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.