I am a mathematics student at Cardiff University and a friendly individual who is very patient. I have been teaching classical guitar for a year to a range of ages from 8 to 15 and so have experience as a teacher.

I am a mathematics student at Cardiff University and a friendly individual who is very patient. I have been teaching classical guitar for a year to a range of ages from 8 to 15 and so have experience as a teacher.

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Parametric equations are a set of equations which both depend on the same variable, such as t. An example of this would be:

x = 2t^{2}+1 and y = t^{4}-2

As the value of t changes the equations will give you seperate values for x and for y which can be plotted on a coordinate grid.

To differentiate a parametric equation you must first differentiate both the equation for x and for y seperately with respect to t. So in this case it would be:

dx/dt = 4t and dy/dt = 4t^{3}

We now have dx/dt and dy/dt. By simply divding dy/dt by dx/dt we get dy/dx as the dt cancels in the division (Since dividing is the same as multiplying by the reciprocal so (dy/dt)/(dx/dt) = (dy/dt)x(dt/dx) = dy/dx).

So for our example:

(dy/dt)/(dx/dt) = 4t^{3}/4t = t^{2} = dy/dx.

Parametric equations are a set of equations which both depend on the same variable, such as t. An example of this would be:

x = 2t^{2}+1 and y = t^{4}-2

As the value of t changes the equations will give you seperate values for x and for y which can be plotted on a coordinate grid.

To differentiate a parametric equation you must first differentiate both the equation for x and for y seperately with respect to t. So in this case it would be:

dx/dt = 4t and dy/dt = 4t^{3}

We now have dx/dt and dy/dt. By simply divding dy/dt by dx/dt we get dy/dx as the dt cancels in the division (Since dividing is the same as multiplying by the reciprocal so (dy/dt)/(dx/dt) = (dy/dt)x(dt/dx) = dy/dx).

So for our example:

(dy/dt)/(dx/dt) = 4t^{3}/4t = t^{2} = dy/dx.

A quadratic expression is of the form ax^{2}+bx+c. It is often useful to know what values of x will make the expression equal to 0; these are known as the roots of the equation. An easy way of seeing what these roots are is by factorisation of the equation (note that quadratic equations cannot always be factorised).

Starting with the general equation ax^{2}+bx+c = 0 we must look for a pair of numbers whose sum is equal to b and whose product is equal to c. So in the example x^{2}+2x+1 = 0 that would be 1 and 1, as 1+1 = 2 = b and 1x1 = 1 = c.

Once you have determined what the pair of numbers are you can put them into brackets like so: x^{2}+2x+1 = (x+1)(x+1).

Now for a second example where one of the numbers is negative. If x^{2}+2x-8 = 0 then the pair of numbers which add to 2 and multiply to give -8 is 4 and -2. We can now rewrite this in brackets as x^{2}+2x-8 = (x+4)(x-2).

A quadratic expression is of the form ax^{2}+bx+c. It is often useful to know what values of x will make the expression equal to 0; these are known as the roots of the equation. An easy way of seeing what these roots are is by factorisation of the equation (note that quadratic equations cannot always be factorised).

Starting with the general equation ax^{2}+bx+c = 0 we must look for a pair of numbers whose sum is equal to b and whose product is equal to c. So in the example x^{2}+2x+1 = 0 that would be 1 and 1, as 1+1 = 2 = b and 1x1 = 1 = c.

Once you have determined what the pair of numbers are you can put them into brackets like so: x^{2}+2x+1 = (x+1)(x+1).

Now for a second example where one of the numbers is negative. If x^{2}+2x-8 = 0 then the pair of numbers which add to 2 and multiply to give -8 is 4 and -2. We can now rewrite this in brackets as x^{2}+2x-8 = (x+4)(x-2).

When you are faced with an integral which is a product, such as *x.cos(x)*, you may be able to integrate it by parts. The statement of the integration by parts is that:

*∫u(dv/dx)dx = uv-∫v(du/dx)dx*

So if you have a function of the form *u(dv/dx)* (such as *x.cos(x)*) it can be presented in the above form. In this case *u = x* and *dv/dx = cos(x). *

When doing an intergration by parts it is useful to draw a grid first and work out what *v* and *du/dx* are (since we already know *u* and *dv/dx*) like so:

*u = x*

*dv/dx = cos(x)*

*v = sin(x)* (this is what you get when you integrate *cos(x)* as* ∫(dv/dx)dx = v*)

*du/dx = 1* (what you get from differentiating *u*)

And so now that we have everything we need we can plug things into the equation:

*∫x.cos(x)dx = uv-∫v(du/dx)dx = x.sin(x) - ∫1.sin(x)dx*

and then to finish we integrate the last bit:

* ∫x.cos(x)dx = x.sin(x)+cos(x)+c* (since this is an indefinite integral we must add a constant of integration *c*).

When you are faced with an integral which is a product, such as *x.cos(x)*, you may be able to integrate it by parts. The statement of the integration by parts is that:

*∫u(dv/dx)dx = uv-∫v(du/dx)dx*

So if you have a function of the form *u(dv/dx)* (such as *x.cos(x)*) it can be presented in the above form. In this case *u = x* and *dv/dx = cos(x). *

When doing an intergration by parts it is useful to draw a grid first and work out what *v* and *du/dx* are (since we already know *u* and *dv/dx*) like so:

*u = x*

*dv/dx = cos(x)*

*v = sin(x)* (this is what you get when you integrate *cos(x)* as* ∫(dv/dx)dx = v*)

*du/dx = 1* (what you get from differentiating *u*)

And so now that we have everything we need we can plug things into the equation:

*∫x.cos(x)dx = uv-∫v(du/dx)dx = x.sin(x) - ∫1.sin(x)dx*

and then to finish we integrate the last bit:

* ∫x.cos(x)dx = x.sin(x)+cos(x)+c* (since this is an indefinite integral we must add a constant of integration *c*).