Differentiation - Rates Of Change

Differentiation – Rates of change Example: The radius of a circular oil slick is increasing at the rate of 1.5 ms-1. a) Find the rate at which the area of the slick is increasing when its radius is 20m. b) Find the rate at which the perimeter of the slick is increasing when its radius is 20m. a) Area of circle, A = πr2. ….Differentiating w.r.t. r gives…. 

𝑑𝐴 𝑑𝑟

= 2πr = 40π, when r = 20m. We want



𝑑𝐴 𝑑𝑡

=

𝑑𝐴 𝑑𝑟

×

𝑑𝑟 𝑑𝑡

increasing 𝑑𝐶 𝑑𝑡

𝑑𝐶

= 𝑑𝑟 ×

𝑑𝑟 𝑑𝑡

the rate at which the area is increasing

= (40π) × (1.5) = 60π (m2 s-1).

b) Perimeter of circle, C = πD = 2πr. We want



𝑑𝐴 , 𝑑𝑡

𝑑𝐶 , 𝑑𝑡

the rate at which the perimeter is

= (2π) × (1.5) = 3π (ms-1).

Exercise: 1. The radius of a circular disc is increasing at a rate of 0.01 mm s-1. Find the rate at which the area is increasing when its radius is 25 mm. 2. Wine is spilled onto a carpet forming a circular stain which increases in area at a rate of 150 mm2 s-1. Find the rate at which the radius is changing when the area of the stain is 1200mm2. 3. A metal cylindrical rod is being heated and is expanding so that the volume of the rod increases at a rate of 200cm3 s-1. After t seconds the length of the rod is ten times the radius of the rod. Find the rate of increase of the radius when the rod is 60cm long. [V = πr2h] 4. A spherical balloon is being blown up so that its radius increases at a constant rate of 0.01m1. 4

a. Find the rate of increase of its volume when the radius is 0.2m. [V = 3πr3]

b. Find the rate of increase of its surface area when the radius is 0.2m. [ A = 4πr2] 5. A domestic bath is modelled as a rectangular tank with dimensions as shown in the diagram. The bath is being filled by a mixer tap which delivers water at a rate of 0.07m3 min-1. At time t the depth of the water is h m. How fast is h changing.

0.6m

0.7m 1.6m

6. A vertical cone whose height (H) is three times its base radius (R) is being filled with oil at a rate of 0.01 litres per second. At time t the depth of oil in the cone is h cm. [V = ⅓πr2h ] a. Show that the volume of oil in the cone at time t is

𝜋

𝑉 = 27 ℎ3

b. Find an expression for the rate of change of h in cm s-1. [1 litre = 1000cm3] 7. A current I amps flows through a resistance R ohms. The power developed is given by P = I2R. Find the rate of change of power for a resistance of 20 ohms if: I = 5 + 2e-3t, where t is time in seconds.

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Differentiation – Rates of change Answers: 1. 0.5π

2. 1.22

3. 0.0589

5. 0.0625

6. 𝒉𝟐 ;

𝝅 𝟗

𝟗𝟎 𝝅𝒉𝟐

4. 0.00016π, 0.016π

7. -240e-3t(5 + 2e-3t)

Solutions 1. A = πr2  𝑑𝐴 = 2𝜋𝑟 = 2 × 𝜋 × 25 when r = 25 𝑑𝑟 Now

𝑑𝐴 𝑑𝑡

=

𝑑𝐴 𝑑𝑟

𝑑𝑟 𝑑𝑡

×

= 50π × 0.01 = 0.5π m2 s-1

5. V = 1.6 × 0.7 × h = 1.12h 

𝑑𝑉 𝑑ℎ

Now 2. A = πr2  𝑑𝐴 = 2𝜋𝑟 … (differentiate wrt r) 𝑑𝑟 Now

𝑑𝑟 𝑑𝑡

=

𝑑𝑟 𝑑𝐴

𝑑𝐴 𝑑𝑡

×

= 1.12

𝑑ℎ 𝑑𝑡

=

𝑑ℎ 𝑑𝑉

×

𝑑𝑉 𝑑𝑡

1

= 1.12 × 0.07

= 0.0625 m s-1

1

= 2𝜋𝑟 × 150

We need to find r from A 6. (a) V = 1 𝜋𝑟 2 ℎ … h = 3r …  r = (ℎ) 3 3

As A = πr2  1200 = πr2 1200 𝜋

 r2 = 𝑑𝑟

𝑑𝑡 =

...r=√

1 2×𝜋×19.5441

1200 𝜋

= 19.5441

× 150 = 1.2215 …



-1

= 1.22 mm s

𝑑𝑉 𝑑𝑟

Now =

= 30𝜋𝑟 2 = 30 × π × 62 = 1080π (when r = 6)

𝑑𝑟 𝑑𝑡

=

1 1080𝜋

4. (a) 𝑉 =

𝑑𝑟 𝑑𝑉

×

𝑑𝑉 𝑑𝑡

× 200 = 0.0589 cm s 4 𝜋𝑟 3 3

Now



𝑑𝑉 𝑑𝑡

𝑑𝑉 𝑑𝑟

=

Now

𝑑𝐴 𝑑𝑡

𝑑𝑉 𝑑ℎ

𝜋

𝜋

= 3 (27) ℎ2 = 9 ℎ2 litres per

=

𝑑𝑉 𝑑𝑟

4 𝜋(3𝑟 2 ) 3

×

= 4𝜋𝑟

=

𝑑ℎ 𝑑𝑣

×

𝑑𝑣 𝑑𝑡

=

9 𝜋ℎ 2

× 0. 01 ×

90

= 𝜋ℎ2 cm s-1 7. P = I2R … diff… wrt I, where R = constant = 2IR = 40I … when R = 2. 𝑑𝑃 𝑑𝑡

=

𝑑𝑃 𝑑𝐼

×

𝑑𝐼 𝑑𝑡

But I = 5 + 2e-3t

𝑑𝑟 𝑑𝑡

= 8𝜋𝑟 = 8 × π × 0.2 = 1.6π

=

𝑑𝐴 𝑑𝑟

𝑑𝑟 𝑑𝑡

𝑑ℎ 𝑑𝑡

1000

Now

2

𝑑𝐴 𝑑𝑟

×

(b) Now

𝑑𝑃 𝑑𝐼

-1

= 4π × 0.22 × 0.01 = 0.00016π m3 s-1 (b) A = 4πr2 

𝜋

second

3. V = πr2h . . . and h = 10r . . . We need to replace ‘h’ with ‘r’ as we want the rate of change of ‘r’.Note: when h = 60cm, r = 60/10 = 6cm.  V = πr2(10r) = 10πr3 . . . differentiating … 

ℎ 2

1

 V = 3 𝜋 (3 ) ℎ …  … V = 27 ℎ3

= 1.6π × 0.01

𝑑𝐼

 𝑑𝑡 = 2[𝑒 −3𝑡 (−3)] = −6𝑒 −3𝑡 ∴

𝑑𝑃 = 2(5 + 2𝑒 −3𝑡 ) × 20 × (−6𝑒 −3𝑡 ) 𝑑𝑡

= -240e-3t(5 + 2e-3t)

= 0.016π m2 s-1

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