Revision:Uses of Differentiation

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Stationary Points

Stationary points are points on a graph where the gradient is zero and are of three types:

• maximum,
• minimum or
• point of inflexion.

The three are illustrated here:

(diagram missing)

Example

Find the coordinates of the stationary points on the graph .

We know that at stationary points: .

By differentiating:

.

Therefore the stationary points on this graph occur when , which is when .

When , therefore the coordinates of the stationary point are (0,0).

Maximum, minimum or point of inflexion?

At all the stationary points, the gradient is the same (= zero) but it is often necessary to know which is which. Therefore the gradient at either side of the stationary point needs to be looked at.

• At maximum points, the gradient is positive just before the maximum, it is zero at the maximum and it is negative just after the maximum.

• At minimum points, the gradient is negative, zero then positive.

• Finally at points of inflexion, the gradient can be positive, zero, positive or negative, zero, negative.

This is illustrated here:

(diagram missing)

Example

Find the stationary points on the graph of and state their nature (ie whether they are maxima, minima or points of inflexion).

At stationary points,

Therefore

Therefore

Therefore either or

Therefore .

When

When .

Looking at the gradient either side of :

When negative.

When

When positive.

So the gradient goes -ve, zero, +ve, which shows a minimum point.

Looking at the gradient either side of :

When +ve

When

When -ve

So the gradient goes +ve, zero, -ve, which shows a maximum point.

Therefore there is a maximum point at (-1/3 , 2/3) and a minimum point at (0,0).

Solving Practical Problems

This method of finding maxima and minima is very useful and can be used to find the maximum and minimum values of all sorts of things.

Example

Find the least area of metal required to make a cylindrical container from thin sheet metal in order that it might have a capacity of 2000pcm³.

The total surface area of the cylinder,

The volume,

Therefore .

Therefore

Therefore

So we have an expression for the surface area.

To find when the surface area is a minimum, we need to find .

When :

Therefore

So

So

So

You should then check that this is indeed a minimum using the technique above. So the minimum area occurs when r = 10. This minimum area is found by substituting into the equation for the area the value of r = 10.

Therefore the minimum amount of metal required is 600p cm².

Comments

I question the title of the page - it's not obvious what is included in here.