We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
from (i) u found dy/dx to be "-tant"
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 nowfrom (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
find the stationary point from the double derivative. use equation to find the remaining coordinate of the stationary point.
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
I've understood it well..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
hope this helps bro >>>
thankkss alot totallyyy get it man awesome!! (Y)
thanks
OMG!!
when you expand ... add all cofficients of x ... and equate with zero to find a
Ohh thank you!!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 ..
:')
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now
