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C3 Help please! - EDEXCEL

Given that x = cos y^2,

find dy/dx in terms of y.

I think the prob is how would I differentiate cosy^2, I got -2ysiny^2 but that's wrong I think.

Help please!

Thanks! :smile:
Reply 1
Avatar for -James-
-James-
OP
In 5c) why is b taken as 15.

http://www.carlgauss.co.uk/PDFs/A-Level%20Papers/C3/C3%20-%20Mock%20Paper.pdf

?
Reply 2
Avatar for raheem94
raheem94
13
Original post by -James-
Given that x = cos y^2,

find dy/dx in terms of y.

I think the prob is how would I differentiate cosy^2, I got -2ysiny^2 but that's wrong I think.

Help please!

Thanks! :smile:


x=cos(y2) \displaystyle x = cos (y^2)

Differentiate with respect to y y

dxdy=2y×−sin(y2)=−2ysin(y2) \displaystyle \frac{dx}{dy} = 2y \times -sin(y^2) = -2ysin(y^2)

Find the reciprocal to get the final answer,
dydx=−12ysin(y2)=−cosec(y2)2y \displaystyle \frac{dy}{dx} = - \frac{1}{2ysin(y^2)} = - \frac{cosec(y^2)}{2y}
Reply 3
Avatar for -James-
-James-
OP
Original post by raheem94
x=cos(y2) \displaystyle x = cos (y^2)

Differentiate with respect to y y

dxdy=2y×−sin(y2)=−2ysin(y2) \displaystyle \frac{dx}{dy} = 2y \times -sin(y^2) = -2ysin(y^2)

Find the reciprocal to get the final answer,
dydx=−12ysin(y2)=−cosec(y2)2y \displaystyle \frac{dy}{dx} = - \frac{1}{2ysin(y^2)} = - \frac{cosec(y^2)}{2y}


cheers mate! :smile:

Would you be able to look at my 2nd q cause I don't understood why they took b as 15? and why not another other number :s-smilie:
(edited 11 years ago)
Reply 4
Avatar for raheem94
raheem94
13
Original post by -James-
wait, what happend to the y at the bottom ?


I made a typo before, sorry if you saw my unedited post.
Reply 5
Avatar for raheem94
raheem94
13
Original post by -James-


We know A=75 A = 75

We need to decide on a B B .

We have to use it find A+B and A−B A + B \ and \ A -B

A+B=…  ⟹  75+B=…A−B=…  ⟹  75−B=… A + B = \ldots \implies 75 + B = \ldots \\ A - B = \ldots \implies 75-B = \ldots

The question hints at the use of sin⁡60 and cos⁡60 \sin 60 \ and \ \cos 60

So how can we make one of the equation equal to 60 60

If we sub in B=−15 B = -15 , we get, A+B=75−15=60 A + B = 75 - 15 = 60 we know sin⁡60 and cos⁡60 \sin60 \ and \ \cos60 so it is fine till here, and if we sub it in the other equation A−B=75+15=90 A - B = 75 +15 = 90 we also know sin⁡90 and cos⁡90 \sin90 \ and \ \cos90 hence it also works.

So both B=15 or B=−15B = 15 \ or \ B = -15 can be used, it is just the case of thinking that which value of B B will give you values of A+B and A−BA+B \ and \ A - B for which you know the exact of values of sin⁡ and cos⁡ \sin \ and \ \cos .

Hope it makes sense.
Reply 6
Avatar for YMAS.46
YMAS.46
What paper is this do you know by any chance?!

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