Consider the part of the curve \(4{x^2} + {y^2} = 4\) shown in the diagram below.

(a)     Find an expression for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) in terms of x and y .

(b)     Find the gradient of the tangent at the point \(\left( {\frac{2}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right)\).

(c)     A bowl is formed by rotating this curve through \(2\pi \) radians about the x-axis.

Calculate the volume of this bowl.

(a)     \(8x + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)     M1A1

Note: Award M1A0 for \(8x + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4\) .

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{{4x}}{y}\)     A1

(b)     – 4     A1

(c)     \(V = \int {\pi {y^2}{\text{d}}x} \) or equivalent     M1

\(V = \pi \int_0^1 {(4 - 4{x^2}){\text{d}}x} \)     A1

\( = \pi \left[ {4x - \frac{4}{3}{x^3}} \right]_0^1\)     A1

\( = \frac{{8\pi }}{3}\)     A1

Note: If it is correct except for the omission of \(\pi \) , award 2 marks.

[8 marks]

The first part of this question was done well by many, the only concern being the number that did not simplify the result from \( - \frac{{8x}}{{2y}}\). There were many variations on the formula for the volume in part c), the most common error being a multiple of \(2\pi \) rather than \(\pi \). On the whole this question was done well by many.