The curves meet at {jatex options:inline}x+4=0 \rightarrow x=-4, \; x+2=0 \rightarrow x=-2{/jatex}.

The gradients functions of the curves at this point are {jatex options:inline}\frac{dy}{dx}=2x-5, \; \frac{dy}{dx}=4x-1{/jatex} and the gradients at {jatex options:inline}x=-2{/jatex} are {jatex options:inline}2 \times -2-5=-9{/jatex} and {jatex options:inline}4 \times -2+1=-7{/jatex}.

The angle between the curves at {jatex options:inline}x=-2{/jatex} is {jatex options:inline}tan^{-1}(-7)-tan^{-1}(-7)=1,79^o{/jatex}.]]>

Example: Find the angle between the vectorsand

We can use the same formula to find the angle between vectors in two, three or more dimensions:

Example: Find the angle between the vectorsand

To find the angle between two lines we use the same formula. The vectorsandare the tangent vectors, which are easily identifiable since the general equation of a line is given by the equationwhereis a point on the line andis the tangent vector.

Example: Find the angle between the linesand

Takeand

If the equation of the line is in column vector form, the process is almost identical:

then

The required angle is 180-56.51=123.49 degrees.

]]>HenceFor each of these three components we make the parameter the subject:

Each of these expressions are equal toso they are all equal to each other:

This expression is the cartesian form of a line in three dimensions.

Conversely, given the cartesian form of a line we can write out down the vector form:

Each term is not constant sinceandare variables, so put them equal to a common parameterand write down separate equations for each component, solving them forand

Then

]]>The parametric equationbecomes the single cartesian equation

Example: From the parametric equationsfind a cartesian equation that relatesand

Because there areandterms in the parametric equations, we look for an equation that relatesandWe can rearrangeto giveand to givehence Expanding and simplifying gives

Example: From the parametric equationsfind a cartesian equation that relatesand

We can makethe subject of the first equation and substitute it into the second.

There is a slight complication hanging over from the parametric equation which is not visible in the cartesian equations.since we must be taking the square root of a non negative number, andIf we consider the cartesian equation in isolation, we can substitute any value ofWe must have the condition inherited from the parametric equations that

Example: From the parametric equationsfind a cartesian equation that relatesand

Because there areandterms in the parametric equations, we look for an equation that relatesandWe can rearrangeto giveand to givehence Expanding and simplifying gives

]]>We can then integrate both sides:

Example: Solve the differential equation

Multiply byand divide byto giveWe can now integrate:

is a product which we integrate by parts obtainingTo find the constantwe need what is called a boundary condition. Suppose then that we have that whenSubstitute these values into (1) to obtain

hence

Example: Solve

Factorise the right hand side intoto givewhich is separable.

Multiply byand divide byto giveWe can now integrate:(1)

If we are to makethe subject we exponentiate both sides, raisingto the power of both sides:

whereNotice how the constant termin (1) becomes the constant factorwhen we exponentiate.

]]>The last expression is of the formwhich differentiates toApplying this example towe obtainDifferentiatingonly introduces another factorto giveWe can then find tangents and normals in the usual way.

Example: Find the equation of the tangent and normal to the curveat the point with coordinates

For the tangent

so at the point

For the normal

]]>Quotient Rule

Chain Rule

For each of these rules you can complete the table

u |
v |

u' |
v' |

Then sub into the appropriate formula.

Examples

Differentiate

u |
v |

u' 1 |
v' |

Differentiate

u |
v x |

u' |
v' 1 |

Differentiate

u |
v |

u' |
v' |

{jatex options:inline}f(x)=f(-x){/jatex}

respectively.

When you differentiate these equations you get

{jatex options:inline}f'(x)=--f'(-x)=f'(-x){/jatex} (1)

{jatex options:inline}f'(x)=-f'(-x){/jatex} (2)

Differentiating an odd function gives an even function and vice versa. The same is not true when integrating in general, because of the role of the arbitrary constant .Integrating(1)and(2)gives

{jatex options:inline}f(x)=-f(-x)+c{/jatex}

{jatex options:inline}f(x)=f(-x)+c{/jatex}

so the odd and even conditions fail without the further condition that {jatex options:inline}c=0{/jatex}.]]>

Example: A plane passes through the three pointsandFind the equation of the plane.

Substituting the first pointinto the equation of the planegives

Similarly the second and third giveandWe solve the simultaneous equations,

(1)

(2)

(3)

(1)+(3) gives

Subinto (2) to give

Subandinto (1) to give

The equation of the plane is thenCancel the factorto giveand clear all the fractions to give the final answer

There is an alternative form for the equation of a plane to terms of vectors:whereandare parameters andis a point in the plane. For the plane give above we can findandby subtracting points in the plane from each other:

and

.

The vector form is not unique since any points in the plane can be used.

]]>Examples of odd functions - {jatex options:inline}sin x, \; x^3, \; {/jatex}.

Even functions have the property that {jatex options:inline}f(x)=f(-x){/jatex} meaning that the graph of {jatex options:inline}f(x){/jatex} has reflectional symmetry in the {jatex options:inline}y{/jatex} axis. Every function that {jatex options:inline}f(x){/jatex} that can be written as a function of {jatex options:inline}x^2{/jatex} is even e.g. {jatex options:inline}f(x)=2-x^2{/jatex}.

Examples of even functions - {jatex options:inline}cosx, \; sin^2 x{/jatex}.

An even function can be construction from an odd function.If {jatex options:inline}f(x){/jatex} is odd then {jatex options:inline}f(x^2){/jatex} and {jatex options:inline}(f(x))^2{/jatex} are both even.

A function is neither odd nor even if neither of the above conditions holds for a function to be either odd or even.

Examples of functions that are neither odd nor even - {jatex options:inline}e^x, \; (x-1)^2{/jatex}.

An odd function {jatex options:inline}g(x){/jatex} can be constructed from a function {jatex options:inline}f(x){/jatex} that is neither odd nor even - {jatex options:inline}g(x)=f(x)-f(-x){/jatex}, and an even function {jatex options:inline}g(x){/jatex} can be constructed using {jatex options:inline}g(x)=f(x)+f(-x){/jatex}.]]>