# Applications

### 4.5.1 Kinematics

Kinematics deals with the movement of bodies over time. When you are given one function to calculate displacement, velocity or acceleration you can use differentiation or integration to determine the functions for the other two.

The derivative represents the rate of change, i.e. the gradient of a graph. So, velocity is the rate of change in displacement and acceleration is the rate of change in velocity.

A diver jumps from a platform at time t = 0 seconds. The distance of the diver above water level at time t is given by s(t) = −4.9t2 + 4.9t + 10, where s is in metres. Find when velocity equals zero. Hence find the maximum height of the diver.
1. Find an equation for velocity by differentiating equation for distance

v(t) = −9.8t + 4.9

2. Solve for v(t) = 0

−9.8t + 4.9 = 0, t = 0.5

3. Put value into equation for distance to find height above water

s(0.5) = −4.9(0.5)2 + 4.9(0.5) + 10            = 11.225 m

### 4.5.2 Optimization

We can use differentiation to find minimum and maximum areas/volumes of various shapes. Often the key skill with these questions is to find an expression using simple geometric formulas and rearranging in order to differentiate.

Finding the minimum/maximum area or volume

The sum of height and base of a triangle is 40 cm. Find an expression for its area in terms of x, its base length. Hence find its maximum area.

1. Find expressions for relevant dimensions of the shape

length of the base (b) = x

height + base = 40

so h + x = 40

area of triangle A= ½ xh

2. Reduce the number of variables by solving the simultaneous equations

Since h = 40 − x , substitute h into A:

A = ½ x(40 − x) =−½ x2 + 20x

3. Differentiate

f ′(x) = −x + 20

4. Find x when f ′(x) = 0

−x + 20 = 0 ⇒ x = 20

5. Plug x value in f (x)

−½ 202 + 20(20) = −200 + 400 = 200cm2

Note: If an expression is given in the problem, skip to step 2 (e.g. cost/profit problems)