Trevor Pythagoras

You can now subscribe to Trevor Pythagoras for free by email, just click the link below. Subscribe to Trevor Pythagoras by email

Stationary Points (Maximum and Minimums) and Differentiation

December 5, 2009 David Woodford Leave a comment

On a graph a stationary point is any point where the gradient is 0 so where the graph is flat. For example the graph y=x² has one stationary point at the origin.

Finding the Stationary Points

We know that stationary point occur when the gradient is 0 so when the derivative of the graph is 0, so in order to find the stationary points we but first differentiate the curve.

For example lets consider the graph $y = 3x^2 + 2x - 7$ . We cab differentiate this to find
$\frac{dy}{dx} = 6x + 2$

We must then equate the derivative to 0 and solve the resulting equation. This is because we are trying to find the points where the gradient is zero and these point occur exactly at the solutions of the equation we have formed.

So in our example we form the equation
$6x + 2 = 0$
by equating our expression for $\frac{dy}{dx}$ , $6x + 2$ , to 0
Solving this equation we find that stationary points occur exactly when
$x = \frac{2}{6} = \frac{1}{3}$
Note that there can be more than solution to this equation, each of which is a valid stationary point.

Finally we should also find the y co-ordinate for the stationary point by putting this value of x into the initial equation. So for this example $y= 3 \cdot \frac{1}{3}^2 + 2 \cdot \frac{1}{3} - 7 = -6$
So the only stationary point is at $(\frac{1}{3},-6)$

Nature of Stationary Points

The nature of a stationary point simply means what the graph is doing around it and are characterised by the second derivative, $\frac{d^{2}y}{ dx^2}$ (found by differentiating the derivative). There are three types of stationary point:

Maximum Points: These are stationary points where the graph is sloping down on either side of the stationary point (a sad face type of curve).
Here ${d^{2}y}{dx^2} < 0$
Minimum Points: These are stationary where the graph is sloping upwards on either side of the point (a happy face)

Here ${d^{2}y}{dx^2} > 0$
Point of Inflection: Here the direction of the slope of the graph is the same either side of the stationary point, it can be in either direction.

At a point of inflection ${d^{2}y}{dx^2} = 0$ but ${d^{2}y}{dx^2} = 0$ isn’t enough to ensure that a point really is a point of inflection as it could still be a maximum or minimum point

Checking the nature of a Stationary Point when ${d^{2}y}{dx^2} = 0$
In this case the easiest thing to do is look a small distance either side of the point and see whether the y value is greater than or less than that of the stationary point. You can then draw yourself a picture to see what it is. For example if they are both greater than the stationary point you know it is a minimum point, but if one is greater and one is less than it is a point of inflection

Warning: checking points either side does not guarantee the correct result as there may be another stationary point or a break in the graph between where you are checking and the stationary point so you should always check using the derivatives if possible

Categories: algebra, calculus Tags: differentiation, maximum, minimum, stationary points

The Chain Rule

November 21, 2009 David Woodford Leave a comment

The chain rule allows you to differentiate composite functions (functions of other functions) ie) f(g(x)) such as sin(3x²) or (5x³+2x+3)². The rule is as follows
$\frac{d}{dx}(f(g(x)) = \frac{dg}{dx}\frac{df}{dg}(g(x)) = g'(x)f'(g(x))$
or to understand it more simply you differentiate the inner function and multiply it by the derivative of the outer function (leaving what’s inside alone).

Differentiating brackets raised to a power

The chain rule can be a great short cut to differentiating brackets raised to a power as it doesn’t require you to multiply them all out, it also enables you to differentiate brackets raised to an unknown power.
Consider $\frac{d}{dx}((ax + b)^n)$
This is the composite of the functions ax+b and tⁿ. So we differentiate them both to get a and nt^n-1 and then apply the formula to get
$\frac{d}{dx}((ax + b)^n) = an(ax+b)^{n-1}$
Notice how we multiplied the derivative of the inner function, a, by the derivative of the outer function nt^n-1 but substituted ax+b back in for t.

To generalise we can replace the ax+b with f(x) and by applying the above get
$\frac{d}{dx}((f(x))^n) = f'(x)n(f(x))^{n-1}$

Differentiating Trigonometric functions

We can also use the chain rule when differentiating sin(f(x)) and cos(f(x)) since we know how to differentiate sin(x) and cos(x).
Using the chain rule we get
$\frac{d}{dx}(sin(f(x)) = f'(x)cos(f(x))$
and
$\frac{d}{dx}(cos(f(x)) = -f'(x)sin(f(x))$

Categories: calculus, maths Tags: chain rule, cos, differentiation, sin

Find equation of tangent to a curve

November 19, 2009 David Woodford Leave a comment

The tangent to a curve is a line which touches the curve at a point without intersecting it at that point so the gradient of the curve at that point and the gradient of the tangent are the same. So we can work out the point the tangent passes though and the gradient of the tangent from the equation of the curve, which will give us enough information to find the equation of the tangent.

Example y=x²
Find the equation of the tangent to the curve $y=x^2$ when x=4?

To do this we first need to find the gradient of the curve which we can do by differentiating it.
$\frac{d}{dx}(x^2) = 2x$
so at the point x=t the gradient is 2t.

From this we can get a general equation for the tangent using the equation for the gradient of a straight line
grad = $latex \frac{y – y_1}{x – x_1}
to get the general equation for the tangent at the point x=t by substituting x₁=t, y₁=t^sup>2 and m=2t
$2t = \frac{y - t^2}{x - t} \rightarrow 2xt - 2t^2 + t^2 = y \rightarrow y=2xt-t^2$

Then we can substitute in t=4 to find the equation of the tangent when x=4 to get
$y=8x-16$
which is our final answer.

Categories: algebra, co-ordinate geometry, maths

Inequalities

November 1, 2009 David Woodford Leave a comment

An inequality(or inequation) is similar to an equation accept for instead of saying both side of the inequality are equal we say one side is greater than (or equal to depending upon the type of inequality) the other, this is done using the greater than ( $>$ ), greater than or equal to ( $\geq$ ), less than ( $<$ ) and less than or equal to ( $\leq$ ).

Examples

Some simple examples which contain only one variable are:

$5x > 3$
$2x-7 < x+5$
$x^2 - 4 \leq x$

Solving and Manipulating Inequalities

Inequalities can be solved by rearranging them and isolating the variable you want to find in a similar way to normal equation (see the post quadratic inequalities to see how to solve quadratics). However, rather than getting an exact value such as x=3 we get a range (open or closed) of values such as x<2 or -3<-1.

Much of the manipulation is the same though there are slight variations when dividing or multiplying by negative numbers or taking the reciprocal. The important thing to remember is that like normal equations we must do the same to both sides.

Addition and Subtraction

Addition and subtraction are exactly the same to equalities. We can add or subtract whatever we like as long as we do the same to both the sides. This enables us to take expressions “to the other side” by reversing their sign. For example all the following manipulations are valid.

$x + 3 < 4 \Leftrightarrow x < 4-3 = 1$
$x - 3 < 4 \Leftrightarrow x < 4+3 = 7$
$x < 4 \Leftrightarrow x + 3 < 4+3 = 7$

Multiplication and Division

Again we can perform multiplication and division in a similar way to the way we perform it with equalities by doing the same to both sides. However, if we are multiplying or dividing by a negative number we must reverse the direction of the inequality since
$-x < y \Leftrightarrow x>-y$
This means we must be careful when diving by an unknown since by definition we don’t know whether or not it is positive or negative. If this has to be done you should consider both the cases it is positive and negative separately and if it is only positive or negative then the other inequality should lead to a contradiction which can easily be spotted such as x<0 and x>3.

Examples of valid manipulation are below:
$2x < 6 \Leftrightarrow x<3$
$-2x < 6 \Leftrightarrow x>3$
$\frac{4}{x} < 3 \Leftrightarrow \frac{4}{3} < x$ for $x \geq 0$ and/or $\frac{4}{3} > x$ for $x<0$

Reciprocals

When taking the reciprocal or “one over” of an expression you must reverse the inequality so
$x < y \Leftrightarrow \frac{1}{x} > \frac{1}{y}$

Categories: algebra Tags: inequality, inequation, manipulate, solve

Fundamental Theorem of Calculus

October 31, 2009 David Woodford Leave a comment

This theorem forms much of the basis of calculus and the uses of differentiation and integration. It basically states that differentiation and integration are opposites so if you differentiate and integral you’ll get the function you started with. This can be stated as follows:

if $F(x) = \int_a(x)^b(x) \! f(t) \, dx$ then $\frac{dF}{dx} = f(a(x))\frac{da}{dx} - f(b(x))\frac{db}{dx}$

or in the more simple case

if $F(x) = \int_0^x \! f(t) \, dx$ then $\frac{dF}{dx} = f(x)- f(0)$

It is this idea that allows us to know, for example,
$\int \! \frac{1}{1+x^2} \, dx = tan^-1(x) + c$
from the knowledge that
\ $frac{d(tan^-1(x))}{dx} = \frac{1}{1+x^2}$

This makes much of integration easier as it is often much easier to work out the derivative a function than work out the integral of one so we can look for functions which when differentiated give us the function that we want to integrate and then know that the integral is that function plus a constant.

Categories: calculus, maths Tags: calculus, differentiation, fundamental theorem, integration

Integrating Fractions – using the natrual logarithm – Example tan(x)

October 27, 2009 David Woodford Leave a comment

From result found be differentiating the natural logarithm,
$\frac{d}{dx} (ln(f(x))) = \frac{f'(x)}{f(x)}$
for some function f(x),

and the fundamental theorem of calculus we cay say that

$\int \! \frac{f'(x)}{f(x)} \, dx = ln|f(x)| + c$ where c is the integration constant

Simple Example

The most basic example of this is the integration of 1/x,

$\int \! \frac{1}{x} \, dx = ln|x| + c$

More complex example: Integration of tan(x)

A slightly more complicated example of this is the integration of tan(x). To do this we must remember that $tan(x) = \frac{sin(x)}{cos(x)}$ and notice that $\frac{d}{dx}(cos(x)) = -sin(x)$ . This means that -tan(x) is of the form $\frac{f'(x)}{f(x)}$ as required. Using this we can get

$\int \! tan(x) \, dx = \int \! \frac{sin(x)}{cos(x)} \, dx = lan|cos(x)| + c$

Trick for using this identity

Sometimes we get integrals that are almost in this form but not exactly, eg) $\int \! \frac{x}{5 + x^2} \, dx$ , however to solve these we can often factorise a constant so that it is in the required form. In this example we can take out a 2 so we get $\frac{1}{2} \int \! \frac{2x}{ 5 + x^2} \, dx = ln|5 + x^2| + c$

Categories: calculus Tags: differentiation, integration, logarithm, tan(x)

Exponential Functions

October 24, 2009 David Woodford Leave a comment

Exponential functions are any function of the form
$y = a^{bx}$ latex for some constants a and b.

If a and b are both positive then the graph will be an upward curve which tends to infinity as x tends to infinity and tends to 0 as x tends to negative infinity and looks something like the below. Note that all exponential graphs cut the y axis at 1.

The graph of y = 2^x (y equals 2 to the power x)

If a is positive and b is negative the graph is simply a reflection of this about the y axis to give the following graph:

The graph of y=2^-x (y equals 2 to the power of minus x)

The most import exponential graph is $y=e^x$ because the gradient of this graph is always equal to the value of $e^x$ at that point.

Categories: algebra, calculus Tags: exponential, function, graph

Taylor Series with example cos(x)

October 13, 2009 David Woodford Leave a comment

The Taylor series is the general case of the Maclaurin Series for calculating the value of a function. It enables you calculate the value of a function at any point if you can find the value of the function and and all its derivatives at any point. This is done as a power series. The series is as follows:

$f(x) = \sum^{\infty}_{n=0} \frac{f^{(n)}(x-a)^n}{n!} = f(a) + \frac{f^{(1)}(a)(x-a)}{1} + \frac{f^{(2)}(a)(x-a)^2}{2!} + \frac{f^{(3)}(a)(x-a)^3}{3!} + ...$

The series is said to be taken about $a$ meaning we calculate the derivatives of the function at the point a and then from these we find the value of the function. Because the series is infinite we can never find the value of the function exactly but we can give it to any required degree of accuracy by taking the first i terms of the series. The Maclaurin Series is just the Taylor series about 0.

Example cos(x)

An example of the Taylor series is to find a power series for cos(x). We can choose to do this about any value of a so in this example we will use $a = \frac{\pi}{2}$ . In this example we will look at the first 2 no zero terms.

The first we need to find the value of the derivatives and the function at pi/2.
The first one is:
$cos(\frac{\pi}{2}) = 0$

We now need to differentiate the function to get:
$\frac{d}{dx} cos(x) = -sin(x)$
and then take its value at $\frac{\pi}{2}$ which is:
$-sin({\pi}{2}) = -1$

Similarly for the second term we differentiate again to find the second derivative is:
$\frac{d}{dx} (-sin(x)) = -cos(x)$
so at pi/2 this is
$-cos(\frac{\pi}{2}) = 0$

Since the last term was zero we need to find the next one which is:
$\frac{d}{dx} (-cos(x)) = sin(x)$
so $sin(\frac{\pi}{2}) = 1$

Now we have found all the values of the function and its derivative we need for the level of accuracy required we can simply put these values into the series to get

$cos(x) = 0 + (-1)(x-\frac{\pi}{2}) + \frac{0(x-\frac{\pi}{2})^2}{2!} + \frac{1(x-\frac{\pi}{2})^3}{3!} + ...$
$cos(x) = -(x-\frac{\pi}{2}) + \frac{(x-\frac{\pi}{2})^3}{3!} - ...$

To test this we can try $x = \frac{\pi}{3}$
$cos(\frac{\pi}{3}) = -(\frac{\pi}{3}-\frac{\pi}{2}) + \frac{(\frac{\pi}{3}-\frac{\pi}{2})^3}{3!} - ...$
$cos(\frac{\pi}{3}) = 0.49967$
which is close to the 1/2 it really is.

Categories: algebra, calculus, maths Tags: maclaurin series, taylor series

Fermat’s Last Theorem

October 1, 2009 David Woodford Leave a comment

Fermat was a 17th century mathematician who provided a number of theorems and some of their proofs. The most intriguing of hi theorems is Fermat’s Last Theorem which is as follows:

the equation
$x^n + y^n = z^n$
has no integer (whole number) solutions for n>2

For example a solution for n=2 is x=3, y=4, z = 5 since
$3^2 + 4^2 = 9 + 16 = 25 = 5^2$
however the theorem states that for any n larger than 2 a set of integer solutions such as these cannot be found.

Despite the simplicity of this theorem it took 300 years until 1994 for it to be solved by Andrew Wiles using advanced maths.

A fascinating book on the problem is called Fermat’s Last Theorem (by Simon Singh) which goes through the history of the problem and many of the people who have attempted to solve it.

Categories: maths Tags: andrew wiles, Fermat, fermats last theorem

Compound tan – tan(A+B)

September 24, 2009 David Woodford Leave a comment

We can use expressions for cos(A+B) and sin(A+B) to help us find tan(A+B).

Using the identity tanx = sinx / cosx we can write

$tan(A+B) = \frac{sin(A+B)}{cos(A+B)}$

we can now substitue in
sin(A+B) = sinAcosB + sinBcosA
and
cos(A+B) = cosAcosB – sinAsinB
to get

$tan(A+B) = \frac{sinAcosB + sinBcosA}{cosAcosB - sinAsinB}$

We can now divide both the top and bottom by cosAcosB to get

$tan(A+B) = \cfrac{\cfrac{sinAcosB + sinBcosA}{cosAcosB}}{\cfrac{cosAcosB - sinAsinB}{cosAcosB}}$
or
$tan(A+B) = \cfrac{\cfrac{sinAcosB}{cosAcosB} + \cfrac{sinBcosA}{cosAcosB}}{\cfrac{cosAcosB}{cosAcosB} - \cfrac{sinAsinB}{cosAcosB}}$

We can now simplify this by cancelling any cosA and cosB to get

$tan(A+B) = \cfrac{\cfrac{sinA}{cosA} + \cfrac{sinB}{cosB}}{1 - \cfrac{sinA sinB}{cosA cosB}}$

finally by substituting the identity $tan(x) = \frac{sinx}{cosx}$ we find our result

$tan(A+B) = \cfrac{ tanA + tanB}{1 - tanAtanB}$

And it can be shown that this result can be extended to

$tan(A \pm B) = \cfrac{tanA \pm tanB}{1 \mp tanAtanB}$

Categories: maths, trigonometry Tags: compuond tan, proof, tan, tan(A+B)

Older Entries

	Safe Steroid Store on Intergration using a Reduction…
	Yoko Housman on How to Create a Facebook …
	cortadoras de tubos on Radians and Degrees \|\| Sectors…
	Pedro Vitor on Radians and Degrees \|\| Sectors…
	adrian on Area And Circumference of a Ci…

This blog has moved to
trevorpythag.co.uk

Trevor Pythagoras

Stationary Points (Maximum and Minimums) and Differentiation

Finding the Stationary Points

Nature of Stationary Points

The Chain Rule

Differentiating brackets raised to a power

Differentiating Trigonometric functions

Find equation of tangent to a curve

Inequalities

Examples

Solving and Manipulating Inequalities

Addition and Subtraction

Multiplication and Division

Reciprocals

Fundamental Theorem of Calculus

Integrating Fractions – using the natrual logarithm – Example tan(x)

Simple Example

More complex example: Integration of tan(x)

Trick for using this identity

Exponential Functions

Taylor Series with example cos(x)

Example cos(x)

Fermat’s Last Theorem

Compound tan – tan(A+B)

Blog Has Moved

Tags

Top Posts

Comments

Top Clicks

Blog Stats

This blog has moved to trevorpythag.co.uk

Finding the Stationary Points

Nature of Stationary Points

Differentiating brackets raised to a power

Differentiating Trigonometric functions

Examples

Solving and Manipulating Inequalities

Addition and Subtraction

Multiplication and Division

Reciprocals

Simple Example

More complex example: Integration of tan(x)

Trick for using this identity

Example cos(x)

Blog Has Moved

Tags

Top Posts

Comments

Top Clicks

Blog Stats

This blog has moved to
trevorpythag.co.uk