Date | May 2014 | Marks available | 5 | Reference code | 14M.2.hl.TZ2.10 |
Level | HL only | Paper | 2 | Time zone | TZ2 |
Command term | Find | Question number | 10 | Adapted from | N/A |
Question
Consider the curve with equation (x2+y2)2=4xy2(x2+y2)2=4xy2.
Use implicit differentiation to find an expression for dydx.
Find the equation of the normal to the curve at the point (1, 1).
Markscheme
METHOD 1
expanding the brackets first:
x4+2x2y2+y4=4xy2 M1
4x3+4xy2+4x2ydydx+4y3dydx=4y2+8xydydx M1A1A1
Note: Award M1 for an attempt at implicit differentiation.
Award A1 for each side correct.
dydx=−x3−xy2+y2xy2−2xy+y3 or equivalent A1
METHOD 2
2(x2+y2)(2x+2ydydx)=4y2+8xydydx M1A1A1
Note: Award M1 for an attempt at implicit differentiation.
Award A1 for each side correct.
(x2+y2)(x+ydydx)=y2+2xydydx
x3+x2ydydx+y2x+y3dydx=y2+2xydydx M1
dydx=−x3−xy2+y2yx2−2xy+y3 or equivalent A1
[5 marks]
METHOD 1
at (1, 1), dydx is undefined M1A1
y=1 A1
METHOD 2
gradient of normal =−1dydx=−(yx2−2xy+y3)(−x3−xy2+y2) M1
at (1, 1) gradient =0 A1
y=1 A1
[3 marks]