4.1 - Related Rates

ex: Air is being pumped into a spherical balloon so that its volume increases at a rate of 100cm3/s. How fast is the radius of the balloon increasing when the diameter is 50cm?

Step 1: Collect problem info…

Known Unknown Snapshot

€

dV

dt=100

€

dr

dt= ? D = 50

Step 2: Write formula that relates known and unknown…

€

Vsphere =4

3πr3

Step 3: Take derivative of formula on both sides w/respect to ‘t’ …

Step 4: Plug-in knowns and snapshot…

€

100 = 4π 25( )2 dr

dt

⎛ ⎝ ⎜

⎞ ⎠ ⎟

Step 5: Solve for unknown…

€

100

4π 252( )

=dr

dt

€

⇒100

2500π=dr

dt

€

⇒dr

dt=

1

25π

€

dV

dt=100

Diameter = 50

implicit derivative(placeholder)

€

dV

dt=

4

3π

⎛ ⎝ ⎜

⎞ ⎠ ⎟3r2 dr

dt

⎛ ⎝ ⎜

⎞ ⎠ ⎟

ex: A 15 foot ladder is resting against the wall. The bottom is initially 10 feet away from the wall and is being pushed towards the wall at a rate of ¼ ft/sec.  How fast is the top of the ladder moving up the wall 12 seconds after we start pushing?

x

y

z

= 10 (starting distance)

Step 1: Diagram and Label (Cartesian) Step 2: Collect Problem Info

€

x =10

y = ?

z =15

€

dx

dt= −.25

dy

dt= ?

dz

dt= 0

Snapshot: t = 12 sec

Values after snapshot

€

x =10 − .25*12[ ] = 7

€

z =15€

y = ?

€

x2 + y2 = z2

€

y2 = z2 − x2

€

y = z2 − x2

€

y = 152 − 72

€

y = 225 − 49

€

y = 176

€

y = 4 11

3) Find a formula to relate the known & unknown quantities

€

x2 + y2 = z2

4) Differentiate both sides of the formula w/respect to t:

€

2xdx

dt+ 2y

dy

dt= 2z

dz

dt

€

⇒ xdx

dt+ ydy

dt= zdz

dt

5) Fill in knowns and snapshot – isolate unknown:

€

7(−.25) + 4 11dy

dt=15(0)

€

⇒ −1.75 + 4 11dy

dt= 0

€

⇒ 4 11dy

dt=1.75

€

⇒dy

dt=

1.75

4 11≈ 0.132 ft /sec

IC 4.1A – pg 272# ’s 1-7 all

Derivative

€

dV

dt=

1

3π • 2r

dr

dt•dh

dt

A water tank has the shape of an inverted circular cone with base radius 2m and height 4m. If water is being pumped into the tank at a rate of 2 m3/min, find the rate at which the water level is rising when the water is 3m deep.

€

dV

dt= 2

Known Unknown

€

dh

dt

Snapshot

€

h = 3

Formula

€

V(cone) =1

3πr2h

Diagram:

r = 2

h = 4

Replacing variables

• What is the relationship between r and h?

€

h = 2r

€

⇒ r =h

2

€

V(cone) =1

3π r2h

€

⇒ V =1

3π •

h

2

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

• h

€

⇒ V =1

3π •

h

4

2 ⎛

⎝ ⎜

⎞

⎠ ⎟• h

€

⇒ V =1

3πh3

4

€

⇒ V =π

12h3 • Try the derivative again…

€

V =π

12h3

€

⇒dV

dt=π

12• 3h2 dh

dt

€

⇒dV

dt=πh2

4

dh

dt

⎛ ⎝ ⎜

⎞ ⎠ ⎟

• Fill-in knowns and snapshot…

€

dV

dt= 2

Known Snapshot

€

h = 3

€

⇒ 2 =π 32

4

dh

dt

⎛ ⎝ ⎜

⎞ ⎠ ⎟

• Solve for unknown…

€

⇒2

9π

4

⎛ ⎝ ⎜

⎞ ⎠ ⎟=dh

dt

⎛ ⎝ ⎜

⎞ ⎠ ⎟

€

⇒8

9π=dh

dt

⎛ ⎝ ⎜

⎞ ⎠ ⎟≈ 0.28m /min

A man walks along a straight path at a speed of 4 ft/s. A rotating searchlight located 20 ft away at point perpendicular to the path follows the man. At what rate is the light rotating when the man is 15 feet from the point on the path that is perpendicular to the light?

θ

20

x Known Unknown

Snapshot Formula

• Probably better to NOT have a quotient in our formula…

€

dx

dt= 4

€

dθ

dt= ?

x = 15

€

tanθ =x

20

Not needed at the moment…

€

tanθ =x

20

€

⇒ 20tanθ = x

• Now do derivative…

€

20sec2θdθ

dt=dx

dt

• Fill-in knowns and snapshot…

4=dt

dxKnown Snapshot

15=x

€

20sec2θdθ

dt= 4 • Solve for unknown…

€

dθ

dt=

4

20sec2θ

€

⇒dθ

dt=

1

5sec2θ• Recall that:

€

secθ =1

cosθ=hyp

adj

€

dθ

dt=

1

5sec2θ

€

⇒dθ

dt=

1

5cos2θ

• Need cos to solve this…

• Use snapshot…

Snapshot15=x

22 43 +=

(3)

(4)

= 5θ

20

15

€

cosθ =adj

hyp=

4

5

€

⇒ cos2θ =4

5

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

€

⇒1

5cos2θ =

1

5

4

5

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

€

=16

125= 0.128 rad /s

IC 4.1B – pg 272# ’s 14, 17,18,21,22

Calc 4.1 - Steps for solving related rates problems

1. Identify knowns, unknowns and snapshot.

2. Find a formula that relates knowns and unknowns.

3. Take the derivative of the formula on both sides w/respect to t.

4. Plug known values and snapshot into derivative.

5. Solve for unknown.