Mathematics: Post your doubts here!

[daredevil]

from (i) u found dy/dx to be "-tant"
you can use (y - y1) = m(x - x1) to find the equation of the tangent
where y1 = a(sin^3)t , x1 = a(cos^3)t and m = -tant = -sint/cost

so,
y - a(sin^3)t = -sint/cost(x - a(cos^3)t) [multiply by cost]
ycost - acost(sin^3)t = -xsint + asint(cos^3)t
xsint + ycost = asint(cos^3)t + acost(sin^3)t
xsint + ycost = asintcost (cos^2t + sin^2t) [cos^2t + sin^2t =1]
xsint + ycost = asintcost <==equation of tangent

hope u understood

OOoo thankkuu soo much!! I was totally mixing up normal and tangent and I HAVE No IDEA how i was doing that xD thankss i get it now

 

[m7aug]

please help me in question visible above (part ii) it is paper 1 from 12 n/o variant 13

 

[daredevil]

View attachment 33547
please help me in question visible above (part ii) it is paper 1 from 12 n/o variant 13

find the stationary point from the double derivative. use equation to find the remaining coordinate of the stationary point.
the stationary point with a negative double derivative is the maximum point. one stationary point has y coordinate 0 so the other one will be at b. i hope u get it. consult the marking scheme along with the explanation. maybe u'll get it. if not then tell me and i'll elaborate further

 

[tanady9]

View attachment 33547
please help me in question visible above (part ii) it is paper 1 from 12 n/o variant 13

You would have noticed that where the x coordinate is b, there is the maximum point of the curve. You must know that where there is maximum or minimum point on the curve,that is it's stationary point. Meaning that the gradient at that point is zero. As we know that dy/dx of an equation is its gradient at any point, differentiate the equation, substitute b in place of x and equate this equation to zero. Then you will get the value of b. Hope i helped

 

[daredevil]

You would have noticed that where the x coordinate is b, there is the maximum point of the curve. You must know that where there is maximum or minimum point on the curve,that is it's stationary point. Meaning that the gradient at that point is zero. As we know that dy/dx of an equation is its gradient at any point, differentiate the equation, substitute b in place of x and equate this equation to zero. Then you will get the value of b. Hope i helped

View attachment 33547
please help me in question visible above (part ii) it is paper 1 from 12 n/o variant 13

that was actually more helpful than what i jibber jabbered about

 

[salvatore]

5 ii)
here they told us about the gradient, that means we need to differentiate that equation.

6e^2x + ke^y + e^2y = c
ke^y + e^2y = c - 6e^2x (we differentiate each term with respect to x)
ke^y(dy/dx) + 2e^2y(dy/dx) = -12e^2x
dy/dx(ke^y + 2e^2y) = -12e^2x
hence dy/dx = -12e^2x/(ke^y + 2e^2y)

you're given that at point P the gradient is -6 and that the point P has coordinates (ln3 , ln2)
substitute these into the expression for gradient ( dy/dx = -12e^2x/(ke^y + 2e^2y) ) and u'll find the value of 'k'
use this value and the equation from 5(i) to find 'c'
in the end u should get 'k = 5' and 'c = 68'

hope you've understood

I've understood it well..
Thanks a lot for your help

 

[daredevil]

plz can anyonesolve these for me!!! thankss

 

[daredevil]

View attachment 33549

plz can anyonesolve these for me!!! thankss

 

[biscuitbiscuit]

How to find the coefficients of x and x^2 in the expansion of (4-x)(2-4x)^6

 

[yousef]

View attachment 33549

plz can anyonesolve these for me!!! thankss

hope this helps bro >>>

 

[yousef]

How to find the coefficients of x and x^2 in the expansion of (4-x)(2-4x)^6

using the binomial expansion formulae >> in your textbook .... say your expansion is 1 + 2ax + 6bx^2 >> here the cofficient of x is ( 2a ) ... ie all numbers with x ...
and coffiecient of x^2 is ( 6b )

 

[daredevil]

hope this helps bro >>>

thankkss alot totallyyy get it man awesome!! (Y)
except the ii
see here they have inserted (pie)/4 in the markking scheme:....

 

[biscuitbiscuit]

using the binomial expansion formulae >> in your textbook .... say your expansion is 1 + 2ax + 6bx^2 >> here the cofficient of x is ( 2a ) ... ie all numbers with x ...
and coffiecient of x^2 is ( 6b )

thanks

 

[biscuitbiscuit]

How to do 6 (ii), http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_12.pdf and what does 'no term in x in the expansion....' means

 

[yousef]

thankkss alot totallyyy get it man awesome!! (Y)
except the ii
see here they have inserted (pie)/4 in the markking scheme:....

OMG!!


To explain this , if you integrate cos(ax) form .. it will be 1/a sin(ax)... same with integrating sin(ax) which is -1/a cos (ax ) since integral of sinx is -cosx ....

Now when it comes to limits .. we got the equation as 1/2 sin2(theta) .... right ?
so when substituting the limits .. in this ..it should be in theta not as x ...as its given in equation as 1 and o ..
so we know that x=tan(theta) ... substituting both gives :
1 =tan(theta) & 0=tan(theta) .. so you get theta as tan-1( x) .. which is 0 and 45 degree .. and we know that 45 degree in radians is 1/4pi ..

 

[yousef]

How to do 6 (ii), http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_12.pdf and what does 'no term in x in the expansion....' means

when you expand ... add all cofficients of x ... and equate with zero to find a

 

[periyasamy]

Guys need a help here.

 

[daredevil]

OMG!!


To explain this , if you integrate cos(ax) form .. it will be 1/a sin(ax)... same with integrating sin(ax) which is -1/a cos (ax ) since integral of sinx is -cosx ....

Now when it comes to limits .. we got the equation as 1/2 sin2(theta) .... right ?
so when substituting the limits .. in this ..it should be in theta not as x ...as its given in equation as 1 and o ..
so we know that x=tan(theta) ... substituting both gives :
1 =tan(theta) & 0=tan(theta) .. so you get theta as tan-1( x) .. which is 0 and 45 degree .. and we know that 45 degree in radians is 1/4pi ..

Ohh thank you!! :')

 

[daredevil]

Guys need a help here.

m contains A and B means the vector AB is part of the plane m.
find the vector AB by : OB - OA
and if m is parellel to OC then vector of m should be a multiple of vector OC.
see if you get it... otherwise i'll try to explain further. sorry can't solve it right now.

 

[biscuitbiscuit]

t

when you expand ... add all cofficients of x ... and equate with zero to find a

thanks