Date | November 2017 | Marks available | 8 | Reference code | 17N.1.hl.TZ0.7 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Determine | Question number | 7 | Adapted from | N/A |
Question
The folium of Descartes is a curve defined by the equation x3+y3−3xy=0, shown in the following diagram.
Determine the exact coordinates of the point P on the curve where the tangent line is parallel to the y-axis.
Markscheme
x3+y3−3xy=0
3x2+3y2dydx−3xdydx−3y=0 M1A1
Note: Differentiation wrt y is also acceptable.
dydx=3y−3x23y2−3x (=y−x2y2−x) (A1)
Note: All following marks may be awarded if the denominator is correct, but the numerator incorrect.
y2−x=0 M1
EITHER
x=y2
y6+y3−3y3=0 M1A1
y6−2y3=0
y3(y3−2)=0
(y≠0)∴ A1
x = {\left( {\sqrt[3]{2}} \right)^2}{\text{ }}\left( { = \sqrt[3]{4}} \right) A1
OR
{x^3} + xy - 3xy = 0 M1
x({x^2} - 2y) = 0
x \ne 0 \Rightarrow y = \frac{{{x^2}}}{2} A1
{y^2} = \frac{{{x^4}}}{4}
x = \frac{{{x^4}}}{4}
x({x^3} - 4) = 0
(x \ne 0)\therefore x = \sqrt[3]{4} A1
y = \frac{{{{\left( {\sqrt[3]{4}} \right)}^2}}}{2} = \sqrt[3]{2} A1
[8 marks]