· web viewdx revision of graphs 1 st drawing a graph when given a function to draw do out a table...
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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