## Sketching derivative graphs (cubics)

The important thing to remember here is that we are doing a sketch. It does not have to be accurate, but id does have to conatian all the key points.

Hints:

- A cubic differentiates to a quadratic (parabola)
- A the turning points have a gradient of zero and will move to the x axis on the derivative sketch
- A point with a positive gradient will be above the x axis in the sketch
- A point with a pnegative gradient will be below the x axis in the sketch

Examples

Describe and sketch the derivative graph for the function below

The critical points are , because this is where the turning points occur.

These points will be on the x axis in the sketch.

For the gradient is positive.

These points will be above the x axis in the sketch.

For the gradient is negative .

These points will be below the x axis in the sketch.

The graph is a cubic so the sketch will be a parabola

### Practice

## Integration: Area under a curve proof

Consider the diagram below

Let = the area from the beginning to the end of the brown section.

The blue section is a rectangle and has an area

For small

Rearranging gives

As the right hand side

So

## Integration: the area under a curve

### Definite Integration

When we work out a definite integral we end up with a value as opposed to a function

A definite integral is in the form

To calculate

We integrate (find the inverse derivative)

We substitute in into this new function

Then substitute the second value from the first

#### Notation

### Area under a curve

To find the area under a curve we find the definite integral between the two bounds (ends)

#### Proof

Examples

Find the area (A) under the curve and the lines

This is to be expected as all we had was rectangle that was 4 high and 3 wide

Find the area (A) under the curve and the lines(note: we do not require the constant of integration because it will be cancelled out in the subtraction)

Find the area (A) under the curve and the lines

### Areas below the x axis

Let’s look at an example

Here, we have the same rectangle as in the first example, but this time it is below the x axis. The result is that the area is given as negative.

We would report the area as positive, but understand that it is below the axis.

Example

Find the area (A) between the x axis the lines and the curve

### When areas are above and below the axis

Let’s consider the line and go from

If we look at the graph we can see that there is an area, so what has happened.

Well the area below the axis has cancelled out the area above.

To calculate the desired area we must find the area from 0 to 2 and then from 2 to 4.

So has an area of 4 (area is positive)

So has an area of 4

Examples

Find the area between the x axis, and the curve

First we find if the function equals zero between

let

Since it crosses the axis at we need to integrate from 0 to 1 and then from 1 to 2. We will call these areas .

Since

So the area required

Find the area between the x axis, and the curveFirst we find if the function equals zero between

let

Since it crosses the axis at we need to integrate from 0 to 1 and then from 1 to 4. We will call these areas .

So the area required

## Questions: Differentiation

- Find the coordinates of the point where the curve has a gradient of 4.
- Find the gradient of the curve at .
- Find the x coordinate where the curves and have the same gradient.
- Find the gradients at the points of intersection of the two curves and
- Find the coordinates of the points where the curve has a gradient of 0.
- Find the gradient of the curve with equation at the point
- Differentiate and hence show that the curve never has a gradient of zero.
- Show that the curve can be written in the form and find the gradient at
- Find the gradient of the curve at the point
- Find the three gradients where the curve cuts the axes

## Integration, the inverse of differentiation

Integration can be used to reverse the effect of differentiation.

The rule for differentiation is: if then

So we multiply by the power and then take one off the power.

If we reverse this process we get add one to the power and divide by this new power.

So if then

Lets consider what happens when we differentiate a constant i.e.

This is the same as

So if we differentiate a constant we get zero which means if we integrate zero we get a constant.

So then

Examples

, what is a general solution for y?

To get y we need to integrate

So

This is a general solution because could have any value.

Here I have given values that range for -1 to 3, but there were an infinite other possibilities. The important thing to note is that if we differentiate any of these curves they all give the same expression.

The gradient function of a curve is , the curve passes through the point .

What is the particular solution.

Integrating gives

Since the curve goes through we can substitute in these values to get

This gives the particular solution

A particular solution is one where the value of c has been found so that now we can only draw one particular curve.

### Notation

means integrate . The means that the variable we are considering is x.

Example

Find the general solution of

## Finding the turning points of a polynomial

The turning points are sometimes called the stationary points or maximums and minimums.

Lets look at what happens to the gradient at a turning point

- If we consider the maximum, you can see that before it has a positive gradient and after it has a negative gradient. During this transition it has a gradient of zero. So at a maximum
- If we consider the minimum, you can see that before it has a negative gradient and after it has a positive gradient. During this transition it has a gradient of zero. So at a minimum
- So we can see that at a turning point (maximum or minimum)

Examples

### Find the maximum of the curve

- We need to find the gradient function
- We let , because this is when the maximum will occur
- We solve the equation to find the value of x where the maximum occurs
- We substitute this value in the original equation to get the value for y

Let

If

the maximum value is

### Find the coordinates of the turning points of

- We need to find the gradient function
- We let , because this is when the turning points will occur
- We solve the equation to find the values of x where the turning points occurs
- We substitute the values into the original equation to get the values for y

Let

So of

If then and if then

So the coordinates of the turning points are and

### Given that the curve has a stationary point at find the value of a

- We need to find the gradient function
- We let , because this is when the turning points will occur and we will let
- We will solve the resulting equation to find the value of a

when

### Find the equation of the tangent to the curve at

- We need to find the gradient function , because to get the equation of a line we need the gradient and a point it passes through
- Substitute into the equation for y to get the required point the line passes through
- Substitute into the equation for to get the gradient of the line
- Use to get the equation of the line

At

At

So substituting into gives

## Differentiating Polynomials from first principles

The gradient of a curve is always changing. To calculate the gradient at a point we can consider the gradient of a chord going through that point and gradually make the length of the chord shorter. As the length gets closer to zero the gradient of the chord should get closer to the gradient of the tangent at the point.

When we are finding the gradient function we are differentiating

Consider the diagram

You can see from the graph that as gets smaller so does and that the gradient of the chord gets closer to that of the tangent.

So as , the gradient at x

We refer to the gradient at x as

### Finding the gradient function of

Let

The point that is across from has the coordinates

Substituting into gives

Since we get

If we now divide through by we get

Now if we let we get

### Finding the gradient function of

Note that

Let

The point that is across from has the coordinates

Substituting into gives

So

Since

we get

If we now divide through by

we get

Now if we let we get

You should be able to see that when we differentiate that we multiply by the power and then subtract one from the power.

Examples