Calculus+topic+page

EXCELLENCE QUESTIONS???? Click [|here] for some worked examples (some beyond Level 2)

Gradient of a curve? Download the powerpoint demonstrations to see them again. 1. gradient of a curve - [|Gradient at a point.ppt] 2. Topic overview (bit of everything) -[|Gradient function.ppt]

media type="youtube" key="rAof9Ld5sOg" height="344" width="425" =Calculus Notes=

Firstly, you need to know some notation! f(x) means "function of x". It is another name for "y". That means f(4) is "function of 4" - in other words, the y value you get from an x value of 4.

Try and keep your notation consistent.

If you call your original function y, call its derivative and call the second derivative If you call your original function f(x), call its derivative f ' (x) and call the second derivative f '' (x)

What is a derivative?
A derivative is a rule for all the gradients on a graph. Linear graphs like y = 2x + 5 only have one gradient, so their derivative is a single number. More complex curves have a lot of gradients! e.g. To get a derivative you need to do two things: multiply each term by its exponent and then reduce the exponents by one. Because any constant (lone number) terms are x to the power of zero, they disappear when you differentiate.

e.g.

Questions involving derivatives
The most likely questions involving derivatives are: e.g. y = x2 + 5x. What is the gradient at the point (2, 14)? The derivative is 2x + 5. Use the point's x value, 2, to get the gradient. 2x 5 + 5 = 15. At (2, 14) the gradient is 15.
 * Find the gradient for a certain point.** To do this, differentiate to get the gradient rule. Substitute in the x value of the point.

e.g. where does the function f(x) = x2 - 3x + 4 have the gradient 2? The derivative is 2x - 3. We need to solve the equation 2x - 3 = 2. This gives an x value of 2.5 Substitute the x value 2.5 into f(x). This gives the y value 2.75 Answer: the function f(x) = x2 - 3x + 4 has the gradient 2 at the point (2.5, 2.75)
 * Find where a function has a particular gradient.** To do this, differentiate to get the gradient rule. Turn the rule into an equation, with the desired gradient as its answer. Solve the equation to get an x value. Get a y value by substituting into the original function.

e.g. where is the maximum point of the graph of y = -x2 + 4x - 6? The derivative is -2x + 4. Find the solution to -2x + 4 = 0. Answer: x = 2 To find the maximum point, we need the y value also. Substitute the x value of 2 into the original rule. This gives us the point (2, -2)
 * Finding minimums and maximums.** Same as above - but set the gradient function to equal zero.

Merit/Excellence applications:
This often involves There are 3 ways to do this
 * finding turning points of cubic functions.** (when you differentiate, you still have a quadratic. This means you will need to factorise or use the quadratic formula to find 2 x values).
 * Working out whether turning points are minimums or maximums**
 * know the graph shape (positive cubics have a maximum first and a minimum second)
 * Find a point on either side. If they have a lower y value, the turning point must be a maximum.
 * Find the second derivative i.e. differentiate again. If you get a negative answer, it's a maximum. If positive, its a minimum.
 * Optimisation** - this is another way of saying find a maximum. Often it involves a word question where you write the equation for yourself.
 * Kinematics**. This involves thinking of gradient as a rate. The gradient of a distance/time equation is a speed. The gradient of a speed/time graph is an acceleration. Confusingly, the notation is s for distance, v for speed (velocity) and a for acceleration.

e.g. find the acceleration of a particle after 3 seconds that has this movement: d = 5t3 - 2t2 + 4 where d is in m and t is in seconds. Differentiate to get the speed rule: v = 15t2 - 4t Then differentiate to get acceleration a = 30t. After 3 seconds, acceleration is 30x3 = 90m/s

=Integration =

Differentiation and integration are opposites. Therefore, integration is the opposite process:
 * increase the powers by one.
 * divide each term by its **new** power.

Why do indefinite integrals end with + c?
f(x) = 3x + 4 f(x) = 3x + 5 f(x) = 3x - 103 ... all of these have a gradient of 3. If you are asked to integrate the number 3, the answer is 3x + c because any constant (number) could have been in the original function. c is called the constant of integration. It stands for a number that MAY have been lost through differentiation.

What's this dx thing? [[image:f4066.gif]]
You don't have to worry about it until Year 13. It means that x is the variable we are concentrating on, that x is the variable whose exponent will be increased.

(the answer would be )
 * In Year 13,** you might get a question such as [[image:integrals.gif width="8" height="19"]] xyz dy. This means that out of the 3 variables, the one that is affected is y.

This year - just ignore it!

Definite Integrals
For definite integrals, two x value limits are given.
 * integrate the expression
 * substitute each x value (one at a time) to give 2 answers
 * Subtract the two answers.

Questions involving Integration:
At achieved level, the most likely questions are: Here, the shaded area is the same as integrating the function with the limits b and a
 * Calculate an area**. A simple area is just a definite integral


 * Calculate an original function, given a derivative and a point.**

This means, not only do you integrate, you also use some values to work out what c is. e.g.

=Merit/Excellence Applications=

These are usually
 * (a) Kinematics or (b) more complex areas**

e.g.
 * Kinematics** - differentiation in reverse. Integrate acceleration to get speed. Integrate speed to get distance. Each time, the constant of integration must be calculated from the information given.

**Complex areas - see this [|website] for good Excellence examples**
Areas above the x axis appear as positive values. Areas below the x axis appear as negative values. If you want an area that goes above and below, you must break it into separate parts.

In this picture, integrating from a to c would give a wrong answer. The green area would be subtracted from the yellow. To get a total area, integrate from a to b. Integrate from b to c. Make both answers positive, then add them.

Excellence area questions
Excellece questions often involve finding the area BETWEEN two graphs. This involves finding an area from under each graph and then subtracting. What makes it difficult is that you are usually not told the exact intersection points of the graphs. You have to calculate them using simultaneous equations. You have examples of all area question types in your workbook (p161-163)


 * GOOD LUCK**