Lesson 6, Topic 4
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Rate of Change

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The derivative of a function is simply referred to as the rate of change of the function. If a quantity y depends on and varies with a quantity x, then the rate of change of y with respect to x is $$\frac{dy}{dx}$$

For instance, the rate of change of pressure p with respect to height h is

$$\frac{dp}{dh}$$

Example:

1. Suppose the amount of water in a holding tank at time t is given by

v(t) = 2t2 – 16t + 35

Determine;

a. if the volume of water in the tank is increasing or decreasing at time t = 1m

b. if the volume of water of in the tank is increasing or decreasing at time   t = 5m

c. time at which the volume of water in the tank is not changing.

Solution

a. v(t) = 2t2 – 16t + 35

$$\frac{dv}{dt}$$ = 4t – 16

$$\frac{dv}{dt}$$ = 4(1) – 16

$$\frac{dv}{dt}$$ = -12

At t = 1, $$\frac{dv}{dt}$$ = -12

Rate of change is negative, therefore the volume is decreasing.

b. $$\frac{dv}{dt}$$ = 4t – 16

At t = 5

$$\frac{dv}{dt}$$ = 4(5) – 16

$$\frac{dv}{dt}$$ = 20 – 16

$$\frac{dv}{dt}$$ = 4

Rate of change is positive, therefore the volume of water is decreasing.

c. When the volume of water is constant,

$$\frac{dv}{dt}$$ = 4t – 16 = 0

4t – 16 = 0

4t = 16

t = $$\frac {16}{4}$$

t = 4m

At time 4m there was no change in volume of water in the tank.

Example

2. The distance x meters moved by a car in a time t seconds is given by

x = 3t3 – 2t2 + 4t – 1

Determine the velocity and acceleration when;

i. t = 0
ii. t = 1.5s

Solution:

x = 3t3 – 2t2 + 4t – 1

Velocity, v = $$\frac {dx}{dt} \scriptsize = 9t^2 \; – \; 4t + 4$$

Acceleration, a = $$\frac {dv}{dt} \scriptsize = 18t \; – \; 4$$

i. when t = 0

velocity, v = 9(0) – 4(0) + 4 = 4m/s

acceleration, a = 18(0) – 4 = -4ms-2

ii. when t = 1.5s

velocity, v = 9(1.5)2 – 4(1.5) + 4 = 18.25m/s

acceleration, a = 18(1.5) – 4 = 23ms-2

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