8/3/2019 Application of Diffrentiation
1/5
APPLICATION OF DIFFRENTIATION
Find the stationary points on the graph of y = 2x2
+ 4x3
and state their nature (i.e. whether they are
maxima, minima or points of inflexion).
dy/dx = 4x + 12x2
At stationary points, dy/dx = 0
Therefore 4x + 12x2 = 0 at stationary points
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 = 2x2 + 4x3 = 2(-1/3)2 + 4(-1/3)3 = 2/9 - 4/27
= 2/27
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 , 2/27) 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 closed
cylindrical container from thin sheet metal in
order that it might have a capacity of 2000p cm3.
The total surface area of the cylinder, S, is 2pr2 + 2prh
The volume = pr2h = 2000p
Therefore pr2h = 2000p.
Therefore h = 2000/r2
8/3/2019 Application of Diffrentiation
2/5
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 = 4000p
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 cm
http://khvmathematics.blogspot.com/2009/06/application-of-differentiation.html
8/3/2019 Application of Diffrentiation
3/5
A 20 m ladder leans against a wall. The top slides down at a
rate of 4 ms-1
. How fast is the bottom of
the ladder moving when it is 16 m from the wall?
Steps:
1. Make a sketch of the problem2. Identify constant and variable
quantities3. Establish relationship between quantities.4.
Differentiate w.r.t time.5. Evaluate at point of interest.
Following is a demonstration of how to solve this problem in
LiveMath.
You obviously still need to understand the mathematics!
Answer
Now the relation between xand yis:
x2
+ y2
= 202
Now, differentiating throughout w.r.t time:
That is:
