the room. This time is the sum of the rates of Felicia and Katy, or:

[latex]\begin{array}{rcrl} \text{Total time: } &T \left(\dfrac{1}{4}\text{ h}+\dfrac{1}{12}\text{ h}\right)&=&1\text{ room} \\ \\ \text{This can also be written as: }&\dfrac{1}{4}\text{ h}+\dfrac{1}{12}\text{ h}&=&\dfrac{1 \text{ room}}{T} \\ \\ \text{Solving this yields:}&0.25+0.083&=&\dfrac{1 \text{ room}}{T} \\ \\ &0.333&=&\dfrac{1 \text{ room}}{T} \\ \\ &t&=&\dfrac{1}{0.333}\text{ or }\dfrac{3\text{ h}}{\text{room}} \end{array}[/latex]

Example 9.10.1

Karl can clean a room in 3 hours. If his little sister Kyra helps, they can clean it in 2.4 hours. How long would it take Kyra to do the job alone?

The equation to solve is:

[latex]\begin{array}{rrrrl} \dfrac{1}{3}\text{ h}&+&\dfrac{1}{K}&=&\dfrac{1}{2.4}\text{ h} \\ \\ &&\dfrac{1}{K}&=&\dfrac{1}{2.4}\text{ h}-\dfrac{1}{3}\text{ h}\\ \\ &&\dfrac{1}{K}&=&0.0833\text{ or }K=12\text{ h} \end{array}[/latex]

Example 9.10.2

Doug takes twice as long as Becky to complete a project. Together they can complete the project in 10 hours. How long will it take each of them to complete the project alone?

[latex]\begin{array}{rrl} \dfrac{1}{R}+\dfrac{1}{2R}&=&\dfrac{1}{10}\text{ h,} \\ \text{where Doug's rate (} \dfrac{1}{D}\text{)}& =& \dfrac{1}{2}\times \text{ Becky's (}\dfrac{1}{R}\text{) rate.} \\ \\ \text{Sum the rates: }\dfrac{1}{R}+\dfrac{1}{2R}&=&\dfrac{2}{2R} + \dfrac{1}{2R} = \dfrac{3}{2R} \\ \\ \text{Solve for R: }\dfrac{3}{2R}&=&\dfrac{1}{10}\text{ h} \\ \text{which means }\dfrac{1}{R}&=&\dfrac{1}{10}\times\dfrac{2}{3}\text{ h} \\ \text{so }\dfrac{1}{R}& =& \dfrac{2}{30} \\ \text{ or }R &= &\dfrac{30}{2} \end{array}[/latex]

This means that the time it takes Becky to complete the project alone is [latex]15\text{ h}[/latex].

Since it takes Doug twice as long as Becky, the time for Doug is [latex]30\text{ h}[/latex].

Example 9.10.3

Joey can build a large shed in 10 days less than Cosmo can. If they built it together, it would take them 12 days. How long would it take each of them working alone?

[latex]\begin{array}{rl} \text{The equation to solve:}& \dfrac{1}{(C-10)}+\dfrac{1}{C}=\dfrac{1}{12}, \text{ where }J=C-10 \\ \\ \text{Multiply each term by the LCD:}&(C-10)(C)(12) \\ \\ \text{This leaves}&12C+12(C-10)=C(C-10) \\ \\ \text{Multiplying this out:}&12C+12C-120=C^2-10C \\ \\ \text{Which simplifies to}&C^2-34C+120=0 \\ \\ \text{Which will factor to}& (C-30)(C-4) = 0 \end{array}[/latex]

Cosmo can build the large shed in either 30 days or 4 days. Joey, therefore, can build the shed in 20 days or −6 days (rejected).

The solution is Cosmo takes 30 days to build and Joey takes 20 days.

Example 9.10.4

Clark can complete a job in one hour less than his apprentice. Together, they do the job in 1 hour and 12 minutes. How long would it take each of them working alone?

[latex]\begin{array}{rl} \text{Convert everything to hours:} & 1\text{ h }12\text{ min}=\dfrac{72}{60} \text{ h}=\dfrac{6}{5}\text{ h}\\ \\ \text{The equation to solve is} & \dfrac{1}{A}+\dfrac{1}{A-1}=\dfrac{1}{\dfrac{6}{5}}=\dfrac{5}{6}\\ \\ \text{Therefore the equation is} & \dfrac{1}{A}+\dfrac{1}{A-1}=\dfrac{5}{6} \\ \\ \begin{array}{r} \text{To remove the fractions, } \\ \text{multiply each term by the LCD} \end{array} & (A)(A-1)(6)\\ \\ \text{This leaves} & 6(A)+6(A-1)=5(A)(A-1) \\ \\ \text{Multiplying this out gives} & 6A-6+6A=5A^2-5A \\ \\ \text{Which simplifies to} & 5A^2-17A +6=0 \\ \\ \text{This will factor to} & (5A-2)(A-3)=0 \end{array}[/latex]

The apprentice can do the job in either [latex]\dfrac{2}{5}[/latex] h (reject) or 3 h. Clark takes 2 h.

Example 9.10.5

A sink can be filled by a pipe in 5 minutes, but it takes 7 minutes to drain a full sink. If both the pipe and the drain are open, how long will it take to fill the sink?

The 7 minutes to drain will be subtracted.

[latex]\begin{array}{rl} \text{The equation to solve is} & \dfrac{1}{5}-\dfrac{1}{7}=\dfrac{1}{X} \\ \\ \begin{array}{r} \text{To remove the fractions,} \\ \text{multiply each term by the LCD}\end{array} & (5)(7)(X)\\ \\ \text{This leaves } & (7)(X)-(5)(X)=(5)(7)\\ \\ \text{Multiplying this out gives} & 7X-5X=35\\ \\ \text{Which simplifies to} & 2X=35\text{ or }X=\dfrac{35}{2}\text{ or }17.5 \end{array}[/latex]

17.5 min or 17 min 30 sec is the solution

For Questions 1 to 8, write the formula defining the relation. Do Not Solve!!

• Bill’s father can paint a room in 2 hours less than it would take Bill to paint it. Working together, they can complete the job in 2 hours and 24 minutes. How much time would each require working alone?
• Of two inlet pipes, the smaller pipe takes four hours longer than the larger pipe to fill a pool. When both pipes are open, the pool is filled in three hours and forty-five minutes. If only the larger pipe is open, how many hours are required to fill the pool?
• Jack can wash and wax the family car in one hour less than it would take Bob. The two working together can complete the job in 1.2 hours. How much time would each require if they worked alone?
• If Yousef can do a piece of work alone in 6 days, and Bridgit can do it alone in 4 days, how long will it take the two to complete the job working together?
• Working alone, it takes John 8 hours longer than Carlos to do a job. Working together, they can do the job in 3 hours. How long would it take each to do the job working alone?
• Working alone, Maryam can do a piece of work in 3 days that Noor can do in 4 days and Elana can do in 5 days. How long will it take them to do it working together?
• Raj can do a piece of work in 4 days and Rubi can do it in half the time. How long would it take them to do the work together?
• A cistern can be filled by one pipe in 20 minutes and by another in 30 minutes. How long would it take both pipes together to fill the tank?

For Questions 9 to 20, find and solve the equation describing the relationship.

• If an apprentice can do a piece of work in 24 days, and apprentice and instructor together can do it in 6 days, how long would it take the instructor to do the work alone?
• A carpenter and his assistant can do a piece of work in 3.75 days. If the carpenter himself could do the work alone in 5 days, how long would the assistant take to do the work alone?
• If Sam can do a certain job in 3 days, while it would take Fred 6 days to do the same job, how long would it take them, working together, to complete the job?
• Tim can finish a certain job in 10 hours. It takes his wife JoAnn only 8 hours to do the same job. If they work together, how long will it take them to complete the job?
• Two people working together can complete a job in 6 hours. If one of them works twice as fast as the other, how long would it take the slower person, working alone, to do the job?
• If two people working together can do a job in 3 hours, how long would it take the faster person to do the same job if one of them is 3 times as fast as the other?
• A water tank can be filled by an inlet pipe in 8 hours. It takes twice that long for the outlet pipe to empty the tank. How long would it take to fill the tank if both pipes were open?
• A sink can be filled from the faucet in 5 minutes. It takes only 3 minutes to empty the sink when the drain is open. If the sink is full and both the faucet and the drain are open, how long will it take to empty the sink?
• It takes 10 hours to fill a pool with the inlet pipe. It can be emptied in 15 hours with the outlet pipe. If the pool is half full to begin with, how long will it take to fill it from there if both pipes are open?
• A sink is ¼ full when both the faucet and the drain are opened. The faucet alone can fill the sink in 6 minutes, while it takes 8 minutes to empty it with the drain. How long will it take to fill the remaining ¾ of the sink?
• A sink has two faucets: one for hot water and one for cold water. The sink can be filled by a cold-water faucet in 3.5 minutes. If both faucets are open, the sink is filled in 2.1 minutes. How long does it take to fill the sink with just the hot-water faucet open?
• A water tank is being filled by two inlet pipes. Pipe A can fill the tank in 4.5 hours, while both pipes together can fill the tank in 2 hours. How long does it take to fill the tank using only pipe B?

Answer Key 9.10

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Related Rates

Calculus related rates problem solving strategy.

We will use the steps outlined below to solve each Related Rates problem on this site, step-by-step, every single time. We hope that this will help you see the strategy we’re using so you can learn it too, and then be able to apply it to all of your problems, especially those on your exams. That’s as opposed to learning just how to solve a particular problem on your homework, say, since you may well never see that specific problem again.

• Draw a picture of the physical situation.
• Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.
• a simple geometric fact (like the relation between a sphere’s volume and its radius, or the relation between the volume of a cylinder and its height); or
• a trigonometric function (like $\tan{\theta}$ = opposite/adjacent); or
• similar triangles; or
• the Pythagorean theorem.
• Take the derivative with respect to time of both sides of your equation. Remember the chain rule.
• Solve for the quantity you’re after.

On our Related Rates page , we solve an example of each of the approaches listed in 2.B above: i. Geometric fact. Typical problems: A circle’s radius changes, a snowball melts, a rectangle’s height and/or width changes. ii. Trig function. Typical problems: A searchlight rotates, a rocket takes off, a kite travels horizontally. iii. Similar triangles. Water fills a cone or trough, sand falls onto a conical pile, person walks away from a light pole that casts a shadow. iv. Pythagorean theorem. Typical problems: Cars/ships/joggers move along 90 degree paths, baseball players run along the diamond, boat is pulled toward a dock. To see an example of each type, please visit our Related Rates page .

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Study Guides > Calculus Volume 1

Related rates, 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.

We have seen that for quantities that are changing 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, if a balloon is being filled with air, both the radius of the balloon and the volume of the balloon 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.

Setting up Related-Rates Problems

In many real-world applications, related quantities are changing with respect to time. For example, if we consider the balloon example again, we can say that the rate of change in the volume, [latex]V[/latex], is related to the rate of change in the radius, [latex]r[/latex]. In this case, we say that [latex]\frac{dV}{dt}[/latex] and [latex]\frac{dr}{dt}[/latex] are related rates because [latex]V[/latex] is related to [latex]r[/latex]. 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.

Inflating a Balloon

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

The volume of a sphere of radius [latex]r[/latex] centimeters is

Since the balloon is being filled with air, both the volume and the radius are functions of time. Therefore, [latex]t[/latex] seconds after beginning to fill the balloon with air, the volume of air in the balloon is

Differentiating 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 equation

The balloon is being filled with air at the constant rate of 2 cm 3 /sec, so [latex]V^{\prime}(t)=2 \, \text{cm}^3 / \sec[/latex]. Therefore,

which implies

When the radius [latex]r=3 \, \text{cm}[/latex],

[latex]\frac{1}{72\pi} \, \text{cm/sec}[/latex], or approximately 0.0044 cm/sec

[latex]\frac{dr}{dt}=\frac{1}{2\pi r^2}[/latex]

Before looking at other examples, let’s outline the problem-solving strategy we will be using to solve related-rates problems.

Problem-Solving Strategy: Solving a Related-Rates Problem

• Assign symbols to all variables involved in the problem. Draw a figure if applicable.
• State, in terms of the variables, the information that is given and the rate to be determined.
• Find an equation relating the variables introduced in step 1.
• Using the chain rule, differentiate both sides of the equation found in step 3 with respect to the independent variable. This new equation will relate the derivatives.
• Substitute all known values into the equation from step 4, then solve for the unknown rate of change.

Note that when solving a related-rates problem, it is crucial not to substitute known values too soon. For example, if the value for a changing quantity is substituted into an equation before both sides of the equation are differentiated, then that quantity will behave as a constant and its derivative will not appear in the new equation found in step 4. We examine this potential error in the following example.

Examples of the Process

Let’s now implement the strategy just described 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 plane and the rate at which the distance between the plane and a person on the ground is changing.

An Airplane Flying at a Constant Elevation

An airplane is flying overhead at a constant elevation of [latex]4000[/latex] ft. A man is viewing the plane from a position [latex]3000[/latex] 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 [latex]600[/latex] ft/sec, at what rate is the distance between the man and the plane increasing when the plane passes over the radio tower?

Step 1. Draw a picture, introducing variables to represent the different quantities involved.

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

Step 2. Since [latex]x[/latex] denotes the horizontal distance between the man and the point on the ground below the plane, [latex]dx/dt[/latex] represents the speed of the plane. We are told the speed of the plane is 600 ft/sec. Therefore, [latex]\frac{dx}{dt}=600[/latex] 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 [latex]ds/dt[/latex] when [latex]x=3000[/latex] ft.

Step 3. From the figure, we can use the Pythagorean theorem to write an equation relating [latex]x[/latex] and [latex]s[/latex]:

Step 4. Differentiating this equation with respect to time and using the fact that the derivative of a constant is zero, we arrive at the equation

Step 5. Find the rate at which the distance between the man and the plane is increasing when the plane is directly over the radio tower. That is, find [latex]\frac{ds}{dt}[/latex] when [latex]x=3000[/latex] ft. Since the speed of the plane is [latex]600[/latex] ft/sec, we know that [latex]\frac{dx}{dt}=600[/latex] ft/sec. We are not given an explicit value for [latex]s[/latex]; however, since we are trying to find [latex]\frac{ds}{dt}[/latex] when [latex]x=3000[/latex] ft, we can use the Pythagorean theorem to determine the distance [latex]s[/latex] when [latex]x=3000[/latex] and the height is [latex]4000[/latex] ft. Solving the equation

for [latex]s[/latex], we have [latex]s=5000[/latex] ft at the time of interest. Using these values, we conclude that [latex]ds/dt[/latex] is a solution of the equation

Note : When solving related-rates problems, it is important not to substitute values for the variables too soon. For example, in step 3, we related the variable quantities [latex]x(t)[/latex] and [latex]s(t)[/latex] by the equation

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 [latex]x(t)=3000[/latex] into the equation before differentiating, our equation would have been

After differentiating, our equation would become

As a result, we would incorrectly conclude that [latex]\frac{ds}{dt}=0[/latex].

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

[latex]500[/latex] ft/sec

[latex]\frac{ds}{dt}=300[/latex] ft/sec

We now return to the problem involving the rocket launch from the beginning of the chapter.

Chapter Opener: A Rocket Launch

Step 1. Draw a picture introducing the variables.

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

Step 2. We are trying to find the rate of change in the angle of the camera with respect to time when the rocket is [latex]1000[/latex] ft off the ground. That is, we need to find [latex]\frac{d\theta}{dt}[/latex] when [latex]h=1000[/latex] ft. At that time, we know the velocity of the rocket is [latex]\frac{dh}{dt}=600[/latex] ft/sec.

Step 3. Now we need to find an equation relating the two quantities that are changing with respect to time: [latex]h[/latex] and [latex]\theta[/latex]. How can we create such an equation? Using the fact that we have drawn a right triangle, it is natural to think about trigonometric functions. Recall that [latex]\tan \theta[/latex] is the ratio of the length of the opposite side of the triangle to the length of the adjacent side. Thus, we have

This gives us the equation

Step 4. Differentiating this equation with respect to time [latex]t[/latex], we obtain

Step 5. We want to find [latex]\frac{d\theta}{dt}[/latex] when [latex]h=1000[/latex] ft. At this time, we know that [latex]\frac{dh}{dt}=600[/latex] ft/sec. We need to determine [latex]\sec^2 \theta[/latex]. Recall that [latex]\sec \theta[/latex] 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 [latex]5000[/latex] ft. To determine the length of the hypotenuse, we use the Pythagorean theorem, where the length of one leg is [latex]5000[/latex] ft, the length of the other leg is [latex]h=1000[/latex] ft, and the length of the hypotenuse is [latex]c[/latex] feet as shown in the following figure.

We see that

and we conclude that the hypotenuse is

Therefore, when [latex]h=1000[/latex], we have

Recall from step 4 that the equation relating [latex]\frac{d\theta}{dt}[/latex] to our known values is

When [latex]h=1000[/latex] ft, we know that [latex]\frac{dh}{dt}=600[/latex] ft/sec and [latex]\sec^2 \theta =\frac{26}{25}[/latex]. Substituting these values into the previous equation, we arrive at the equation

Therefore, [latex]\frac{d\theta}{dt}=\frac{3}{26}[/latex] rad/sec.

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 [latex]4000[/latex] ft from the launch pad and the velocity of the rocket is [latex]500[/latex] ft/sec when the rocket is [latex]2000[/latex] ft off the ground?

[latex]\frac{1}{10}[/latex] rad/sec

Find [latex]\frac{d\theta}{dt}[/latex] when [latex]h=2000[/latex] ft. At that time, [latex]\frac{dh}{dt}=500[/latex] ft/sec.

In the next 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.

Water Draining from a Funnel

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

Step 1: Draw a picture introducing the variables.

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

Step 2: We need to determine [latex]\frac{dh}{dt}[/latex] when [latex]h=\frac{1}{2}[/latex] ft. We know that [latex]\frac{dV}{dt}=-0.03 \text{ft}^3 / \text{sec}[/latex].

Step 3: The volume of water in the cone is

From the figure, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the same. Therefore, [latex]\frac{r}{h}=\frac{1}{2}[/latex] or [latex]r=\frac{h}{2}[/latex]. Using this fact, the equation for volume can be simplified to

Step 4: Applying the chain rule while differentiating both sides of this equation with respect to time [latex]t[/latex], we obtain

Step 5: We want to find [latex]\frac{dh}{dt}[/latex] when [latex]h=\frac{1}{2}[/latex] ft. Since water is leaving at the rate of [latex]0.03 \, \text{ft}^3 / \text{sec}[/latex], we know that [latex]\frac{dV}{dt}=-0.03 \, \text{ft}^3 / \text{sec}[/latex]. Therefore,

It follows that

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

[latex]-0.61[/latex] ft/sec

We need to find [latex]\frac{dh}{dt}[/latex] when [latex]h=\frac{1}{4}[/latex].

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.
• In terms of 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.

For the following exercises, find the quantities for the given equation.

1.  Find [latex]\frac{dy}{dt}[/latex] at [latex]x=1[/latex] and [latex]y=x^2+3[/latex] if [latex]\frac{dx}{dt}=4[/latex].

2.  Find [latex]\frac{dx}{dt}[/latex] at [latex]x=-2[/latex] and [latex]y=2x^2+1[/latex] if [latex]\frac{dy}{dt}=-1[/latex].

3.  Find [latex]\frac{dz}{dt}[/latex] at [latex](x,y)=(1,3)[/latex] and [latex]z^2=x^2+y^2[/latex] if [latex]\frac{dx}{dt}=4[/latex] and [latex]\frac{dy}{dt}=3[/latex].

[latex]\frac{13}{\sqrt{10}}[/latex]

For the following exercises, sketch the situation if necessary and used related rates to solve for the quantities.

4. [T] If two electrical resistors are connected in parallel, the total resistance (measured in ohms, denoted by the Greek capital letter omega, [latex]\Omega[/latex]) is given by the equation [latex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/latex]. If [latex]R_1[/latex] is increasing at a rate of [latex]0.5 \Omega / \text{min}[/latex] and [latex]R_2[/latex] decreases at a rate of [latex]1.1 \Omega / \text{min}[/latex], at what rate does the total resistance change when [latex]R_1=20 \Omega[/latex] and [latex]R_2=50 \Omega[/latex]?

5.  A 10 ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the bottom moving along the ground when the bottom of the ladder is 5 ft from the wall?

[latex]2\sqrt{3}[/latex] ft/sec

6.  A 25 ft ladder is leaning against a wall. If we push the ladder toward the wall at a rate of 1 ft/sec, and the bottom of the ladder is initially [latex]20[/latex] ft away from the wall, how fast does the ladder move up the wall [latex]5[/latex] sec after we start pushing?

7.  Two airplanes are flying in the air at the same height: airplane [latex]A[/latex] is flying east at 250 mi/h and airplane [latex]B[/latex] is flying north at [latex]300[/latex] mi/h.  If they are both heading to the same airport, located 30 miles east of airplane [latex]A[/latex] and 40 miles north of airplane [latex]B[/latex], at what rate is the distance between the airplanes changing?

The distance is decreasing at [latex]390[/latex] mi/h.

8.  You and a friend are riding your bikes to a restaurant that you think is east; your friend thinks the restaurant is north. You both leave from the same point, with you riding at 16 mph east and your friend riding [latex]12[/latex] mph north. After you traveled [latex]4[/latex] mi, at what rate is the distance between you changing?

9.  Two buses are driving along parallel freeways that are [latex]5[/latex] mi apart, one heading east and the other heading west. Assuming that each bus drives a constant [latex]55[/latex] mph, find the rate at which the distance between the buses is changing when they are [latex]13[/latex] mi apart, heading toward each other.

The distance between them shrinks at a rate of [latex]\frac{1320}{13}\approx 101.5[/latex] mph.

10.  A 6-ft-tall person walks away from a 10 ft lamppost at a constant rate of [latex]3[/latex] ft/sec. What is the rate that the tip of the shadow moves away from the pole when the person is [latex]10[/latex] ft away from the pole?

11.  Using the previous problem, what is the rate at which the tip of the shadow moves away from the person when the person is 10 ft from the pole?

[latex]\frac{9}{2}[/latex] ft/sec

12.  A 5-ft-tall person walks toward a wall at a rate of 2 ft/sec. A spotlight is located on the ground 40 ft from the wall. How fast does the height of the person’s shadow on the wall change when the person is 10 ft from the wall?

13.  Using the previous problem, what is the rate at which the shadow changes when the person is 10 ft from the wall, if the person is walking away from the wall at a rate of 2 ft/sec?

It grows at a rate [latex]\frac{4}{9}[/latex] ft/sec

14.  A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/sec. You are running on the ground starting directly under the helicopter at a rate of 10 ft/sec. Find the rate of change of the distance between the helicopter and yourself after 5 sec.

15.  Using the previous problem, what is the rate at which the distance between you and the helicopter is changing when the helicopter has risen to a height of 60 ft in the air, assuming that, initially, it was 30 ft above you?

The distance is increasing at [latex]\frac{135\sqrt{26}}{26}[/latex] ft/sec

For the following exercises, draw and label diagrams to help solve the related-rates problems.

16.  The side of a cube increases at a rate of [latex]\frac{1}{2}[/latex] m/sec. Find the rate at which the volume of the cube increases when the side of the cube is 4 m.

17.  The volume of a cube decreases at a rate of 10 m/sec. Find the rate at which the side of the cube changes when the side of the cube is 2 m.

[latex]-\frac{5}{6}[/latex] m/sec

18.  The radius of a circle increases at a rate of 2 m/sec. Find the rate at which the area of the circle increases when the radius is 5 m.

19.  The radius of a sphere decreases at a rate of 3 m/sec. Find the rate at which the surface area decreases when the radius is 10 m.

[latex]240\pi \, \text{m}^2[/latex]/sec

20.  The radius of a sphere increases at a rate of 1 m/sec. Find the rate at which the volume increases when the radius is 20 m.

21.  The radius of a sphere is increasing at a rate of 9 cm/sec. Find the radius of the sphere when the volume and the radius of the sphere are increasing at the same numerical rate.

[latex]\frac{1}{2\sqrt{\pi}}[/latex] cm

22.  The base of a triangle is shrinking at a rate of 1 cm/min and the height of the triangle is increasing at a rate of 5 cm/min. Find the rate at which the area of the triangle changes when the height is 22 cm and the base is 10 cm.

23.  A triangle has two constant sides of length 3 ft and 5 ft. The angle between these two sides is increasing at a rate of 0.1 rad/sec. Find the rate at which the area of the triangle is changing when the angle between the two sides is [latex]\pi /6[/latex].

The area is increasing at a rate [latex]\frac{(3\sqrt{3})}{8} \, \text{ft}^{2} / \text{sec}[/latex].

24.  A triangle has a height that is increasing at a rate of 2 cm/sec and its area is increasing at a rate of 4 [latex]\text{cm}^2 / \text{sec}[/latex]. Find the rate at which the base of the triangle is changing when the height of the triangle is 4 cm and the area is 20 [latex]\text{cm}^2[/latex].

For the following exercises, consider a right cone that is leaking water. The dimensions of the conical tank are a height of 16 ft and a radius of 5 ft.

25.  How fast does the depth of the water change when the water is 10 ft high if the cone leaks water at a rate of 10 [latex]\text{ft}^3[/latex]/min?

The depth of the water decreases at [latex]\frac{128}{125\pi}[/latex] ft/min.

26.  Find the rate at which the surface area of the water changes when the water is 10 ft high if the cone leaks water at a rate of 10 [latex]\text{ft}^3[/latex]/min.

27.  If the water level is decreasing at a rate of 3 in./min when the depth of the water is 8 ft, determine the rate at which water is leaking out of the cone.

The volume is decreasing at a rate of [latex]\frac{(25\pi )}{16}{\text{ft}}^{3}\text{/min}.[/latex]

28.  A vertical cylinder is leaking water at a rate of 1 [latex]\text{ft}^3[/latex]/sec. If the cylinder has a height of 10 ft and a radius of 1 ft, at what rate is the height of the water changing when the height is 6 ft?

29.  A cylinder is leaking water but you are unable to determine at what rate. The cylinder has a height of 2 m and a radius of 2 m. Find the rate at which the water is leaking out of the cylinder if the rate at which the height is decreasing is 10 cm/min when the height is 1 m.

The water flows out at rate [latex]\frac{2\pi}{5} \, \text{m}^3[/latex]/min.

30.  A trough has ends shaped like isosceles triangles, with width 3 m and height 4 m, and the trough is 10 m long. Water is being pumped into the trough at a rate of [latex]5 \, \text{m}^3[/latex]/min. At what rate does the height of the water change when the water is 1 m deep?

31.  A tank is shaped like an upside-down square pyramid, with base of 4 m by 4 m and a height of 12 m (see the following figure). How fast does the height increase when the water is 2 m deep if water is being pumped in at a rate of [latex]\frac{2}{3} \, \text{m}^3[/latex]/sec?

[latex]\frac{3}{2}[/latex] m/sec

For the following problems, consider a pool shaped like the bottom half of a sphere, that is being filled at a rate of 25 [latex]\text{ft}^3[/latex]/min. The radius of the pool is 10 ft.

32.  Find the rate at which the depth of the water is changing when the water has a depth of 5 ft.

33.  Find the rate at which the depth of the water is changing when the water has a depth of 1 ft.

[latex]\frac{25}{19\pi}[/latex] ft/min

34.  If the height is increasing at a rate of 1 in/sec when the depth of the water is 2 ft, find the rate at which water is being pumped in.

35.  Gravel is being unloaded from a truck and falls into a pile shaped like a cone at a rate of 10 [latex]\text{ft}^3[/latex]/min. The radius of the cone base is three times the height of the cone. Find the rate at which the height of the gravel changes when the pile has a height of 5 ft.

[latex]\frac{2}{45\pi}[/latex] ft/min

36.  Using a similar setup from the preceding problem, find the rate at which the gravel is being unloaded if the pile is 5 ft high and the height is increasing at a rate of 4 in/min.

For the following exercises, draw the situations and solve the related-rate problems.

37.  You are stationary on the ground and are watching a bird fly horizontally at a rate of 10 m/sec. The bird is located 40 m above your head. How fast does the angle of elevation change when the horizontal distance between you and the bird is 9 m?

The angle decreases at [latex]\frac{400}{1681}[/latex] rad/sec.

38.  You stand 40 ft from a bottle rocket on the ground and watch as it takes off vertically into the air at a rate of 20 ft/sec. Find the rate at which the angle of elevation changes when the rocket is 30 ft in the air.

39.  A lighthouse, [latex]L[/latex], is on an island 4 mi away from the closest point, [latex]P[/latex], on the beach (see the following image). If the lighthouse light rotates clockwise at a constant rate of 10 revolutions/min, how fast does the beam of light move across the beach 2 mi away from the closest point on the beach?

[latex]100\pi[/latex] mi/min

40.  Using the same setup as the previous problem, determine at what rate the beam of light moves across the beach 1 mi away from the closest point on the beach.

41.  You are walking to a bus stop at a right-angle corner. You move north at a rate of 2 m/sec and are 20 m south of the intersection. The bus travels west at a rate of 10 m/sec away from the intersection – you have missed the bus! What is the rate at which the angle between you and the bus is changing when you are 20 m south of the intersection and the bus is 10 m west of the intersection?

The angle is changing at a rate of [latex]\frac{21}{25}[/latex] rad/sec.

For the following exercises, refer to the figure of baseball diamond, which has sides of 90 ft.

42. [T] A batter hits a ball toward third base at 75 ft/sec and runs toward first base at a rate of 24 ft/sec. At what rate does the distance between the ball and the batter change when 2 sec have passed?

43. [T] A batter hits a ball toward second base at 80 ft/sec and runs toward first base at a rate of 30 ft/sec. At what rate does the distance between the ball and the batter change when the runner has covered one-third of the distance to first base? ( Hint : Recall the law of cosines.)

The distance is increasing at a rate of 62.50 ft/sec.

44. [T] A batter hits the ball and runs toward first base at a speed of 22 ft/sec. At what rate does the distance between the runner and second base change when the runner has run 30 ft?

45. [T] Runners start at first and second base. When the baseball is hit, the runner at first base runs at a speed of 18 ft/sec toward second base and the runner at second base runs at a speed of 20 ft/sec toward third base. How fast is the distance between runners changing 1 sec after the ball is hit?

The distance is decreasing at a rate of 11.99 ft/sec.

46. [T] A building that is 225 feet tall casts a shadow of various lengths [latex]x[/latex] as the day goes by. An angle of elevation [latex]\theta[/latex] is formed by lines from the top and bottom of the building to the tip of the shadow, as seen in the following figure. Find the rate of change of the angle of elevation [latex]\frac{d\theta}{dx}[/latex] when [latex]x=272[/latex] feet.

47. [T] A pole stands 75 feet tall. An angle [latex]\theta[/latex] is formed when wires of various lengths of [latex]x[/latex] feet are attached from the ground to the top of the pole, as shown in the following figure. Find the rate of change of the angle [latex]\frac{d\theta}{dx}[/latex] when a wire of length 90 feet is attached.

Answer: -0.0168 radians per foot

48. [T] A television camera at ground level is 2000 feet away from the launching pad of a space rocket that is set to take off vertically, as seen in the following figure. The angle of elevation of the camera can be found by [latex]\theta = \tan^{-1}(\frac{x}{2000})[/latex], where [latex]x[/latex] is the height of the rocket. Find the rate of change of the angle of elevation after launch when the camera and the rocket are 5000 feet apart.

49. [T] A local movie theater with a 30-foot-high screen that is 10 feet above a person’s eye level when seated has a viewing angle [latex]\theta[/latex] (in radians) given by [latex]\theta = \cot^{-1}(\frac{x}{40})- \cot^{-1}(\frac{x}{10})[/latex],

where [latex]x[/latex] is the distance in feet away from the movie screen that the person is sitting, as shown in the following figure.

• Find [latex]\frac{d\theta}{dx}[/latex].
• Evaluate [latex]\frac{d\theta}{dx}[/latex] for [latex]x=5,10,15[/latex], and 20.
• Interpret the results in b.
• Evaluate [latex]\frac{d\theta}{dx}[/latex] for [latex]x=25,30,35[/latex], and 40
• Interpret the results in d. At what distance [latex]x[/latex] should the person sit to maximize his or her viewing angle?

Answer: a. [latex]\frac{d\theta}{dx}=\frac{10}{100+x^2}-\frac{40}{1600+x^2}[/latex] b. [latex]\frac{18}{325}, \, \frac{9}{340}, \, \frac{42}{4745}, \, 0[/latex] c. As a person moves farther away from the screen, the viewing angle is increasing, which implies that as he or she moves farther away, his or her screen vision is widening. d. [latex]-\frac{54}{12905}, \, -\frac{3}{500}, \, -\frac{198}{29945}, \, -\frac{9}{1360}[/latex] e. As the person moves beyond 20 feet from the screen, the viewing angle is decreasing. The optimal distance the person should sit for maximizing the viewing angle is 20 feet.

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GMAT Math : Rate Problems

Study concepts, example questions & explanations for gmat math, all gmat math resources, example questions, example question #1 : rate problems.

A group of students are making posters to advertise for a bake sale. 12 large signs and 60 small signs are needed. It takes 10 minutes to paint a small sign and 30 minutes to paint a large sign. How many students will be needed to paint all of the signs in 2 hours or less?

Example Question #226 : Problem Solving Questions

Jason is driving across the country. For the first 3 hours, he travels 60 mph. For the next 2 hours he travels 72 mph. Assuming that he has not stopped, what is his average traveling speed in miles per hour?

In the first three hours, he travels 180 miles.

In the next two hours, he travels 144 miles.

for a total of 324 miles.

Divide by the total number of hours to obtain the average traveling speed.

Example Question #1 : Calculating Rate

Tom runs a 100m race in a certain amount of time.  If John runs the same race, he takes 2 seconds longer.  If John ran at 8m/s, approximately how fast did Tom run?

Tom runs a 100m race in a certain amount of time.  If John runs the same race, he takes 2 seconds longer.  If John ran at 8m/s, how fast did Tom run?

Jim and Julia, a married couple, work in the same building.

One morning, both left at 9:00, but in different cars. Jim arrived at 10:10; Julia arrived 10 minutes later. If Jim's average speed was 54 miles per hour, what was Julia's average speed (nearest whole number)?

or, rounded, 47 miles per hour.

Example Question #229 : Problem Solving Questions

Andy and his wife Donna both work at the same building. 

One morning, Andy left home at 8:00; Donna left 5 minutes later. Each arrived at their common destination at 8:50. Andy drove at an average speed of 45 miles per hour; what was Donna's average speed, to the nearest mile per hour?

Example Question #230 : Problem Solving Questions

Kenny and Marie, a married couple, work in the same building.

One morning, both left at 9:00, but in different cars. Kenny arrived at 10:10; Marie arrived 10 minutes later. If Kenny's average speed was 6 miles per hour faster than Marie's, how far is their work place from their home (nearest whole mile)?

Jerry took a car trip of 320 miles. The trip took a total of six hours and forty minutes; for the first four hours, his average speed was 60 miles per hour. What was his average speed for the remaining time?

If Sally drives q miles in 3 hours, her rate is 3/q miles per hour.  Plug this rate into the distance equation and solve for the time:

A cat runs at a rate of 12 miles per hour. How far does he run in 10 minutes?

None of the other answers are correct.

We need to convert hours into minutes and multiply this by the 10 minute time interval:

Example Question #4 : Rate Problems

In order to qualify for the next heat, the race-car driver needs to average 60 miles per hour for two laps of a one mile race-track.  The driver only averages 40 miles per hour on the first lap.  What must be the driver's average speed for the second lap in order to average 60 miles per hour for both laps?

90 miles per hour

80 miles per hour

120 miles per hour

240 miles per hour

100 miles per hour

If the driver needs to drive two laps, each one mile long, at an average rate of 60 miles per hour. To find the average speed, we need to add the speed for each lap together then divide by the number of laps. The equation would be as follows:

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Part of the Depression Grand Challenge:

Using the problem-solving framework

It’s normal to be worried. Worry can draw our attention and energy to urgent problems that need solving. But worry becomes an issue when it stops us from solving problems or when we spend time focusing on things we can’t control.  Here, we share STAND skills for responding to worry over a solvable problem , reducing the time we spend worrying and how much those worries affect us.

First, when you notice your thoughts returning to negative outcomes, ask yourself two questions:

• What am I worried could happen? Be in touch with what is causing the worry.
• Do I have control over what could happen? In other words, is it solvable?

These two questions help us know which strategies might work better.

STAND Tip: Managing worry over a problem you can't solve

Problem-solving framework

Just because a problem can be solved doesn’t mean it’s easy to solve. We can even put off taking action because we aren’t sure what to do. This problem-solving framework created by Raphael Rose can help us decide what we should do:

• Write a clear, specific description of the problem.  Identify the who, what, where, and when.
• Brainstorm solutions, as many as you can. All options should be considered.
• Rate the solutions.  Consider pros and cons: time, effort, cost, possible negative consequences and the need to rely on others.
• Pick a solution. Choose one with the fewest cons and that can be done quickly.
• Make an action plan.  Detail steps including the who, what, when, where, and how.

On your own: Walk through our problem-solving worksheet

Employ this problem-solving framework when you realize your worry concerns something solvable, as often as you need. Remember, worry over these kinds of problems is normal and able to be overcome with the help of this framework.

Downloadable resources to use on your own

Information Sheet

Managing Worry

Fillable Activity

Problem-Solving Worksheet

Need help now? Call the National Suicide Prevention Lifeline by dialing 988 .

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