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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<x<\frac{4}{3} 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

Categories: Differentiation

Simultaneous Equations: Substitution Method

November 30, 2010 Leave a comment

This is exactly the same method we use to find where two lines intersect.

Process:

  1. Rearrange one of the equations to make one of the variables a subject
  2. Substitute for this variable into the other equation
  3. Solve the resulting equation
  4. Substitute these solutions into one of the original equations to find the values of the other variable

Example

Solve the system of equations

y=4x-3\rightarrow 1

3y+x=7\rightarrow 2

Substitute eqn 1 into eqn 2

\implies 3(4x-3)+x=7

\displaystyle \implies 12x-9+x=7\implies 13x=16\implies \frac{16}{13}

Substituting into eqn 1 gives

\displaystyle y=4\left(\frac{16}{13}\right)-3=\frac{25}{13}


Solve the system of equations

y-3=x^2+4x\rightarrow 1

2x=y-6\rightarrow 2

Rearrange eqn 2 to make y the subject

\implies y=2x+6\rightarrow 3

Substitute eqn 3 into eqn 1

\implies (2x+6)-3=x^2+4x\implies x^2+2x-3=0

Factorising gives

(x+3)(x-1)=0\implies x=-3\mbox{ or }x=1

Substituting these values into eqn 3 gives

y=0\mbox{ and }y=8

So the solutions are (-3,0)\mbox{ and }(1,8)


Solve the system of equations

3x+2y=13\rightarrow 1

2x-y=-3\rightarrow 2

Rearranging eqn 2 to make y the subject gives

y = 2x+3\rightarrow 3

Substituting eqn 3 into eqn 1 gives

3x+2(2x+3)=13\implies 7x+6=13\implies x=1

Substituting the value of x into eqn 3 gives

y=5

Categories: Equations

Simultaneous equations: Elimination method

November 30, 2010 Leave a comment

Simultaneous equations are sytems of equations where there are a number of variables.

Our task is to find the values of the variables that satisfy all the equations.
For this method we will solve equations of the type

\displaystyle ax+by=c\rightarrow 1

\displaystyle dx+ey=f\rightarrow 2

With these equations we can do one of three things

  1. Multiply by a scaler.
  2. Add the equations together
  3. Subtract one equation from another

Multiplying by a scaler

eqn 1 multiplied by 6 gives

\displaystyle 6ax+6by=6c\rightarrow 3

Adding two equations

eqn 1 + eqn 2 gives

\displaystyle (a+d)x+(b+e)y=c+f\rightarrow 4

Subtract one equation from another

eqn 1 – eqn 2 gives

\displaystyle (a-d)x+(b-e)y=c-f\rightarrow 5

Type 1 Simultaneous equations

With this type of problem the y terms are the same size but different signs.

To solve these we:

  1. Add the equations together (this will cancel out the y terms)
  2. Solve the resulting equation to find the value of x
  3. Substitute the x value into one of the original equations
  4. Solve the resulting equation to find the value of y
  5. Check by substituting x and y into the other equation

Examples

Solve

\displaystyle 3x+4y=18\rightarrow 1

\displaystyle 7x-4y=2\rightarrow 2

This is a type 1 problem because we have two 4y’s, but one is positive and the other is negative.

So eqn 1 + eqn 2 gives

\displaystyle 10x=20

\displaystyle \implies x=2

Substitute x=2 into eqn 1

gives \displaystyle 6+4y=18

\implies 4y=12

\implies y=3

Check in eqn 2

14-12=2 true


Solve

\displaystyle 2x-5y=13\rightarrow 1

\displaystyle 3x+5y=7\rightarrow 2

This is a type 1 problem because we have two 5y’s, but one is positive and the other is negative.

So eqn 1 + eqn 2 gives

\displaystyle 5x=20

\displaystyle \implies x=4

Substitute x=4 into eqn 2 (it is easier to deal with positives)

gives \displaystyle 12+5y=7

\implies 5y=-5

\implies y=-1

Check in eqn 1

8--5=13 true

Type 2 Simultaneous equations

With this type of problem the y terms are the same size and sign.

To solve these we:

  1. Subtract one equation from the other (this will cancel out the y terms)
  2. Solve the resulting equation to find the value of x
  3. Substitute the x value into one of the original equations
  4. Solve the resulting equation to find the value of y
  5. Check by substituting x and y into the other equation

Examples

Solve

\displaystyle 4x+2y=20\rightarrow 1

\displaystyle 7x+2y=23\rightarrow 2

This is a type 2 problem because we have two +2y’s.

So eqn 2 – eqn 1 gives

\displaystyle 3x=3

\implies x=1

Substitute x=1 into eqn 1

gives \displaystyle 4+2y=20

\implies 2y=16

\implies y=8

Check in eqn 2

\displaystyle 7+16=23 true


Solve

\displaystyle 2x-4y=6\rightarrow 1

\displaystyle 5x-4y=3\rightarrow 2

This is a type 2 problem because we have two -4y’s.

So eqn 2 – eqn 1 gives

\displaystyle 3x=-3\rightarrow 1

\implies x=-1

Substitute x=-1 into eqn 1

gives \displaystyle -2-4y=6

\implies 4y=-8

\implies y=-2

Check in eqn 2

\displaystyle -5--8=3 true

Type 3 Simultaneous equations

With this type of problem the y terms are different.

To solve these we:

  1. Multiply each equation by the other equations positive y coefficient
  2. Solve the resulting type 1 or type 2 problem

Examples

Solve

\displaystyle 3x+4y=-5\rightarrow 1

\displaystyle 5x+2y=1\rightarrow 2

Multiply eqn 1 by 2 (that is the number of y’s in eqn 2)

\displaystyle 6x+8y=-10\rightarrow 3

Multiply eqn 2 by 4 (that is the number of y’s in eqn 1)

\displaystyle 20x+8y=4\rightarrow 4

Equations 3 and 4 now make a type 2 problem

Subtracting eqn 3 from eqn 4 gives

\displaystyle 14x=14

\displaystyle \implies x=1

Substitute x=1 into eqn 2

\displaystyle \implies 5+2y=1

\displaystyle \implies 2y=-4

\displaystyle \implies y=-2

Check in eqn 1

\displaystyle 5-4=1 true


Solve

\displaystyle 4x-2y=7\rightarrow 1

\displaystyle x+3y=10\rightarrow 2

Multiply eqn 1 by 3 (that is the number of y’s in eqn 2)

\displaystyle 12x-6y=21\rightarrow 3

Multiply eqn 2 by 2 (that is the number of y’s in eqn 1)

\displaystyle 2x+6y=20\rightarrow 4

Equations 3 and 4 now make a type 1 problem

Adding eqn 3 and eqn 4 gives

\displaystyle 14x=41\rightarrow 4

\displaystyle x=\frac{41}{14}

Substitute x=\frac{41}{14} into eqn 2

\displaystyle \implies \frac{41}{14}+3y=10

\displaystyle \implies 3y=\frac{99}{14}\implies y=\frac{33}{14}

Check in eqn 1

\displaystyle \frac{164}{14}-\frac{66}{14}=7 true

Categories: Equations

Circle theorems 2: Questions

November 24, 2010 Leave a comment

Set 1

Set 2

Categories: Geometry

Circle Theorems 2

November 23, 2010 Leave a comment

Tangents to circles

Investigation

Theorem 1

The length of the tangents from a point to a circle are of equal length

Theorem 2

The Angle between a tangent and a radius is a right-angle

Theorem 3

The angel between a tangent and a cord is equal to the angle subtended by the cord

Examples

Given that PA is a tangent find the value of x
Since PA is a tangent \angle PAO = 90^o
\implies x = 180^o-90^o-65^o=25^o \because 180^o \mbox{in a }\triangle

Given that PA and PB are tangents find the value of x
Since PA and PB are tangents they must both be of equal length.  Therefore \triangle ABP is isosceles.
\implies \angle ABP =\frac{180^o-49^o}{2}=65.5^o \because base angles in an isosceles triangle are equal.
\implies x=90^o-65.5^o=24.5^o \because the angle between a tangent and a radius is a right-angle.

Find the value of x
Since \triangle ABC \mbox{ is isosceles}\implies \angle ACB = 50^o
\implies x=50^o \because \mbox{Angle subtended by the cord}

Questions

Categories: Geometry

Trigonometry: Right-angled triangles Quesitons

November 22, 2010 Leave a comment

For each set of questions find the value of \mbox{w, x, y and z}

Make sure to write down all the working.  It is a good habit to be rigorous in your answers.

Set 1


Set 2

 


Set 3


Set 4


Set 5

Categories: Triangles

Trigonometry and Right-angled triangles

November 21, 2010 Leave a comment

Conceptual Understanding

If we find the ratio between a pair of sides on a triangle the value will be a constant even on larger triangles provided the triangles are similar to each other.

The right-angled trigonometry triangle

The hypotenuse is the longest side and is opposite the right-angle

The opposite is the side that is opposite the given angle. (Or the angle we wish to find)

The adjacent is the side next to the given angle.  (It is between the given angle and the right angle)

The ratios SohCahToa

 

\displaystyle sin \theta = \frac{opp}{hyp}

\displaystyle cos \theta = \frac{adj}{hyp}

\displaystyle tan \theta = \frac{opp}{adj}

When solving a trigonometry right-angled triangle problem we need two pieces of information and then use the relevant ratio to find the third.

There are three possible scenarios.

  1. We are finding an angle
  2. We are finding a side that is a numerator in the ratio
  3. We are finding a side that is a denominator in the ratio

Examples

Type 1 (The unknown is the numerator)

We have the hypotenuse and we want the opposite side.  The only ratio with hypotenuse and opposite in it is the sine ratio.

\displaystyle sin\theta = \frac{opp}{hyp}

Let’s fill out the equation

\displaystyle sin 35^o=\frac{x}{17}

\displaystyle \implies 0.374=\frac{x}{17}

\implies x=9.75cm


Type 2 (The unknown is the denominator) 

We have the opposite and we want the adjacent side.  The only ratio with adjacent and opposite in it is the tangent ratio.

\displaystyle tan \theta = \frac{opp}{adj}

Let’s fill out the equation

\displaystyle tan 27^o=\frac{28}{x}

\implies x\times tan 27^o=28

\displaystyle \implies x=\frac{28}{tan 27^o}

\therefore x =54.95mm


Type 3 (The angle is unknown) 

We have the hypotenuse and adjacent side.  The only ratio with adjacent and hypotenuse in it is the cosine ratio.

\displaystyle cos \theta = \frac{adj}{hyp}

Let’s fill out the equation

\displaystyle cos x=\frac{15}{32}

\implies \displaystyle x=cos^{-1}\left(\frac{15}{32}\right)

\implies x =62.0^o

Questions

Categories: Triangles