Related RatesRelated Rates

Kirsten MaundKirsten Maund

Dahlia SweeneyDahlia Sweeney

BackgroundBackground

Calculus was invented to predict Calculus was invented to predict phenomena of change: planetary motion, phenomena of change: planetary motion, objects in freefall, varying populations, etc. objects in freefall, varying populations, etc. In many practical applications, several In many practical applications, several related rates vary together. Naturally, the related rates vary together. Naturally, the rates at which they vary are also related to rates at which they vary are also related to each other. With calculus, we can describe each other. With calculus, we can describe and calculate such related rates.and calculate such related rates.

What are related rates?What are related rates?

A related rates problem involves two or A related rates problem involves two or more quantities that vary with time and an more quantities that vary with time and an equation that expresses some relationship equation that expresses some relationship between them.between them.

Typically, the values of these quantities at Typically, the values of these quantities at some instant are given together with all some instant are given together with all their time rates of change but one. The their time rates of change but one. The problem is usually to find the time rate of problem is usually to find the time rate of change this is not given, at some instant change this is not given, at some instant specified in the problem.specified in the problem.

How to Solve Related RatesHow to Solve Related Rates One common method for solving such a One common method for solving such a

problem is to begin with implicit differentiation problem is to begin with implicit differentiation of the equation that relates the given of the equation that relates the given quantities.quantities.

For example, suppose that x and y are each For example, suppose that x and y are each functions of time such that:functions of time such that:

xx^2 + ^2 + yy^2 = ^2 = aa^2 (a is a constant)^2 (a is a constant) Differentiate both sides of this equation with Differentiate both sides of this equation with

respect to time respect to time tt. This produces the equation:. This produces the equation:

22xx dx/dt + 2 dx/dt + 2yy dy/dt = 0 dy/dt = 0

If the values of If the values of xx, , yy, and dx/dt at a certain , and dx/dt at a certain instant instant tt are known, then the last equation are known, then the last equation can be solved for the value of dy/dt at time can be solved for the value of dy/dt at time tt..

Note that it is not necessary to know Note that it is not necessary to know xx and and yy as functions of as functions of tt..

It is typical for a related rates problem to It is typical for a related rates problem to contain insufficient information to express contain insufficient information to express xx and and yy as functions of as functions of tt..

WARNING!WARNING!

The most common error to be avoided The most common error to be avoided is the premature substitution of the is the premature substitution of the given data, before rather than after given data, before rather than after

implicit differentiation.implicit differentiation.

Strategy for SolvingStrategy for Solving

Step 1Step 1: Make a drawing of the : Make a drawing of the situation if possible.situation if possible.

Step 2:Step 2: Use letters to represent the Use letters to represent the variables involved in the situation - variables involved in the situation - say say xx, , yy..

Step 3:Step 3: Identify all rates of change Identify all rates of change given and those to be determined, given and those to be determined, Use the calculus notation (dx/dt, Use the calculus notation (dx/dt, dy,dt, etc) to represent them.dy,dt, etc) to represent them.

Step 4Step 4: Determine an equation that : Determine an equation that involves bothinvolves both The variables from step two The variables from step two The derivative of step threeThe derivative of step three

Step 5Step 5: Differentiate (by implicit : Differentiate (by implicit differentiation) the equation of step differentiation) the equation of step fourfour

Step 6Step 6: Substitute all know values : Substitute all know values into the differentiated equationinto the differentiated equation

Step 7Step 7: Use algebraic : Use algebraic manipulation ,if necessary, to solve manipulation ,if necessary, to solve for the unknown rate or quantityfor the unknown rate or quantity

Formulas You May Need To Formulas You May Need To KnowKnow

3V a 2V r h

V lwh

34

3

rV

2

3

r hV

3

bhV

Example #1Example #1

A ladder 10 feet long is resting against a wall. If A ladder 10 feet long is resting against a wall. If the bottom of the ladder is sliding away from the the bottom of the ladder is sliding away from the wall at a rate of 1 foot per second, how fast is wall at a rate of 1 foot per second, how fast is the top of the ladder moving down when the the top of the ladder moving down when the bottom of the ladder is 8 feet from the wall?bottom of the ladder is 8 feet from the wall?

First, draw the picture:First, draw the picture:

We have dx/dt is one foot per second. We have dx/dt is one foot per second. We want to find dy/dt.We want to find dy/dt.

XX and and y y are related by the are related by the Pythagorean ThereomPythagorean Thereom

Differentiate both sides of this equation Differentiate both sides of this equation with respect to with respect to tt to get to get

When x = 8 ft, we haveWhen x = 8 ft, we have ThereforeTherefore

The top of the ladder is sliding down The top of the ladder is sliding down (because of the negative sign in the (because of the negative sign in the result) at a rate of 4/3 feet per second. result) at a rate of 4/3 feet per second.

Example #2Example #2 A man 6 ft tall walks with a speed of 8 ft A man 6 ft tall walks with a speed of 8 ft

per second away from a street light atop per second away from a street light atop an 8 foot pole. How fast is the tip of his an 8 foot pole. How fast is the tip of his shadow moving along the ground when he shadow moving along the ground when he is 100 feet from the light pole.is 100 feet from the light pole.

18 ft

z - x x

z

6 ft

Let x be the man’s distance from the pole and z Let x be the man’s distance from the pole and z be the distance of the tip of his shadow from be the distance of the tip of his shadow from the base of the pole.the base of the pole.

Even though x and z are functions of t, we do Even though x and z are functions of t, we do not attempt to obtain implicit formulas for either.not attempt to obtain implicit formulas for either.

We are given that dx/dt = 8 (ft/sec), and we We are given that dx/dt = 8 (ft/sec), and we want to find dz/dt when x = 100 (ft).want to find dz/dt when x = 100 (ft).

We equate ratios of corresponding sides of the We equate ratios of corresponding sides of the two similar triangles and find that z/18 = (z-x)/6two similar triangles and find that z/18 = (z-x)/6

Thus 2z = 3xThus 2z = 3x

Implicit differentiation now gives 2 dz/dt = Implicit differentiation now gives 2 dz/dt = 3 dx/dt3 dx/dt

We substitute dx/dt = 8 and find thatWe substitute dx/dt = 8 and find that(dz/dt = 3/2) * (dx/dt = 3/2) * (8) = 12(dz/dt = 3/2) * (dx/dt = 3/2) * (8) = 12

So the tip of the man’s shadow is moving at 12 So the tip of the man’s shadow is moving at 12 ft per second.ft per second.

Try Me!Try Me!

A ladder 25 ft long is leaning against a A ladder 25 ft long is leaning against a vertical wall. If the bottom of the ladder is vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at 3 pulled horizontally away from the wall at 3 ft/sec, how fast is the top of the ladder ft/sec, how fast is the top of the ladder sliding down the wall, when the bottom is sliding down the wall, when the bottom is 15 ft from the wall?15 ft from the wall?

SolutionSolution

tt = the number of seconds in time that has = the number of seconds in time that has elapsed since the ladder started to slide elapsed since the ladder started to slide down the wall.down the wall.

yy = the number of feet in distance from the = the number of feet in distance from the ground to the top of the ladder at t ground to the top of the ladder at t seconds.seconds.

xx = the number of feet in the distance from = the number of feet in the distance from the bottom of the ladder to the wall at t the bottom of the ladder to the wall at t seconds.seconds.

Because the bottom of the ladder is pulled Because the bottom of the ladder is pulled horizontally away from the wall at 3 ft/sec, horizontally away from the wall at 3 ft/sec, dx/dt = 3. We wish to find dy/dt when dx/dt = 3. We wish to find dy/dt when xx = = 15.15.

From the Pythagorean Thereom, we have From the Pythagorean Thereom, we have yy^2 = 625 – ^2 = 625 – xx^2^2

Because x and y are functions of t, we Because x and y are functions of t, we differentiate both sides of equation one differentiate both sides of equation one with respect to t and obtain 2with respect to t and obtain 2yy dy/dt = -2 dy/dt = -2xx dx/dt giving us dy/dt = -x/y dx/dtdx/dt giving us dy/dt = -x/y dx/dt

When When xx = 15, it follows from equation one = 15, it follows from equation one that that yy = 20. = 20.

Because dx/dt = 3, we get from equation Because dx/dt = 3, we get from equation two: dy/dt = (-15/20) * 3 = -9/4two: dy/dt = (-15/20) * 3 = -9/4

Therefore, the top of the ladder is sliding Therefore, the top of the ladder is sliding down the wall at the rate of 2 ¼ ft/sec down the wall at the rate of 2 ¼ ft/sec when the bottom is 15 ft from the wall.when the bottom is 15 ft from the wall.

The significance of the minus sign is that y The significance of the minus sign is that y is decreasing as is decreasing as tt is increasing. is increasing.

Was Your Answer Correct?Was Your Answer Correct?

BibliographyBibliography

http://www.math.dartmouth.edu/~klbhttp://www.math.dartmouth.edu/~klbooksite/2.17/217examples/217ladder.ooksite/2.17/217examples/217ladder.htmhtm

http://www.math.dartmouth.edu/~klbhttp://www.math.dartmouth.edu/~klbooksite/2.17/217examples/217basebooksite/2.17/217examples/217baseball.htmall.htm

© Maund and Sweeney 2011© Maund and Sweeney 2011