Show that there are exactly two points on the curve where the gradient is zero.

Find the equation of the normal to the curve at the point P.

The normal at P cuts the curve again at the point Q. Find the $x$-coordinate of Q.

The shaded region is rotated by 2$\pi$ about the $y$-axis. Find the volume of the solid formed.

differentiating implicitly:       M1

$2xy + {x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - 4{y^3}\frac{{{\text{d}}y}}{{{\text{d}}x}}$     A1A1

Note: Award A1 for each side.

if $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ then either $x = 0$ or $y = 0$       M1A1

$x = 0 \Rightarrow$ two solutions for $y\left( {y =  \pm \sqrt[4]{5}} \right)$      R1

$y = 0$ not possible (as 0 ≠ 5)     R1

hence exactly two points      AG

Note: For a solution that only refers to the graph giving two solutions at  $x = 0$ and no solutions for $y = 0$ award R1 only.

[7 marks]

at (2, 1)  $4 + 4\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - 4\frac{{{\text{d}}y}}{{{\text{d}}x}}$     M1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - \frac{1}{2}$     (A1)

gradient of normal is 2       M1

1 = 4 + c       (M1)

equation of normal is $y = 2x - 3$     A1

[5 marks]

substituting      (M1)

${x^2}\left( {2x - 3} \right) = 5 - {\left( {2x - 3} \right)^4}$ or ${\left( {\frac{{y + 3}}{2}} \right)^2}\,y = 5 - {y^4}$       (A1)

$x = 0.724$      A1

[3 marks]

recognition of two volumes      (M1)

volume $1 = \pi \int_1^{\sqrt[4]{5}} {\frac{{5 - {y^4}}}{y}} {\text{d}}y\left( { = 101\pi  = 3.178 \ldots } \right)$      M1A1A1

Note: Award M1 for attempt to use $\pi \int {{x^2}} {\text{d}}y$, A1 for limits, A1 for ${\frac{{5 - {y^4}}}{y}}$ Condone omission of $\pi$ at this stage.

volume 2

EITHER

$= \frac{1}{3}\pi  \times {2^2} \times 4\left( { = 16.75 \ldots } \right)$     (M1)(A1)

OR

$= \pi \int_{ - 3}^1 {{{\left( {\frac{{y + 3}}{2}} \right)}^2}} {\text{d}}y\left( { = \frac{{16\pi }}{3} = 16.75 \ldots } \right)$     (M1)(A1)

THEN

total volume = 19.9      A1

[7 marks]