CalculusThis chapter has to do with rates of change, slope= speed = dy =f`(x)= differentiate dx

Revision of Graphs 1 st

Drawing a Graph

When given a function to draw do out a table with x’s down on the left and sub into the graph and get y values.

E.g. Draw the graph of f(x) = 3x2 – 2x -7from -2 ≤ x ≤ 3, where x is an element of R!

Revision of Functions and Graphs (continued)

Task

- Find on the graph:-

1. f (2.5) => draw a line as from x = 2.5=> y = 6.5

2. f(x)= 3 => draw a line as from y = 3=> y = 3x = -1.6 x = 2.2

3. Find the min value of the graphy = -7 at x = 0

Look at the min point on the graph!

4. Where is f(x) < 0?-1.2 ≤ x ≤ 2

5. Where is the graph increasing?0 ≤ x ≤ 3

Now onto Differenciation

Simple differentiation

Rule: Bring down the power in front of the letter and reduce the power by 1.

y = x2-2x

dy =2x1-2(1)x0 x0 any value to power of 0 is 1 dxdy = 2x-2dx

Example

f(x) = 3x3 -4x2+5x -7 the constant 7 cannot be differentiated so ignore

f`(x) = 3(3)x2-2(4)x1+5 any term with x on its own is just the number with the x

f`(x) =9x2 – 8x+5 tidy up

where is the graph below zero?

look at the graph and see where it increases!

The Slope of a tangent to a curve

Slope = f`(x) = dy dx

Find the slope and equation of the tangent line to the curve y= 2x2-3x+4 at (1,3)

Y = 2x2-3x+4

dy = 4x-3 this is the slope as it’s the differentiationdx

You know the x value to be x=1 so sub in to get value for the slope

Slope = 4(1)-3 = 4-3Slope/m = 1

Now I have the slope and the point so I can find the equation of the line.m=1 point (1,3) sub into equation of the line

y-y1=m(x-x1)y-3 =1(x-1)y-3 = x-1 Bring everything to the right to make it look like equation and tidy0 = x-y+3-10 = x-y+2

To work backwards

Find the point on the curve y = x2-2x+11 where slope is 6?

This time you are given slope and have to get point.

dy = 2x-2 slope =6dx

6=2x-2 6-2=2x

8=2x

4=x replace x into top equation to get y

Y=(4)2-2(4)+11

Y=16-8+11

Y=19 (4,19) is the point at which the slope is 6

Max and Min turning Points

1.Find dy dx

2. Let dy = 0 dx

3. Solve to find x

4. When you get x sub back in to find y co-ordinate

5. You will get 2 points state which is max and min (point with greater y will be the max)

6. If asked to prove differentiate twice and if you sub in x and get positive value it’s the min and if you get negative value it`s the max.

7. If only one turning point of a quadratic ax2+bx+c (a>0 is min, a<0 is max)or(+x2= min,-x2=max)

Example

y = x3-3x2-4

dy =3x2-6x Step 1 differentiatedx

0 = 3x2-6x Step 2 let equal 0

0 = 3x(x-2) Factorising using HCF

0=3X 0 =X-2 Let both terms equal 0

0 = x 2 = x Solve for x

Let x=0 find its y by subbing back into the function let x=2 find its y by subbing back into the function

y = x3-3x2-4 y = x3-3x2-4

y=(0)3-3(0)2-4 y = (2)3-3(2)2-4

y = -4 y = -8

(0,-4) max (2,-8) min

To prove get 2nd differential of both

dy =3x2-6x dy =3x2-6xdx dx

d 2 x = 6x -6 d 2 x = 6x -6dy2 dy2

x=0 x=2 =6(0)-6 =6(2)-6 =-6 negative so max =6 positive so min

Example:Find the turning point and state if its max or min?

y = 2x2-6x+1 Sub x in to find y =2(3)2-6(3)+1dy = 4x-6 2 2dx

y = -70 = 4x-6 26 = 4x

( 3 ,-7) is point and a min as +x2 so u shaped graph6 = x 2 243 = x2

To prove get 2nd differential

dy =4x-6 d 2 y = 4 positive so a mindx dx2

Speed and Acceleration

ds = speed d 2 s = Accelerationdt dt2

s = distance, t = time, h = height

if s=3-6t+2t3

1. How far does it travel in 2 seconds?S = 3-6(2)+2(2)3

S = 3-12+16S = 7metres

2. Find the speed after 2 seconds?ds = speeddt

= -6+6t2 after t=2 = -6(2)+6(2)2

= -12+24 = 12m/s is the speed

3. Find the acceleration after t = 3secs

d 2 s = 12tdt2

= 12(3)=36m/s2

4. What time does the body come to rest/stops? When speed =0 a body is at rest

ds = 0 -6+6t2 = 0dt +6t2 =6

t2 = 1t= 1second

Same is done with heightExample:

If h =33+8t-t2

(i) Find speed after 2 seconds?dh = 8-2t after t =2dt

=8-2(2)=8-4=4m/s

(ii) When did the ball stop?dh= 0 speed =0dt

0=8-2t2t=8t=4seconds

(iii) What is the max height reached by the ball?When speed =0 that’s the max height when t= 4 seconds.

H=33+8t-t2

H=33+8(4)-(4)2

H=33+32-16H=49m

Example:

Using Graphs to figure out slopes

1.Look at max and min these are where the slope graph cuts the x-axis2. If the graph is decreasing between these points the slope graph is below xaxis if increasing its above xaxis

Example :In the picture below you can see the graph y=-2x+3 and slope graph y=-2

Example:

Example:Here the quadratic graph –x2+3x+4 the slope graph is y=-2x+3You can see max point is 1.5 and the slope graph cuts through x-axis at 1.5.The quadratic decreases from 1.5 and so does the slope graph under axis