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Geometrical Applications of Differentiation

Board of Studies

We have already looked at some basic curve sketching. From your prior knowledge, think about what you may know about:

1. Where is the curve sloping upwards, where is it sloping downwards, andwhere does it have any maximum or minimum values?

2. Where is the curve concave up, where is it concave down, and where does itchange from one concavity to the other?

brainstorm

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Geometrical Applications of Differentiation

Definitions:

Discuss the curve below with respect to these points.

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Differentiate f(x) = x3 - 12x. Hence find whether the curve y = f(x) is increasing,decreasing or stationary at the point where:

(a) x = 5 (b) x = 2 (c) x = 0

Differentiating, f (x) = 3x2 - 12.

(a) Hence f (5) = 75 - 12 > 0, so the curve is increasing at x = 5,

(b) and f (2) = 12 - 12 = 0, so the curve is stationary at x = 2,

(c) and f (0) = - 12 < 0, so the curve is decreasing at x = 0.

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For what value(s) of x is the curve y = x4 - 4x stationary?

Differentiating, y' = 4x3 - 4= 4(x3 - 1).

Put y' = 0 to find where the curve is stationary.Then x3 = 1

x = 1.

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(a) Differentiate y = (x 2)(x 4).

(b) Hence find the values of x where the curve is stationary, and where it is decreasing.

(a) Expanding, y = x2 - 6x + 8,and differentiating, y' = 2x − 6

= 2(x − 3).

(b) Since y' = 0 when x = 3, the curve is stationary at x = 3.

Since y' < 0 when x < 3, the curve is decreasing for x < 3.

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(a) Show that f(x) = x3 + x - 1 is always increasing.

(b) Find f(0) and f(1), and hence explain why the curve has exactly one x-intercept.

(a) Differentiating, f ' (x) = 3x2 + 1.Since squares can never be negative, f ' (x) can never be less than 1, so the function is increasing for every value of x.(b) Substituting, f(0) = - 1 and f(1) = 1.Since f(0) is negative and f(1) is positive, and the curve is continuous, the curve must cross the x-axis somewhere between 0 and 1.Because the function is increasing for every value of x, it can never go back and cross the x-axis at a second point.

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Geometrical Applications of Differentiation

Majorly Important Point...

The function y = f(x) has a stationary point (also called a turning point) at f ' (x) = 0

Condition for a MAXIMUM turning point at point P:f ' (x) = 0

f'(x) > 0 f ' (x) < 0

(you also must check that the gradient is positive before P and negative after P)

Condition for a MINIMUM turning point at point P:

f'(x) < 0 f ' (x) > 0

f ' (x) = 0(you also must check that the gradient is negative before P and positive after P)

NOTE: A maximum turning point is not necessarily the highest value anywhere on the curve - which is why we normally refer to it as a 'local maximum'; similarly with a minimum turning point.

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Geometrical Applications of Differentiation

Stationary Points and Points of InflexionStationary points can be classified into four different types, as pictured below:

+ ++

+- -

-

-

Remember that a turning point only occurs if the gradient changes from positive to negative (or vice versa) on either side of the turning point. In the third and fourth diagrams above, there is no turning point. In the third diagram, the curve is increasing on both sides of the stationary point, and in the fourth, the curve is decreasing on both sides.

Instead, the curve flexes around the stationary point, changing concavity fromdownwards to upwards, or from upwards to downwards.

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Geometrical Applications of Differentiation

You need to take PARTICULAR CARE with points of inflexion!

POINTS OF INFLEXION: A point of inflexion is a point on the curve where the tangent crosses the curve. This means that the concavity changes from upwards todownwards, or from downwards to upwards, around the point.

STATIONARY POINTS OF INFLEXION: A stationary point of inflexion is a point of inflexion where the tangent is horizontal. This means that it is both a point ofinflexion and a stationary point.

To determine the nature of a stationary point get into the habit of always checking what happens with the gradient close by...never take it for granted. Show this checking process as part of your working.

With this in mind, discuss the nature of each of the points in the diagram below:

question:

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Geometrical Applications of Differentiation

Step by step procedure:

geogebra activity'guessing' derivatives

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Geometrical Applications of DifferentiationFind the stationary points of the cubic y = x3 - 6x2 + 9x - 4, determine their nature, and sketch the curve.SOLUTION:dydx = 3x2 - 12x + 9

= 3(x2 - 4x + 3)= 3(x - 1)(x - 3),

so y ' has zeroes at x = 1 and 3, and no discontinuities.

When x = 1, y = 1 - 6 + 9 - 4 = 0,

and when x = 3, y = 27 - 54 + 27 - 4 = - 4.

Hence (1, 0) is a maximum turning point, and (3, - 4) is a minimum turning point.

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Geometrical Applications of Differentiation

Find the stationary points of the quintic f(x) = 3x5 - 20x3 , determine their nature, and sketch the curve.SOLUTION:f ' (x) = 15x4 - 60x2

= 15x2(x2 - 4)= 15x2(x - 2)(x + 2),

so f ' (x) has zeroes at x = - 2, x = 0 and x = 2, and has no discontinuities:

When x = 0, y = 0 - 0 = 0, when x = 2, y = 96 - 160 = - 64,and when x = - 2, y = - 96 + 160 = 64.Hence ( - 2, 64) is a maximum turning point, (2, - 64) is a minimum turning point,and (0, 0) is a stationary point of inflexion.

Note: This function f(x) = 3x5 - 20x3 is odd, and it has as its derivativef ' (x) = 15x4 - 60x2 , which is even. In general, the derivative of an even functionis odd, and the derivative of an odd function is even. This provides a useful check.

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Geometrical Applications of Differentiation

The graph of the cubic f(x) = x3 + ax2 + bx has a stationary point at A(2, 2).Find a and b.

SOLUTION:To find the two unknown constants, we need two independent equations.Since f(2) = 2, 2 = 8 + 4a + 2b (substituting into f(x) )

2a + b = - 3. (1)

Differentiating, f ' (x) = 3x2 + 2ax + b and since f ' (2) = 0, 0 = 12 + 4a + b

4a + b = - 12. (2)

Subtracting (1) from (2), 2a = - 9 a = - 412

,and substituting into (1), - 9 + b = - 3

b = 6.

What do you do if you don't know the value of some coefficients?

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Geometrical Applications of Differentiation

Second Derivatives (and further...)

The derivative of the derivative of a function is called the second derivative of the function. As for the derivative, there is a variety of notations, including

d2y and f'' (x) and f(2) (x) and y'' and y(2)

dx2

What is the second derivative used for? VERY useful for checking the concavity of a curve

geogebra activity

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Geometrical Applications of Differentiation

Another way to visualise the relationship between the function and its derivatives:

function f(x)

first derivative f'(x)

second derivative f"(x)

gradient is positive for f(x)gradient is negative for f(x)

f(x) is concave up

f(x) is concave down

drag them

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Geometrical Applications of Differentiation

Find any points of inflexion of f(x) = x5 - 5x4 and the gradients of the inflexionaltangents, and describe the concavity. Find any turning points, and sketch.

SOLUTION:Here, f(x) = x5 - 5x4 = x4(x - 5)

f ' (x) = 5x4 - 20x3 = 5x3(x - 4)

f " (x) = 20x3 - 60x2 = 20x2(x - 3).

First, f ' (x) has zeroes at x = 0 and x = 4, and no discontinuities: so (0, 0) is a maximum turning point, and (4, - 256) is a minimum turning point.

Secondly, f " (x) has zeroes at x = 0 and x = 3, and no discontinuities, so (3, - 162) is a point of inflexion, but (0, 0) is not.

Since f ' (3) = - 135, the inflexional tangent has gradient - 135.The graph is concave down for x < 0 and 0 < x < 3, and concave up for x > 3.Note: The example given above is intended to show that f " (x) = 0 is NOT a sufficient condition for a point of inflexion - the sign of f " (x) must also change around the point.

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Geometrical Applications of Differentiation

For what values of b is y = x4 - bx3 + 5x2 + 6x - 8 concave down when x = 2?

SOLUTION:Differentiating, y ' = 4x3 - 3bx2 + 10x + 6and differentiating again, y " = 12x2 - 6bx + 10,

so when x = 2, y " = 48 - 12b + 10 = 58 - 12b.

In order for the curve to be concave down at x = 2,58 - 12b < 0 12b > 58

b > 45

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What happens when a coefficient is unknown?

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Geometrical Applications of Differentiation

You may find this useful when sketching:

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Geometrical Applications of Differentiation

Here are some of many practical applications of maximisation and minimisation.• Maximise the volume of a box built from a rectangular sheet of cardboard.• Minimise the fuel used in a flight.• Maximise the profits from manufacturing and selling an article.• Minimise the amount of metal used in a can of soft drink.Such problems can be solved using calculus, provided that a clear functionalrelationship can first be established.

Real World Applications

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Geometrical Applications of Differentiation

An open rectangular box is to be made by cutting square corners out of a square piece of cardboard measuring 60 cm × 60 cm, and folding up the sides. What is the maximum volume of the box, and what are its dimensions then?

SOLUTION:Step 1: DRAW A DIAGRAM!!!!Let V be the volume of the box, and let x be the side lengths of the cut-out squares. Then the box is x cm high, with base a square of side length 60 - 2x,so V = x(60 - 2x)2 ,

= 3600x - 240x2 + 4x3 , where 0 ≤ x ≤ 30.

Differentiating, V ' = 3600 - 480x + 12x2

= 12(x - 30)(x - 10),so V ' has zeroes at x = 10 and x = 30, and no discontinuities.Furthermore, V " = - 480 + 24xso V " (10) = - 240 < 0 and V " (30) = 240 > 0.

Hence (10, 16 000) is the global maximum in the domain 0 ≤ x ≤ 30,and the maximum volume is 16 000cm3 when the box is 10 cm × 40 cm × 40 cm.

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Geometrical Applications of Differentiation

PrimitivesConsider the derivatives of each of these expressions:

These functions are all the same apart from a constant term. This is true generally— any two functions with the same derivative differ only by a constant.

From the examples above, the various functions whose derivatives are 2x must all be of the form f(x) = x2 + C, where C is a constant. By taking different values of the constant C, these functions form an infinite family of curves, each consisting of the parabola y = x2 translated upwards or downwards.

If we know also that the curve must pass through a particular point, say (2, 7), then we can evaluate the constant C by substituting the point into

f(x) = x2+C: 7 = 4+C.

Thus C = 3 and hence f(x) = x2 + 3 In place of the infinite family of functions, there is now a single function.

Such an extra condition is called a boundary condition. It is also called an initial condition if it involves the value of y when x =0, particularly when x is time.

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Geometrical Applications of Differentiation

Definition of Primitive:

Find the primitives of: (a) x3 + x2 + x + 1 (b) 5x3 + 7rub and reveal

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Geometrical Applications of Differentiation

Summary:

• Be able to find first, second and third derivatives of functions and have an understanding of how each is used in sketching curves• Create your summary of this topic, including steps to graphing• Be able to find the primitive of a function• Search out HSC questions on this topic, and complete them focusing on setting out work.