## Applications Differentiation Calculus can be used to solve a range of types of practical problems. Some involve integration studied later in topics 18 − 27.

### Rates of Change

If a car travels 100 km in 2 hours then the car has an average speed of 50 km/h. Speed here is a measure of the average rate of change of distance against time.

Differentiation give a measure of a rate of change at a particular instant. It gives aninstantaneous rate of change.

 Example 1 Example 2 Find the rate of change of volume with respect to time at the instant when t = 5 given that V = 1.6t2 + 5 where V is the volume of oil leaking from a tanker in m3 and t is the time in minutes. DIfferentiating , V ' = 3.2t When t = 5 V '(5) = 3.2 x 5 = 16 The rate of change, after 5 minutes, of volume with respect to time is 16 m3/min. Nico's bubble-blowing machine blows spherical bubbles whose surface area increases at the rate of 25 cm2 /second. What is the rate of increase of the radius of a bubble when its radius is 3 cm? The rate of change of the radiusis 0.33 cm/s (to 2 sig. fig.)

### Maximum and Minimum Problems

In previous topics, calculus was used to find the coordinates of maximum and minimum points on graphs such as parabolas and cubic functions. These techniques are also useful for solving real-life type problems where values need to be maximised or minimised. The problems usually involve similar steps.

Step 1 A diagram may help.
Step 2 Find the function to be differentiated. Sometimes this will be provided and sometimes it has to be worked out from the words of the problem This may involve a substitution to reduce the function to one variable.
Step 3 Differentiate.
Step 4 Put the differentiated function equal to zero and solve the equation to find the maximum or minimum.
Step 5 Verify that the answer is the required turning point by the second derivative test.
Step 6 Substitute solution(s) back into original equation to find the required values and write out the answer in a sentence.

 Example Solution A farmer has 200 metres of fencing. He wants to enclose a rectangular paddock. Find the maximum possible area he can enclose. Step 1 Draw a sketch. w w Step 2 Write down an expression for the area, A, of the paddock. Let the width = w 2w + 2x = 200w = 100 − x Area = length x width A = x(100 − x)A = 100x − x2 Step 3 Differentiate A. A' = 100 − 2x Step 4 To find maximum or minimum value, put A' = 0 and solve it. 100 − 2x = 02x = 100x = 50 Step 5 Carry out the second derivative test. A'' = -2 This is negative, which means it is a MAXIMUM value. Step 6 Substitute into original equation to solve problem and write out the answer in a sentence. When x = 50 w = 100 − 50w = 50 So maximum area = length x width = x.w = 50 x 50 = 2500 m2 The maximum area of the paddock is 2500 m2

### Distance, Velocity and Acceleration Problems

The study of problems involving distance, velocity and acceleration is called kinematics

Kinematics was studied in detail in Year 11 (Year 12 in NZ) in Topic 35.

Kinematics problems often use calculus.