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Mathematics: Post your doubts here!

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wow, its been ten days and no ones answered my question

i just found out the answer myself

thankyou xtremepapers....for nothing :(
 
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Find out the derivative of y using product rule....!! then equate derivate to zero..which gives u various x coordinates.....so..write down those coordinates which lie between that interval to 3dp..!!
 
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kamina1 said:
i hve found the derivative but it is not further simplifying...
dy/dx= e^-3x { sec^2x -3tanx } i hope m not wrong...then..u cannot get any values from e^-3x.....now...take secx and tanx in sinx and cosx..u will get it..!! otherwise..i will give u the solution after an hour..!! m goin out right now..!
 
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The parametric equations of a curve are
x = a cos3 t, y = a sin3 t,

(i) Express
dy
dx
in terms of t. [3]
(ii) Show that the equation of the tangent to the curve at the point with parameter t is
x sin t + y cos t = a sin t cos t. [3]
(iii) Hence show that, if this tangent meets the x-axis at X and the y-axis at Y, then the length of XY
is always equal to a. [2]

thanx.. its june2009, Question6
 
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NehaKush said:
The parametric equations of a curve are
x = a cos3 t, y = a sin3 t,

(i) Express
dy
dx
in terms of t. [3]
(ii) Show that the equation of the tangent to the curve at the point with parameter t is
x sin t + y cos t = a sin t cos t. [3]
(iii) Hence show that, if this tangent meets the x-axis at X and the y-axis at Y, then the length of XY
is always equal to a. [2]

thanx.. its june2009, Question6

(i). d(y)/d(t)=3asin^2t(cost)
d(x)/d(t)=-3a sint cos^2t
d(Y)/d(x)=3a sin^2t cost/-3a sint cos^2t
d(y)/d(x)=-tant

(ii).gradient=-tant , x=acos^3t, y=asin^3t
y-sin^3t= -tant(x-acos^3T)
Y-SIN^3T=-(sint/cost)x-acos^3t
ycost-asin^3tcost=-xsint+asintcos^3t rearrange it
xsint +ycost=asintcost(cos^2t+sin^2t)
xsint+ycost= asintcost

(iii).for y-intercept put y=0,
xsint+o= asintcost
x=acost
for y-intercept put x=0
0+ycost=asintcost
y=asint
coordinates of x are (acost,0)
coordinate of y are (0, asint)
lenght=((acost)^2+(asint)^2)^1/2
=(a^2cost^2+a^2sint^2)^1/2
=(a^2(cos^2t+sin^2t))^1/2
=(a^2)^1/2
=a
 
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y
P


In the diagram the tangent to a curve at a general point P with coordinates (x, y) meets the x-axis at T.
The point N on the x-axis is such that PN is perpendicular to the x-axis. The curve is such that, for all
values of x.
, the area of triangle PTN is equal to tan x, where x is in radians.
(i) Using the fact that the gradient of the curve at P is
PN
TN
, show that
dy
dx
= 1
2y2 cot x. [3]
(ii) Given that y = 2 when x = 1
6π, solve this differential equation to find the equation of the curve,
expressing y in terms of x.

there is a diagram for this.. i cant post it.. its june2008 question8
 
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Please solve thsi question as early as possible
 

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Lol because the question said ALL the Maths books will not be together. That is, 2 Maths books can be together, as well as 3 but ALL 4 of them should not be together. In ur first method, u separated them all dear-that would have been the answer if the question was 'no 2 maths book are together' or 'all Maths books are separate from each other'. U totally missed the possibility of one E, two M, one E, etc... U understand what I'm trying to say? :)
 
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please can you solve one more problem iam aving doubt getting wrong answer if the letter of the word probability are arranged at random, find the probability that the two i,s are separated
 
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9!/2! * 10P2/2! = 181485
total number of arrangements= 11!/ (2!*2!) = 9979200
Probability= 181485/9979200
=0.0182
 
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june 2011 paper 31 please help??
 

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IgoforA said:
june 2011 paper 31 please help??
for 8
i) do a realisation..meaning..multiply u by 6-3i/1-2i and get a form...!! and find modulus and argument
ii) that equation is a straight line starting from u and making an angle of 45 with a horizontal.....and...z is any pt on that line....so.....find the distance from (0,0) to a pt in that line..which is the least...i hope u can find which would b the least one...
iii) it is the circle with centre at (1+i)u with radius 1....so....z is any pt on the circumference of the circle...so find the farthest pt from (0,0) ......and find a distance...

for 9...
put Cos4A as Cos(3A+A)=Cos3A.CosA-Sin3A.SinA and...put the formula SIn3A=3SinA-4Sin^3A...and put the formula of Cos3A=4Cos^3A-3CosA ......put the formula of Cos2A=2Cos^2A-1 ( its difficult to write theta..so use A instead)....

I HOPE that helped..!!
 
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