4.6: Related Rates

34

′y = 1− 2y2x+1

′y = 1−sinyxcosy−1

′y = 3 x+ y( )−1

π2

≤ t ≤ 3π2

First, a review problem:

Consider a sphere of radius 10cm.

If the radius changes 0.1cm (a very small amount) how much does the volume change?

34

3V rπ=

24dV r drπ=

( )24 10cm 0.1cmdV π= ⋅

340 cmdV π=

The volume would change by approximately .340 cmπ

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Now, suppose that the radius is changing at an instantaneous rate of 0.1 cm/sec.

(Possible if the sphere is a soap bubble or a balloon.)

34

3V rπ=

24dV dr

rdt dt

π=

( )2 cm4 10cm 0.1

sec

dV

dtπ ⎛ ⎞= ⋅⎜ ⎟

⎝ ⎠

3cm40

sec

dV

dtπ=

The sphere is growing at a rate of .340 cm / secπ

Note: This is an exact answer, not an approximation like we got with the differential problems.

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Water is draining from a cylindrical tank at 3 liters/second. How fast is the surface dropping?

L3

sec

dV

dt=−

3cm3000

sec=−

Finddh

dt2V r hπ=

2dV dhr

dt dtπ= (r is a constant.)

32cm

3000sec

dhrdt

π− =

3

2

cm3000

secdh

dt rπ=−

(We need a formula to relate V and h. )

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Steps for Related Rates Problems:

1. Draw a picture (sketch).

2. Write down known information.

3. Write down what you are looking for.

4. Write an equation to relate the variables.

5. Differentiate both sides with respect to t.

6. Evaluate.

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Hot Air Balloon Problem:

Given:4

πθ =rad

0.14min

d

dt

θ=

How fast is the balloon rising?

Finddh

dt

tan500

hθ =

2 1sec

500

d dh

dt dt

θθ =

( )2

1sec 0.14

4 500

dh

dt

π⎛ ⎞=⎜ ⎟

⎝ ⎠

h

θ500ft

→

Hot Air Balloon Problem:

Given:4

πθ =rad

0.14min

d

dt

θ=

How fast is the balloon rising?

Finddh

dt

tan500

hθ =

2 1sec

500

d dh

dt dt

θθ =

( )2

1sec 0.14

4 500

dh

dt

π⎛ ⎞=⎜ ⎟

⎝ ⎠

h

θ500ft

( ) ( )2

2 0.14 500dh

dt⋅ =

1

12

4

π

sec 24

π=

ft140

min

dh

dt=

→

4x =

3y =

B

A

5z =

Truck Problem:Truck A travels east at 40 mi/hr.Truck B travels north at 30 mi/hr.

How fast is the distance between the trucks changing 6 minutes later?

r t d⋅ =1

40 410⋅ =

130 3

10⋅ =

2 2 23 4 z+ =29 16 z+ =

225 z=

5 z=→

4x =

3y =

30dy

dt=

40dx

dt=

B

A

5z =

Truck Problem:

How fast is the distance between the trucks changing 6 minutes later?

r t d⋅ =1

40 410⋅ =

130 3

10⋅ =

2 2 23 4 z+ =29 16 z+ =

225 z=

5 z=

2 2 2x y z+ =

2 2 2dx dy dzx y zdt dt dt

+ =

4 40 3 30 5dz

dt⋅ + ⋅ =

250 5dz

dt=

50dz

dt=

miles50

hour

Truck A travels east at 40 mi/hr.Truck B travels north at 30 mi/hr.

π

A police car approaches a 4-way intersection from the north chasing a speeding car that has turned the corner and is moving east. When the cruiser is 0.8 mi north of the intersection and the car is 0.6 mi to the east, the police determine using radar that the distance between the two cars is increasing at 15 mph. If the cruiser is going 60 mph what is the speed of the car?

Example

Given:

dz

dt = 15 mph

dy

dt = -60 mph

x = 0.6, y = 0.8, z = 1

Find: dx

dt

x2 + y2 = z2

2x dxdt

+ 2y dydt

= 2z dzdt

0.6 dxdt

+ 0.8( ) −60( ) = 1 15( )

dxdt

= 105 mph

Related Rate Example

Water runs into a conical tank at the rate of 9 ft3/min. The tank stands point down and has a height of 10 ft and a base radius of 5 ft. How fast is the water level rising when the water is 6 ft deep?

Related Rate ExampleWater runs into a conical tank at the rate of 9 ft3/min. The tank stands point down and has a height of 10 ft and a base radius of 5 ft. How fast is the water level rising when the water is 6 ft deep?

dV

dt = 9, r = ′5 , Find

dhdt

at h = ′6

V = 13 π⋅r2 ⋅h, but

rh =

12, so r =

h2

V = 13 π⋅

h2

⎛⎝⎜

⎞⎠⎟

2

⋅h = 13 π⋅

h3

4 =

π12

⋅h3

dVdt

= π4⋅h2 ⋅

dhdt

9 = π4⋅36⋅

dhdt

dhdt

= 1π = .31831 ft

sec

Related Rate Example

A 13 ft ladder is leaning against a wall. Suppose that the base of the ladder slides away from the wall at the constant rate of 3 ft/sec.

1. Show how the motion of the two ends of the ladder can be represented by parametric equations.

2. What values of t make sense in this problem.

3. Graph using a graphing calculator. State an appropriate viewing window.

4. Use analytic methods to find the rates at which the top of the ladder is moving down the wall at t = .5, 1, 1.5 and 2 sec. How fast is the top of the ladder moving as it hits the ground?

Related Rate Example

A 13 ft ladder is leaning against a wall. Suppose that the base of the ladder slides away from the wall at the constant rate of 3 ft/sec.

1. Show how the motion of the two ends of the ladder can be represented by parametric equations.

2. What values of t make sense in this problem.

0 ≤ t ≤ 133

x1

t( ) = 3t, x1 t( ) = distance from base of wall at time t

y1 t( ) = 132 - 3t( )2 y1 t( ) = height of ladder at time t

Related Rate Example

3. Graph using a graphing calculator. State an appropriate viewing window.

4. Use analytic methods to find the rates at which the top of the ladder is moving down the wall at t = .5, 1, 1.5 and 2 sec. How fast is the top of the ladder moving as it hits the ground?

y2 + x2 = 169

dy

dt = -

x

y dx

dt

dy

dt.5( ) = -.348,

dydt

1( ) = -.712, dydt

1.5( ) = -1.107, dydt

2( ) = -1.561

When ladder hits ground, dydt

(4.3) = - 24.05 ftsec

t: [0,4] x: [0,13] y: [0, 13]