Revision:OCR Core 1 - Applications of differentiation

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Revision:OCR Core 1 - Applications of differentiation

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1 12. Applications of differentiation
2 Also See
3 Comments

12. Applications of differentiation

Derivatives as rates of change

It has been discussed previously (in these notes) that the derivative of a function can be described as the "rate of change" of the function.

This is easy to understand when placed in the context of speed, distance, time, and acceleration. Consider that the gradient of a distance-time graph is the speed, this is the derivative of the function which links the distance, and time, and hence as one might describe the "rate of change" of distance (as speed), the "rate of change" of the function is the gradient of it. Similarly, the second derivative is the "rate of change of the rate of change" of the function, in the context of the distance-time graph, this is the "rate of change" of the speed, the acceleration.

It is useful to be able to understand the terms independent variable and dependent variable. The independent variable is simple the variable that is changing ("x" in most graphs on the Cartesian plane, however any letter can denote this), and the dependent variable is the variable whose value depends upon the value of another variable ("y" is most commonly used in the context of a graph on the Cartesian plane).

A practical example is that the volume of a sphere is the dependent variable, the radius is the independent variable (as the volume depends on the value of the radius).

If one has a linear relationship, one can calculate the "rate of change" using the method discussed in the first section of these notes, namely:

$m = \frac{\delta y}{\delta x}$

This is often called "delta notation".

When one is dealing with a function whose relationship cannot be represented with a straight line, it is not possible to use this method (without some form of other method) to calculate the exact "rate of change". The above equation now can find the gradient of a chord (straight line) that cuts the curve in two places, and therefore can find the average "rate of change" of a function, between two points.

One can relate the derivative to this "average", thus:

$\frac{dy}{dx} = \lim_{\delta x \to 0} \frac{\delta y}{\delta x}$

The limit shows that the chord is tending towards a single point (that point being the one which one would be finding the "rate of change" of the function at).

If one understands the relationship between infinitesimals, and calculus, one can understand the difference (questions can rely on this).

The use of the derivative:

$\frac{dy}{dx}$ ,

is reserved for finding the gradient at an exact point, whereas the "gradient of chord" method:

$\frac{\delta y}{\delta x}$ ,

can find the gradient between two points. (As one has seen before, these are related, but if one is asked for the average rate of change between two points one should not use the derivative).

Example

1. Find the average rate of change between the two points (1, 1), and (4, 16), on the curve $y = x^{2}$ .

Do not take the derivative, one should merely construct a chord, and find the gradient using the straight line method.

Hence:

$m = \frac{\delta y}{\delta x} = \frac{16 - 1}{4 - 1} = \frac{15}{3} = 5$

Second derivatives in practice

Often one will be asked to maximise, or minimise something, and quite possibly verify that the figure that one has attained is indeed a maximum, or minimum. One can see clearly, through the notes made in the previous section that if one is to maximise a function, and verify that the figure obtained is a maximum, one merely must produce the second derivative, and demonstrate that it is negative; if the problem is regarding a minimum, one might demonstrate that the second derivative is positive, and hence there is a minimum.

Maximum and minimum problems

The main type of problem which one will be asked regarding the application of differentiation is this type ("this" being a maximisation, or minimisation problem).

This is mainly an application of the techniques for finding the minima and maxima on a graph, however one does have to work within the constraints of the problem (hence one should not cite negative lengths for the dimensions of a shape), which requires some insight.

It is clear with an example what one must do.

Example

1. In a game, the players must each have a length of string (of a constant length of 30 centimetres), their aim is to enclose the largest possible area within a rectangular bound (the winner is the person with the greatest area). What is the maximum area which can be enclosed? Demonstrate that this is a maximum.

Initially one will have to formulate some form of expression for the area of the shape.

What can the players vary? This is evidently the length of the side (and the area is dependent upon this).

Consider now that if one is to denote the length of one side with "x", one can obtain an expression for the area of the shape:

A = x(15 - x)

One would like to maximise this, so therefore one would like to calculate the points on the graph of y = x(15 - x) which are maxima.

One can use the rule that has previously been used; first find the derivatives (the second derivative will be needed for the second part of the question).

$A = x(15 - x) = 15x - x^{2}$

Hence:

$\frac{dy}{dx} = 15 - 2x$

$\frac{d^{2}y}{dx^{2}} = -2$

(One can immediately explain that for any value of "x" the stationary point is a local maximum as -2 < 0 , so one must only find the stationary point now).

$\frac{dy}{dx} = 15 - 2x = 0$

Hence:

2x = 15

x = 7.5

The question requested the maximum area, and therefore one must use the original formula to calculate this:

$A = x(15 - x) = 7.5 \times (15 - 7.5) = 56.25$

One must remember the units also:

The maximum area that can be enclosed is: $56.25 cm^{2}$ .

The general idea is to formulate an equation where the dependent variable is expressed in terms of the independent variable, and then find the maxima and minima.

The notation of:

$\frac{d^{2}y}{dx^{2}}$