maths p3 problem !!!

[talha196]

hey guys i am having trouble with may 2013 paper32 question 9, kindly help me with this part b.

 

[falak123]

hey guys i am having trouble with may 2013 paper32 question 9, kindly help me with this part b.

in the end use cosine rule to find Z modulus

 

[falak123]

i want help too plz tell oct 13 v31 q5 ii part and q4

 

[waztaz123]

You could just find dx/dt and dy/dt. Next invert dx/dt to find dt/dx. And use chain rule to find that that dy/dx= dy/dt * dt/dx.

 

[waztaz123]

hey guys i am having trouble with may 2013 paper32 question 9, kindly help me with this part b.

in the end use cosine rule to find Z modulus

The shaded diagram is the same as in the picture (shade below the half line and circle) . But I think the key issue is to find the greatest length of modulus of z which might cause issues.
By inspection the greatest length of modulus of z is clearly the point labelled L. We need to find the coordinates of the poinbt labelled L. I used the fact that the half line passes through the center of the circle. And drew a triangle with angle 45 and hypotenuse of length 2 (the radius of the circle). From this triangle I found the x coordinate of point L as 2cos45 and y coordinate as (2 + 2sin45). And next u just find the length of mod of z.

 

[waztaz123]

i want help too plz tell oct 13 v31 q5 ii part and q4

use cosec2θ = 1/2*(cotθ + tanθ). And find instead the integral of 1/2*(cotθ + tanθ) from pie/3 to pie/6.

Write cotθ as 1/tanθ. So 1/2*(cotθ + tanθ)= 1/2*(1/tanθ + tanθ)= ((tanθ)^2 +1)/2tanθ.

Use the identity (tanθ)^2 +1 = (secθ)^2. So ((tanθ)^2 +1)/2tanθ becomes (secθ)^2/ 2tanθ.

Now you could integrate this expression easily as the derivative of tanθ= (secθ)^2 (something u should know).

So the integral of the expression (secθ)^2/2tanθ =1/2* ln|tanθ| from pie/3 to pie/6. Which become 1/2*(ln|√3| - ln|1/√3|) . Using the fact that ln|x| - ln |y| =ln |x/y|. you would get the answer as 1/2ln |√3|

 

[falak123]

i have reacehd till dy/dx= cost-sint/-cost-sint already wat to do next?

You could just find dx/dt and dy/dt. Next invert dx/dt to find dt/dx. And use chain rule to find that that dy/dx= dy/dt * dt/dx.

 

[talha196]

waztaz123 thanks for the help i got it now, heres another question from oct 2013 paper 31. Quest 8 part 2. Please help me with this as well.

 

[itallion stallion]

Can someone plz help me with p3 question, may June 2013 paper 31 q7 part b.thanks in advance!

 

[Sarah_3420]

hi guys i have a problem with vectors. When we calculate the angle between a line and a plane we use ( sin) not cos but sometimes i get negative answer and i don't' know how to solve this problem the marking scheme usually has the same answer but positive. please help

 

[falak123]

when u have to find the angle between line and plane u do use cos and the answer u obtain from cos^-1 is subtracted from 90.

 

[Sarah_3420]

no line and line is cos plane and plane is cos and line and plane is ( sin )