1,151
27
Summary, 5 pages (1300 words)

C4 summary sheet

simplify algebraic fractions by factorising and cancelling factorse. g. (3x+6)/(x^2-4) = 3(x+2)/(x+2)(x-2) = 3/(x-2)add and subtract by finding a common denominatore. g. simplify 2y/x(x+3) + 1/y^2(x+3) +x/y
– find the common denominator (take all of the individual bits on the bottom row and multiply them together) = xy^2(x+3)
-make each of the fractions have the common denominator at the bottom by multiplying the whole fraction by different individual bits
= 2y^3/y^2x(x+3) + x/y^2x(x+3) + x^2y(x+3)/y^2x(x+3)
-combine into one fraction
= 2y^3+x-x^3y-3x^2y/xy^2(x+3)algebraic divisiondegree – the highest power of x in the polynomial (e. g. the degree of 4x^5+6x^2-3x-1 is 5)
divisor – this is the thing you’re dividing by (e. g. if you divide x^2+4x-3 by x+2, the divisor is x+2)
quotient – the but you get when you divide by the divisor (not including the remainder)f(x)= q(x)d(x)+r(x) – way of writing a polynomialq(x)= quotient
d(x)= divisor
r(x)= remainderpartial fractions= opposite of adding fractions
if the fraction is over two different brackets then the partial fraction will be in the form A/bracket 1 +B/bracket 2
if the fraction is over three different brackets then the partial fraction will be in the form A/brackt1 +B/bracket2 +C/bracket3
if the fraction is over one bracket squared and another bracket then the partial fraction will be in the form A/bracket 1 squared + B/bracket 1 +C/bracket 2difference of two squares denominatorse. g. 21x-2/9x^2-4 = 21x-2/(3x-2)(3x+2) = A/(3x-2) + B/(x+2)parametric equationssplit up x and y into separate equations
cartesian equation = single equation linking x and y
parametric = x and y are defined separately with a third variable called a parameter usually t or thetause parametric equations to find where graphs intersectintersect the x axis , y= 0
intersect the y axis , x= 0
intersect the line ay= bx +c , plug the parametric equations into the values of y and x and then rearrange and find the values of the parameter which you then plug into the parametric equations to find the corresponding x and y valuesconverting parametric equations into cartesian equations`1- rearrange one of the equations to make the parameter the subject, then substitute the result into the other equation
or
2-if your equations involve trig functions , use trig identities to eliminate the parameterexamples1- e. g. x= t+1 y= t^2-1
t= x-1
y=(x-1)^2-1
y= x^2-2x

2- e. g. x= 1+sin(x) , y= 1-cos(2x)
use trig function cos(2x)= 1-2sin^2(x)
y= 1-cos(2x)= 1-(1-2sin2(x))= 2sin^2(X)
sin(x)= x-1 so y= 2sin^2(x)
y= 2(x-1)^2 = 2x^2-4x+2

binomial expansionformula in the book
used for when the power is a fraction or is negative
if the constant isn’t 1 then you need to factorise it first ( e. g. (3-x)^4 will become (3(1-x/3))^4 which becomes 81(1-1/3x)^4validityif the power is positive , the expansion is valid for all values of x
if the power is negative or a fraction, the expansion of (p+qx)^n is only valid when | qx/p|<1e. g. 1(1+x)^-1 is valid or when | x|<1e. g. 2 (1+2x)^1/3 is valid if | 2x|<1 > y= cos(2x) becomes y= cos(u) , u= 2x
dy/du= -sin(u) du/dx= 2 so dy/dx = -2sin(2x)
y= sin(x+1) becomes y= sin(u), u= x+1
dy/du = cos(u) du/dx= 1 so dy/dx = cos(x+1)

therefore dy/dx = cos(x+1)-2sin(2x)

e. g. 2 find dy/dx when x= tan(3y)
x= tan(u), u= 3y so dx/du = sec^2(u) du/dy = 3
dx/dy = 3sec^2(3y) so dy/dx = 1/3sec^2(3x) = 1/3cos^2(3y)

dy/dx of inverse reciprocal functionscome from the quotient rule differentiation of 1/sin(x), 1/cos(x) and 1/tan(x)dy/dx of cosec(x)=-cosec(x)cot(x)dy/dx of sec(x)= sec(x)tan(x)dy/dx of cot(x)=-cosec^2(x)use the chain, product and quotient rules with cosec, sec and cote. g. 1 find dy/dx of y= sec(2x^2)
= CHAIN RULE as it is a product of a product
so y= sec(u) u= 2x^2
dy/du= sec(u)tan(u) du/dx = 4x
therefore dy/dx = 4xsec(2x^2)tan(2x^2)

e. g. 2 find dy/dx of y= e^x(cot(x))
= PRODUCT RULE as it is a product of two functions
u= e^x v= cot(x)
du/dx = e^x dv/dx =-cosec^2(x)
dy/dx = u(dv/dx)+v(du/dx)
= e^x(cot(x)-cosec^2(x))

differentiating parametric equationsdy/dx= dy/dt / dx/dt
can use this to find the gradient of tangents and normal (neg reciprocal of tangent)implicit differentiationimplicit relation= an equation thats in the form f(x, y)= g(x, y) rather then y= f(x)steps for implicit differentiation1. differentiate the x^n values with respect to x as usual
2. differentiation the y^n values by d/dy so the result is d/dy of f(y) multiplies by dy/dx
3. use the product rule to do ones with x and y
4. rearrange the equation to make dy/dx the subject

can use this to find the gradient

intergration of sin(x)= -cos)x + cintegration of cos(x)= sin(x) + chow to integrate trig functionsfollow the rule given
if the x of the starting equation’s x has a coefficient that isn’t 1, you just divide by the coeffiecnt when you integratesome fractions integrate to lnS f^-1(x)/f(x) dx = ln| f(x)| + cthis works for some trig functions tootan(x) = sin(x)/cos(x)
dy/dx (cos(x)) =-sin(x) therefore S sin(x)/cos(x) = -ln| cos(x)| +c

S cot(x) dx = ln | sin(x)| +c
S cosec(x) dx = -ln | cosec(x) + cot(x)| +c
S sec(x) dx = ln | sec(x) + tan(x)|

chain rule in reversefind S 6x^5(e^x^6) dx
if you differentiate the bracket you would get 6x^5(e^x^6) therefore as only differentiating the bracket gets this answer so the answer is the bracket
= e^x^6 +cintegration by substitutionon products of two functions
1. find du/dx and rewrite it so dx is on its on
2. rewrite the original integral in terms of u and du
3. integrate as normal
for definite integrals you have to change the limitsintegration by partsS u(dv/dx) dx = uv -S v(du/dx)integrate ln(x)rewrite at 1xln(x)
u= ln(x) dv/dx= 1
du/dx= 1/x v= x
therefore S ln(x) = xln(x) -Sx(1/x) dx = xln(x) -S 1 dx = xln(x) -x +cintegrating partial fractionssplit into partial fractions
when you integrate each fractions they are all going to be Aln(bracket1) +Bln(bracket2)… +c
which you can rewrite as ln((bracket1)^A(bracket2)^B) +cdifferential equations- solving by integrationwrite the differential equation in the form dy/dx = f(x)g(y)
rewrite the equation in the form 1/g(y) dy = f(x) dx
integrate both sides
rearrange into a nice form and remember to +cstarting conditions occur whent= 0decreasing =-k ( rate of decrease is proportional to number of…)vectorshave a size and a direction
scalars are just quantities
length of the vector is the magnitudea, 2a, 3aparallel as they are just multiplied by a scalarposition vectorswhere the point liesa unit vector isany vector with a magnitude of 1 uniti and j unitsi is the x axis
j is the y axis
(k is the z axis)column vectorsway of writing a vectorpythagoras’ theoremfind vector magnitudes
the magnitude of a vector is written | a| the distance of a point from the originsquare root of (x^2 + y^2 + z^2)distance between pointssquare root of (x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2equation of a line through a point and parallel to another vectora straight line which goes through point A and is parallel to vector B has the equation
r= A+tB
where r= position vector of a point on the line
A= position vector of point Aequation of a line passing through two pointsr= c+t(d-c)point of intersection of two linesif two lines intersect then there will be a value of t for each equation which makes the same r
you know if they are parallel if they are multiples of each other
if there is no point of intersection then they are skewscalar producta. b = | a|| b| cos(x)
used to calculate the angle between two lines
rearranges to cos(x)= a. b/| a|| b|
two lines that are perpendicular , the angle will be 90 therefore cos(90) = 0
if they are parallel it would be cos(0) so a. b = | a|| b| a. ba1b1+a2b2+a3b3using the scalar product to find the angle between two vectorse. g. find the angle between the vectors -i-6j and 4i+2j+8k
a=-i-6j b= 4i+2j+8k
scalar product of the vectors
a. b = (-1×4) + (-6×2) + (0x8) = -4-12+0=-16
magnitude = | a| = root 37 | b| = root 84
cos(x)=-16/root 37 x root84 so x= 106. 7 degresif you are finding the angle between two linesyou would use the b bit from r= a+tbproving lines are perpendicularscalar product = 0 ONC4 SUMMARY SHEET SPECIFICALLY FOR YOUFOR ONLY$13. 90/PAGEOrder Now

Thank's for Your Vote!
C4 summary sheet. Page 1
C4 summary sheet. Page 2
C4 summary sheet. Page 3
C4 summary sheet. Page 4
C4 summary sheet. Page 5
C4 summary sheet. Page 6
C4 summary sheet. Page 7
C4 summary sheet. Page 8

This work, titled "C4 summary sheet" was written and willingly shared by a fellow student. This sample can be utilized as a research and reference resource to aid in the writing of your own work. Any use of the work that does not include an appropriate citation is banned.

If you are the owner of this work and don’t want it to be published on AssignBuster, request its removal.

Request Removal
Cite this Summary

References

AssignBuster. (2022) 'C4 summary sheet'. 2 October.

Reference

AssignBuster. (2022, October 2). C4 summary sheet. Retrieved from https://assignbuster.com/c4-summary-sheet/

References

AssignBuster. 2022. "C4 summary sheet." October 2, 2022. https://assignbuster.com/c4-summary-sheet/.

1. AssignBuster. "C4 summary sheet." October 2, 2022. https://assignbuster.com/c4-summary-sheet/.


Bibliography


AssignBuster. "C4 summary sheet." October 2, 2022. https://assignbuster.com/c4-summary-sheet/.

Work Cited

"C4 summary sheet." AssignBuster, 2 Oct. 2022, assignbuster.com/c4-summary-sheet/.

Get in Touch

Please, let us know if you have any ideas on improving C4 summary sheet, or our service. We will be happy to hear what you think: [email protected]