Date | May 2018 | Marks available | 3 | Reference code | 18M.2.hl.TZ1.9 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 9 | Adapted from | N/A |
Question
The following graph shows the two parts of the curve defined by the equation \({x^2}y = 5 - {y^4}\), and the normal to the curve at the point P(2 , 1).
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.
Markscheme
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]