Related Rates - Introductionelyse/100/2016/RelatedRates.pdf · Two dogs are tied with elastic...

Chapter 3: Applications of Derivatives 3.2: Related Rates

Related Rates - Introduction

”Related rates” problems involve finding the rate of change of one quantity, based on therate of change of a related quantity.

Example: RelatedRates 1

Suppose P and Q are quantities that are changing over time, t. Suppose they are relatedby the equation

3P2 = 2Q2 + Q + 3.

IfdP

dt= 5 when P(t) = 1 and Q(t) = 0, then what is

dQ

dtat that time?

6PdP

dt= 4Q

dQ

dt+

dQ

dt

6(1)(5) = 4(0)dQ

dt+

dQ

dt

30 =dQ

dt


Related Rates


A garden hose can pump out a cubic meter of water in about 20 minutes. Supposeyou’re filling up a rectangular backyard pool, 3 meters wide and 6 meters long. How fastis the water rising?

https:

//s.iha.com/7509000004594/Holiday-lettings-Vila-nova-de-oliveirinha-Quinta-da-Mamadeira_4.jpeg

We know the rate of change over time of the volume of water, and we want to know the rate of change overtime of the height of the water. Let V be the volume of water in the cell, and let h be the height of water.Then we relate V and h:

V = 3 · 6 · h = 18h

where V is measured in cubic meters, and h is measured in meters. Then we differentiate both sides withrespect to t:

dV

dt= 18

dh

dtNow, 1

20 = 18 dhdt , so the water level is rising at about 1

20·18 = 1360 meters per minute, or something less than a

third of a centimeter per minute. It’s going to be a lot of work to fill up the pool.

https://s.iha.com/7509000004594/Holiday-lettings-Vila-nova-de-oliveirinha-Quinta-da-Mamadeira_4.jpeg

https://s.iha.com/7509000004594/Holiday-lettings-Vila-nova-de-oliveirinha-Quinta-da-Mamadeira_4.jpeg


Related Rates


A garden hose can pump out a cubic meter of water in about 20 minutes. Supposeyou’re filling up a rectangular backyard pool, 3 meters wide and 6 meters long. How fastis the water rising?

We know the rate of change over time of the volume of water, and we want to know the rate of change overtime of the height of the water. Let V be the volume of water in the cell, and let h be the height of water.Then we relate V and h:

V = 3 · 6 · h = 18h

where V is measured in cubic meters, and h is measured in meters. Then we differentiate both sides withrespect to t:

dV

dt= 18

dh

dt

Now, 120 = 18 dh

dt , so the water level is rising at about 120·18 = 1

360 meters per minute, or something less than athird of a centimeter per minute. It’s going to be a lot of work to fill up the pool.


Solving Related Rates

1. Draw a Picture

2. Write what you know, and what you want to know. Note units.

3. Relate all your relevant variables in one equation.

4. Differentiate both sides (with respect to the appropriate variable!)

5. Solve for what you want.



A cannonball is attached to a rope, which is attached to a pulley on a boat, at waterlevel. The cannonball is taken 8 (horizontal) metres from its attachment point on theboat, then dropped in the water.The cannon ball sinks straight down. The rope is being let out at a constant rate of onemetre per second, and two seconds have passed. How fast is the ball descending?

8



A cannonball is attached to a rope, which is attached to a pulley on a boat, at waterlevel. The cannonball is taken 8 (horizontal) metres from its attachment point on theboat, then dropped in the water.The cannon ball sinks straight down. The rope is being let out at a constant rate of onemetre per second, and two seconds have passed. How fast is the ball descending?

Let R(t) be the length of the rope at time from the ball to the pulley after t seconds; letD(t) be the (vertical) depth of the ball. Then

82 + D2 = R2

and so

2DdD

dt= 2R

dR

dt

After two seconds, R = 8 + 2 = 10, so D = 6. It is given thatdR

dt= 1. Now:

2(6)dD

dt= 2(10)(1)

So,dD

dt=

5

3: the ball is sinking at

5

3metres per second.



You are pouring water into a large jar, through a funnel with an extremely small hole.The funnel lets water into the jar at 100mL per second, and you are pouring water intothe funnel at 300mL per second. The funnel is shaped like a cone with height 20 cm anddiameter at the top also 20 cm. (Ignore the hole in the bottom.) How fast is the heightof the water in the funnel rising when it is 10cm high?

The volume of a cone with radius r and height h is π3r 2h.

”Funnel (PSF)” by Pearson Scott Foresman - Archives of Pearson Scott Foresman, donated to the Wikimedia Foundation. Licensed under Public Domainvia Commons - https://commons.wikimedia.org/wiki/File:Funnel_(PSF).png

https://commons.wikimedia.org/wiki/File:Funnel_(PSF).png





Let V be the volume of water in the funnel, so dVdt = 300− 100 = 200 mL/sec. Let h be the height of water

in the funnel: we’re concerned with the instant when h = 10. Then we need to know r , the radius of the coneshape made by the water in the funnel. By using similar triangles, we see that r

h = 1020 , so r = h

2 . Now,

V =π

3r2h =

π

3

(h

2

)2

h =π

12h3

We differentiate both sides with t as our variable:

dV

dt=π

4h2 dh

dt

So, 200 = π4 (10)2 dh

dt , leavingdh

dt=

8

πcm/sec



A sprinkler is 3m from a long, straight wall. The sprinkler sprays water in a circle, makingthree revolutions per minute. Let P be the point on the wall closest to the sprinkler. Thewater hits the wall at some spot, and that spot moves as the sprinkler rotates. When thespot where the water hits the wall is 1m away from P, how fast is the spot moving?

JJ Harrison https://en.wikipedia.org/wiki/Impact_sprinkler

https://en.wikipedia.org/wiki/Impact_sprinkler


P sprinkler

3

x

θ

Given the labels in the picture above, we know dθdt

= 3(2π) radians per minute, and wecan relate x to θ by tan(θ) = x

3. Then differentiate both sides with respect to t to get:

sec2 θ · dθdt

= 13dxdt

. We only need to find sec θ, which we get from our triangle:

sec θ = hypadj

=√

103

, so dxdt

= 3(√

103

)2

· 6π = 20π meters per minute.



A roller coaster has a track shaped in part like the folium of Descartes: x3 + y 3 = 6xy .When it is at the position (3, 3), its horizontal position is changing at 2 units per secondin the negative direction. How fast is its vertical position changing?

We know dxdt

= −2 when x = y = 3, and we would like to find dydt

at the same time. Therelationship between x and y is given. We differentiate implicitly:

3x2 dx

dt+ 3y 2 dy

dt= 6(x

dy

dt+ y

dx

dt)

3(9)(−2) + 3(9)dy

dt= 6(3

dy

dt+ 3(−2))

dy

dt= 2

So, the roller coaster is moving 2 units per second in the positive y direction.



Two dogs are tied with elastic leashes to a lamp post that is 2 metres from a straightroad. At first, both dogs are on the road, at the closest part of the road to the lamppost. Then, they start running in opposite directions: one dog runs 3 metres per second,and the other runs 2 metres per second. After one second of running, how fast is theangle made by the two leashes increasing?

road

lamp post

doggy 1 doggy 2

θ



Two dogs are tied with elastic leashes to a lamp post that is 2 metres from a straightroad. At first, both dogs are on the road, at the closest part of the road to the lamppost. Then, they start running in opposite directions: one dog runs 3 metres per second,and the other runs 2 metres per second. After one second of running, how fast is theangle made by the two leashes increasing?

road

lamp post

doggy 1 doggy 2

θ

The angle θ made by the leashes is made up of θ1 + θ2, as shown on either side of thedashed line above. tan θ1 = 2t

2= t, so sec2 θ1 · θ′1 = 1, so θ′1 = cos2 θ. When doggy 2 is

two metres away from its starting position, θ + 1′ =(

2

2√

2

)2

= 12. Similarly,

θ′2 = 32

cos2 θ2 = 32

(2√13

)2

= 613

. So, all together, θ′ = 12

+ 613

= 2526

radians per second.



Newton and Leibniz are attending a conference at the same university. Eager not to crosspaths, Newton is one kilometer due east of the math building, heading east, and Leibnizis two kilometers due north of the math building, heading north. If Newton is walking at5kph, and Leibniz is walking 7kph, how fast is the distance between them increasing?

Portrait of Newton, Godfrey Kneller Copy of a potrait of Leibniz by Andreas Scheits


L

Nmath bldg x

y D

We relate all the variables to each other:D2 = x2 + y2

Note D =√

x2 + y2 =√

12 + 22 =√

5.Differentiate with respect to time:

2DdD

dt= 2x

dx

dt+ 2y

dy

dtand fill in what we know so far:

2√

5dD

dt= 2(1)(5) + 2(2)(7)

Then their distance is increasing at 19√5≈ 8.5 kph.



A triangle has one side that is 1cm long, and another side that is 2cm, and the third sideis formed by an elastic band that can shrink and stretch. The two fixed sides are rotatedso that the angle they form, θ, grows by 1.5 radians each second. Find the rate of changeof the area inside the triangle when θ = π/4.

2

1h

θ

sin θ = opphyp

= h1

= h, and A = 12bh, so

A =1

2(2) sin θ = sin θ

Now dAdt

= cos θ · dθdt

= cos(π/4)(1.5) = 3

2√

2

Related Rates - Introductionelyse/100/2016/RelatedRates.pdf · Two dogs are tied with elastic...

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