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Find: rate of change of the radius when the diameter is 50cm.

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(1)

Air is pumped into a spherical balloon:

I

the volume increases with 100 cm

3

/s

Find: rate of change of the radius when the diameter is 50cm.

First step: introduce suggestive notation

I

let V (t) be the volume after time t

I

let r (t) be the radius after time t Then the given problem translates to

V

0

(t) = 100 cm

3

/s Find r

0

(t) when r = 25cm.

How are the volume of a sphere and its radius related?

V = 4

3 πr

3

thus V

0

(t) = d dt

 4 3 πr (t)

3



= 4

3 π · 3r (t)

2

r

0

(t) We solve for r

0

(t):

r

0

(t) = V

0

(t)

4π · r (t)

2

r

0

(t) = 100

4π · 25

2

= 1

25π cm/s

(2)

A ladder of length 10ft rests against a vertical wall.

I

the bottom of the ladder slides away from the wall with 1ft/s How fast is the top sliding when the bottom is 6ft from the wall?

ground wall

10ft ladder y

x

d dt

x = 1

d dt

y =?

Thus

x

2

+ y

2

= 10

2

=

x =6

⇒ 6

2

+ y

2

= 10

2

= ⇒ y = ± p

10

2

− 6

2

= ⇒ y = 8 d

dt (x

2

+ y

2

) = d

dt 10

2

= ⇒ 2x dx

dt + 2y dy dt = 0

= ⇒ dy dt = − x

y dx

dt = ⇒ dy dt = − 6

8 · 1 = − 3 4

The top slides with

34

ft/s when the bottom is 6ft from the wall.

(3)

A water tank has the shape of an inverted circular cone:

I

base radius 2m and the height is 4m,

I

water is pumped into the tank at a rate of 2m

3

/min.

At what rate is the water rising when the water is 3m deep?

h 4 2

r

V = 1 3 πr

2

h How is r related to h?

r h = 2

4 = ⇒ r = 1

2 h V = 1

3 π( 1

2 h)

2

h = 1 12 πh

3

We differentiate both sides with respect to t:

dV dt = d

dt ( 1

12 πh

3

) = 1

12 π3h

2

dh

dt = ⇒ dh dt = 4

πh

2

dV

dt

h=3

= 4

π9 · 2

Thus the water rises with 8/(π9)m/min when its is 3m deep.

(4)

Problem Solving Strategy

Important when solving textual problems:

I

Read the problem carefully.

I

Draw a diagram.

I

Introduce notation, function names for the quantities.

I

Express given information and goal using the notation.

I

Write equations relating the quantities. Eliminate dependent variables (in the previous example we have eliminated the radius as it was dependent on the height).

I

Use the chain rule to differentiate both sides w.r.t. t.

I

Solve for the unknown rate, and substitute the given

information into the resulting formula.

(5)

Two cars are headed for the same road intersection:

I

car A is traveling west with 50mi/h

I

car B is traveling north with 60mi/h At what rate are the cars approaching

when A is 0.3mi and B is 0.4mi from the intersection?

x y

C

z B

A

I

x (t) = distance of A to crossing

I

y (t) = distance of B to crossing

I

z(t) = distance of A to B d

dt x = −50 d

dt y = −60 z

2

= x

2

+ y

2

= ⇒ 2z dz

dt = 2x dx

dt + 2y dy dt dz

dt = x z

dx dt + y

z dy

dt = ⇒ dz dt = 0.3

0.5 (−50) + 0.4

0.5 (−60) = −78

When x = 0.3 & y = 0.4, we get z = 0.5. The answer is 78mi/h.

(6)

Related (Dependent) Rates

We have a right-angled triangle of the form x

20cm φ

The length x increases with 4cm/s.

How fast is the angle φ changing when x = 15cm?

The quantities x and φ are related by:

tan φ = x 20 Differentiating both sides yields:

d

dt tan φ = d dt

x

20 = ⇒ 1

(cos φ)

2

· d φ dt = 1

20 · dx

dt

(7)

We have a right-angled triangle of the form x

20cm H φ

The length x increases with 4cm/s.

How fast is the angle φ changing when x = 15cm?

1

(cos φ)

2

· d φ dt = 1

20 · dx dt

= ⇒ d φ

dt = (cos φ)

2

20 · dx

dt = (cos φ)

2

20 · 4 = (cos φ)

2

5 We have cos φ = 20/H = 20/ √

15

2

+ 20

2

= 20/25 = 4/5.

Thus

d φ dt =  4

5



2

· 1 5 = 4

2

5

3

= 16

125 rad/s

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