 # Uses of differentiation

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:

Example:
Find the coordinates of the stationary points on the graph y = x² .
We know that at stationary points, dy/dx = 0 . By differentiating, dy/dx = 2x. Therefore the stationary points on this graph occur when 2x = 0, which is when x = 0. When x = 0, y = 0, 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:

Example:
Find the stationary points on the graph of y = 2x² + 4x³ and state their nature (ie whether they are maxima, minima or points of inflexion).
dy/dx = 4x + 12x²
At stationary points, dy/dx = 0
Therefore 4x + 12x² = 0
Therefore 4x( 1 + 3x ) = 0
Therefore either 4x = 0 or 3x = -1
Therefore x = 0 or -1/3

When x = 0, y = 0
When x = -1/3, y = 2/3

Looking at the gradient either side of x = 0:
When x = -0.0001, dy/dx = negative
When x = 0, dy/dx = zero
When x = 0.0001, dy/dx = positive
So the gradient goes -ve, zero, +ve, which shows a minimum point.

Looking at the gradient either side of x = -1/3 .
When x = -0.3334, dy/dx = +ve
When x = -0.3333..., dy/dx = zero
When x = -0.3332, dy/dx = -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, S, is 2pr² + 2prh
The volume = p r²h = 2000p
Therefore pr²h = 2000p.
Therefore h = 2000/r²

Therefore S = 2pr² + 2pr( 2000/r² )

= 2pr² + 4000p
r

So we have an expression for the surface area. To find when the surface area is a minimum, we need to find dS/dr
dS = 4pr - 4000p
dr               r²

When dS/dr = 0:
4pr - (4000p)/r² = 0
Therefore 4pr = 4000pp
r²
So 4pr³ = 4000p
So r³ = 1000
So r = 10

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.

S = 2pr² + 4000p
r

= 2p(10)² + 4000p
10
= 200p + 400p
= 600p

Therefore the minimum amount of metal required is 600p cm²

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