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SS3: MATHEMATICS - 2ND TERM

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  1. Matrices I | Week 1
    6 Topics
  2. Matrices II | Week 2
    1 Topic
    |
    1 Quiz
  3. Commercial Arithmetic | Week 3
    7 Topics
    |
    1 Quiz
  4. Coordinate Geometry | Week 4
    8 Topics
    |
    1 Quiz
  5. Differentiation of Algebraic Expressions | Week 5 & 6
    7 Topics
  6. Application of Differentiation | Week 7
    4 Topics
    |
    1 Quiz
  7. Integration | Week 8
    8 Topics
    |
    1 Quiz



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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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