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Archive for the ‘Calculus’ Category

## Sketching derivative graphs (cubics)

December 1, 2010 Leave a comment

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:

1. A cubic differentiates to a quadratic (parabola)
2. A the turning points have a gradient of zero and will move to the x axis on the derivative sketch
3. A point with a positive gradient will be above the x axis in the sketch
4. 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 $x=0\mbox{ and }x=\frac{4}{3}$, because this is where the turning points occur.

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

For $x<0$ the gradient is positive.

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

For $0 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

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Categories: Differentiation

## Integration: Area under a curve proof

November 10, 2010 Leave a comment

Consider the diagram below

Let $A(x)$ = the area from the beginning to the end of the brown section.

The blue section is a rectangle and has an area $\delta x\times f(x)$

For small $\delta x\text{ } A(x+\delta x)\approx A(x)+\delta x\times f(x)$

Rearranging gives $\delta xf(x)\approx A(x+\delta x)-A(x)$

$\implies f(x)\approx\frac{A(x+\delta x)-A(x)}{\delta x}$

As $\delta x\rightarrow 0$ the right hand side $\rightarrow \frac{dA(x)}{dx}$

So $\frac{dA(x)}{dx}=f(x)\implies A(x)=\int f(x)dx$

Categories: Calculus

## Integration: the area under a curve

November 9, 2010 Leave a comment

### 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 $I=\int_a^b f(x)dx$

To calculate $I$

We integrate $f(x)$ (find the inverse derivative)

We substitute in $b\text{ and }a$ into this new function

Then substitute the second value from the first

#### Notation

$\left[F(x)\right]_a^b=F(b)-F(a)$

### 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 $y=4$ and the lines $x=2\text{ and }x=5$

$A=\int_2^5 4dx$

$\implies A=\left[4x\right]_2^5=(4\times 5)-(4\times2)=12$

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 $f(x)=3+2x$ and the lines $x=0\text{ and }x=3$$A=\int_0^3 (3+2x)dx$

$\implies A=\left[3x+x^2\right]_0^3=(3\times 3+3^2)-(3\times 0+0^2)=18$

(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 $y=3x^2+4x+3$ and the lines $x=1\text{ and }x=4$

$A=\int_1^3 (3x^2+4x+3)dx$

$\implies A=\left[x^3+2x^2+3x\right]_1^4=(64+32+12)-(1+2+3)=102$

### Areas below the x axis

Let’s look at an example

$A=\int_2^5 (-4)dx=\left[-4x\right]_2^5=(-4\times 5)-(-4\times2)=-12$

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 $x=0\text{ and } x=4$ and the curve $y=11x-x^2-30$

$A=\int_0^4 (11x-x^2-30)dx=\left[\frac{11x^2}{2}-\frac{x^3}{3}-30x\right]_0^4$

$\implies A=(\frac{11\times 16}{2}-\frac{64}{3}-30\times 4)-(0)=53\frac{1}{3}$

### When areas are above and below the axis

Let’s consider the line $y=2x-4$ and go from $x=0\text{ to } x=4$

$A= \int_0^4 (2x-4)dx=\left[x^2-4x\right]_0^4=(16-16)-(0-0)=0$

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.

$A_1= \int_0^2 (2x-4)dx=\left[x^2-4x\right]_0^2=(-4)-(0-0)=-4$

So $A_1$ has an area of 4 (area is positive)

$A_2= \int_2^4 (2x-4)dx=\left[x^2-4x\right]_2^4=(16-16)-(4-8)=4$

So $A_2$ has an area of 4

$\therefore A=A_1+A_2 = 8$

Examples

Find the area between the x axis, $x=0,x=2$ and the curve $f(x)=x^2-1$

First we find if the function equals zero between $x=0\text{ and }x=2$

let $f(x)=0\implies x^2-1=0\implies x^2=1\implies x=\pm 1$

Since it crosses the axis at $x=1$ we need to integrate from 0 to 1 and then from 1 to 2.  We will call these areas $A_1\text{ and } A_2$.

$A_1=\int_0^1 (x^2-1)dx=\left[\frac{x^3}{3}-x\right]_0^1=(\frac{1}{3}-1)-(0-0)=-\frac{2}{3}$

Since $A_1>0\implies A_1=\frac{2}{3}$

$A_2=\int_1^2 (x^2-1)dx=\left[\frac{x^3}{3}-x\right]_1^2=(\frac{8}{3}-2)-(\frac{1}{3}-1)=\frac{4}{3}$

So the area required $= A_1+A_2=\frac{2}{3} + \frac{4}{3} =2$

Find the area between the x axis, $x=0,x=4$ and the curve $f(x)=x^2+2x-3$

First we find if the function equals zero between $x=0\text{ and }x=4$

let $f(x)=0\implies x^2+2x-3=0\implies (x-1)(x+3)=0\implies x=-1\text{ or }x=3$

Since it crosses the axis at $x=1$ we need to integrate from 0 to 1 and then from 1 to 4.  We will call these areas $A_1\text{ and } A_2$.

$A_1=\int_0^1 (x^2+2x-3)dx=\left[\frac{x^3}{3}+x^2-3x\right]_0^1=(\frac{1}{3}+1-3)-(0+0-0)=-\frac{5}{3}$

$A_2=\int_1^4 (x^2+2x-3)dx=\left[\frac{x^3}{3}+x^2-3x\right]_1^4=(\frac{64}{3}+16-12)-(\frac{1}{3}+1-3)=27$

So the area required $=A_1+A_2=\frac{5}{3}+27=28\frac{2}{3}$

Categories: Integration

## Questions: Differentiation

November 7, 2010 Leave a comment
1. Find the coordinates of the point where the curve $y=x^2+2x-4$ has a gradient of 4.
2. Find the gradient of the curve $y=\frac{x^2}{2}-3x+7$ at $x=3$.
3. Find the x coordinate where the curves $3x^2+2x-7$ and $2x^2-5x$ have the same gradient.
4. Find the gradients at the points of intersection of the two curves $x^2+3x-5$ and $x^2+2x-1$
5. Find the coordinates of the points where the curve $\frac{x^3}{3}+2x^2-21x+4$ has a gradient of 0.
6. Find the gradient of the curve with equation $2x^2-5x+3$ at the point $(3,6)$
7. Differentiate $x^3+3x$ and hence show that the curve $y=x^3+3x$ never has a gradient of zero.
8. Show that the curve $y=(x-1)^2(3-x)$ can be written in the form $y=3-7x+5x^2-x^3$ and find the gradient at $(0,3)$
9. Find the gradient of the curve $y=5x^2+6x-5$ at the point $x=1$
10. Find the three gradients where the curve $y=(2x+1)(4x-1)$ cuts the axes
Categories: Differentiation

## Integration, the inverse of differentiation

November 3, 2010 Leave a comment

Integration can be used to reverse the effect of differentiation.

The rule for differentiation is: if $y=ax^n$ then $\frac{dy}{dx}=anx^{n-1}$

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 $\frac{dy}{dx}=ax^n$ then $y=\frac{ax^{n+1}}{n+1}$

Lets consider what happens when we differentiate a constant i.e. $y=c$

This is the same as $y=cx^0\Rightarrow \frac{dy}{dx}=0\times cx^{-1}=0$

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

So $y=ax^n$ then $\frac{dy}{dx}=anx^{n-1}+c$

Examples

$\frac{dy}{dx}=8x^3+3x^2-5$, what is a general solution for y?

To get y we need to integrate

So $y=\frac{8x^4}{4}+\frac{3x^3}{3}-5x+c$

$\Rightarrow y=2x^4+x^3+5x+c$

This is a general solution because $c$ could have any value.

Here I have given $c$ 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 $\frac{dy}{dx}=6x^2-3$, the curve passes through the point $(1,3)$.

What is the particular solution.

Integrating gives $y=2x^3-3x+c$

Since the curve goes through $(1,3)$ we can substitute in these values to get $3=2-3+c\Rightarrow c=4$

This gives the particular solution $y=2x^3-3x+4$

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

### Notation

$\int f(x) dx$ means integrate $f(x)$.  The $dx$ means that the variable we are considering is x.

Example

Find the general solution of $y=\int (3x+4) dx$

$y=\frac{3x^2}{2}+4x+c$

Categories: Integration

## Finding the turning points of a polynomial

October 29, 2010 Leave a comment

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 $\frac{dy}{dx}=0$
• 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 $\frac{dy}{dx}=0$
• So we can see that at a turning point (maximum or minimum) $\frac{dy}{dx}=0$

Examples

### Find the maximum of the curve $y=15-2x-x^2$

1. We need to find the gradient function $\frac{dy}{dx}$
2. We let $\frac{dy}{dx}=0$, because this is when the maximum will occur
3. We solve the equation to find the value of x where the maximum occurs
4. We substitute this value in the original equation to get the value for y

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

Let $\frac{dy}{dx}=0 \Rightarrow -2-2x=0 \Rightarrow x=-1$

If $x=-1 \Rightarrow y = 15-2(-1)-(-1)^2 = 16$

$\therefore$ the maximum value is $y=16$

### Find the coordinates of the turning points of $y=x^3-6x^2+9x$

1. We need to find the gradient function $\frac{dy}{dx}$
2. We let $\frac{dy}{dx}=0$, because this is when the turning points will occur
3. We solve the equation to find the values of x where the turning points occurs
4. We substitute the values into the original equation to get the values for y

$\frac{dy}{dx}=3x^2-12x+9$

Let $\frac{dy}{dx}=0 \Rightarrow 3x^2-12x+9=0 \Rightarrow 3(x^2-4x+3)=0 \Rightarrow 3(x-1)(x-3)=0$

So $x=1$ of $x=3$

If $x=1$ then $y=4$ and if $x=3$ then $y=-18$

So the coordinates of the turning points are $(1,4)$ and $(3,-18)$

### Given that the curve $y=x^2+ax-5$ has a stationary point at $x = 4$ find the value of a

1. We need to find the gradient function $\frac{dy}{dx}$
2. We let $\frac{dy}{dx}=0$, because this is when the turning points will occur and we will let $x = 4$
3. We will solve the resulting equation to find the value of a

$\frac{dy}{dx}=2x+a$

when $x=4, \frac{dy}{dx}=0 \Rightarrow 0=8+a \Rightarrow a=-8$

### Find the equation of the tangent to the curve $y=2x^3+3x-5$ at $x=1$

1. We need to find the gradient function $\frac{dy}{dx}$, because to get the equation of a line we need the gradient and a point it passes through
2. Substitute $x=1$ into the equation for y to get the required point the line passes through
3. Substitute $x=1$ into the equation for $\frac{dy}{dx}$ to get the gradient of the line
4. Use $y-y_1=m(x-x_1)$  to get the equation of the line

$\frac{dy}{dx}=6x^2+3$

At $x=1, y=0$

At $x=1, \frac{dy}{dx}=9$

So substituting into $y-y_1=m(x-x_1)$ gives $y=9(x-1)$

Categories: Differentiation

## Differentiating Polynomials from first principles

October 28, 2010 Leave a comment

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 $\delta x$ gets smaller so does $\delta y$ and that the gradient of the chord gets closer to that of the tangent.

So as $\delta x \rightarrow 0$, $\frac{\delta y}{\delta x} \rightarrow$ the gradient at x

We refer to the gradient at x as $\frac{dy}{dx}$

### Finding the gradient function of $x^2$

Let $y=x^2$

The point that is $\delta x$ across from $x$ has the coordinates  $(x+\delta x, y+\delta y)$

Substituting into $y=x^2$ gives $y+\delta y=(x+\delta x)^2 \Rightarrow y+\delta y=x^2+2x(\delta x)+(\delta x)^2$

Since  $y=x^2$ we get $x^2+\delta y=x^2+2x(\delta x)+(\delta x)^2 \Rightarrow \delta y=2x(\delta x)+(\delta x)^2$

If we now divide  through by $\delta x$ we get $\frac{\delta y}{\delta x} = 2x+\delta x$

Now if we let $\delta x \rightarrow 0$ we get $\frac{dy}{dx}=2x$

### Finding the gradient function of $x^n$

Note that $(x+a)^n = x^n + nax^{n-1} + \frac{n(n-1)}{2!}a^2x^{n-2} + \frac{n(n-1)(n-2)}{3!}a^3x^{n-3} + ...$

Let $y = x^n$

The point that is $\delta x$ across from $x$ has the coordinates  $(x+\delta x, y+\delta y)$

Substituting into $y=x^n$ gives $y+\delta y=(x+\delta x)^n$

So $y+\delta y = x^n + n\delta x x^{n-1} + \frac{n(n-1)}{2!}(\delta x)^2x^{n-2} + \frac{n(n-1)(n-2)}{3!}(\delta x)^3x^{n-3} + ...$

Since  $y=x^n$

we get $x^n+\delta y = x^n + n\delta x x^{n-1} + \frac{n(n-1)}{2!}(\delta x)^2x^{n-2} + \frac{n(n-1)(n-2)}{3!}(\delta x)^3x^{n-3} + ...$

If we now divide  through by $\delta x$

we get $\frac{\delta y}{\delta x} = n x^{n-1} + \frac{n(n-1)}{2!}\delta xx^{n-2} + \frac{n(n-1)(n-2)}{3!}(\delta x)^2x^{n-3} + ...$

Now if we let $\delta x \rightarrow 0$ we get $\frac{dy}{dx}=n x^{n-1}$

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

Examples

 $y$ $\frac{dy}{dx}$ $x^5$ $5x^4$ $3x^7$ $21x^6$ $5x^{-3}$ $-15x^{-4}$

Questions

Categories: Differentiation