C4 summary sheet

e.

g. (3x+6)/(x^2-4) = 3(x+2)/(x+2)(x-2) = 3/(x-2)

e.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)

degree - 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)

q(x)=quotient d(x)=divisorr(x)=remainder

= 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 2if the fraction is over three different brackets then the partial fraction will be in the form A/brackt1 +B/bracket2 +C/bracket3if 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 2

e.g. 21x-2/9x^2-4 = 21x-2/(3x-2)(3x+2) = A/(3x-2) + B/(x+2)

split 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 theta

intersect the x axis , y=0intersect the y axis , x=0intersect 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 values

`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 parameter

1- e.g.

x=t+1 y=t^2-1 t=x-1y=(x-1)^2-1y=x^2-2x2- 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

formula 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)^4

if 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 = 2|x|<1 = |x|<1/2

= cos(x)

= -sin(x)

= sec^2(x)

only work in radians

e.g1 differentiate y= cos(2x) + sin(x+1)y= cos(2x) becomes y=cos(u) , u=2xdy/du= -sin(u) du/dx= 2 so dy/dx = -2sin(2x)y=sin(x+1) becomes y=sin(u), u=x+1dy/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 =3dx/dy =3sec^2(3y) so dy/dx = 1/3sec^2(3x) = 1/3cos^2(3y)

come from the quotient rule differentiation of 1/sin(x), 1/cos(x) and 1/tan(x)

=-cosec(x)cot(x)

= sec(x)tan(x)

=-cosec^2(x)

e.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^2dy/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))

dy/dx= dy/dt / dx/dtcan use this to find the gradient of tangents and normal (neg reciprocal of tangent)

implicit relation= an equation thats in the form f(x,y)=g(x,y) rather then y=f(x)

1. 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

= -cos)x + c

= sin(x) + c

follow 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 integrate

S f^-1(x)/f(x) dx = ln|f(x)| + c

tan(x) = sin(x)/cos(x) dy/dx (cos(x)) =-sin(x) therefore S sin(x)/cos(x) = -ln|cos(x)| +cS cot(x) dx = ln |sin(x)| +cS cosec(x) dx = -ln |cosec(x) + cot(x)| +cS sec(x) dx = ln |sec(x) + tan(x)|

find S 6x^5(e^x^6) dxif 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 +c

on products of two functions1.find du/dx and rewrite it so dx is on its on 2.rewrite the original integral in terms of u and du3.

integrate as normalfor definite integrals you have to change the limits

S u(dv/dx) dx = uv -S v(du/dx)

rewrite at 1xln(x)u=ln(x) dv/dx=1du/dx=1/x v=xtherefore S ln(x) = xln(x) -Sx(1/x) dx = xln(x) -S 1 dx =xln(x) -x +c

split into partial fractionswhen you integrate each fractions they are all going to be Aln(bracket1) +Bln(bracket2)... +cwhich you can rewrite as ln((bracket1)^A(bracket2)^B) +c

write the differential equation in the form dy/dx = f(x)g(y)rewrite the equation in the form 1/g(y) dy =f(x) dxintegrate both sides rearrange into a nice form and remember to +c

t=0

-k ( rate of decrease is proportional to number of..

.)

have a size and a direction scalars are just quantitieslength of the vector is the magnitude

parallel as they are just multiplied by a scalar

where the point lies

any vector with a magnitude of 1 unit

i is the x axis j is the y axis (k is the z axis)

way of writing a vector

find vector magnitudes the magnitude of a vector is written |a|

square root of (x^2 + y^2 + z^2)

square root of (x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2

a straight line which goes through point A and is parallel to vector B has the equationr=A+tBwhere r= position vector of a point on the line A=position vector of point A

r=c+t(d-c)

if 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 skew

a.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) =0if they are parallel it would be cos(0) so a.

b = |a||b|

a1b1+a2b2+a3b3

e.g. find the angle between the vectors -i-6j and 4i+2j+8ka=-i-6j b=4i+2j+8kscalar product of the vectorsa.b = (-1x4) + (-6x2) + (0x8) = -4-12+0=-16magnitude = |a| = root 37 |b| = root 84cos(x)=-16/root 37 x root84 so x=106.7 degres

you would use the b bit from r=a+tb

scalar product =0