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    Hi can someone please show me how to do these questions attached?

    thanks guys
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    (1)
    Differentiate to give an expression involving \frac{dy}{dx}. Set \frac{dy}{dx}=0 and so obtain a relation between x and y that needs to be satisfied. Substitute this back into the original equation of the curve to eliminate y or x and hence find the coordinates at the stationary points.

    (2)
    I=\int_{0}^{\frac{1}{2}} \frac{1}{(1-x^{2})^{\frac{3}{2}}} dx \\

x=\sin \theta. dx=\cos \theta d\theta. \\

x=\frac{1}{2} \rightarrow \theta =\frac{\pi}{6}  \\

x=0 \rightarrow \theta =0 \\

I=\int_{0}^{\frac{\pi}{6}} \frac{1}{(1-\sin ^{2} \theta)^{\frac{3}{2}}}  . \cos \theta d\theta \\

I=\int_{0}^{\frac{\pi}{6}} \frac{\cos \theta}{\cos ^{3} \theta} d\theta \\

I=\int_{0}^{\frac{\pi}{6}} \sec ^{2} \theta d\theta \\

I=[\tan \theta]_{0}^{\frac{\pi}{6}} \\

I=\tan \frac{\pi}{6} \\

I=\frac{1}{\sqrt{3}}=\frac{\sqrt  {3}}{3}

    (3)
    (a) Use \frac{dy}{dx}=\frac{dy}{dt}.\fra  c{dt}{dx} and \frac{dt}{dx}=\frac{1}{\frac{dx}  {dt}}
    (b) Using (a) and the given results you can calculate the values of x,y,\frac{dy}{dx} at t=\frac{\pi}{4} which is all you need to establish the tangent equation.
    (c) You may find the identity \sin ^{2} x + \cos ^{2} x \equiv 1 useful as it implies 1+\cot ^{2} x \equiv cosec ^{2} x and thus you can eliminate t.
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    I remember those questions alright lol. C4 paper was alright compared to what it could have been. I think I got all of those questions right. :-D
 
 
 
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