202212211654 Solution to 2009-CE-AMATH-II-6

$C$ is a curve with equation $y^3+x^3y=10$ .

(a) Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ .
(b) Find the equation of the tangent to the curve $C$ at the point $(1,2)$ .

Roughwork.

$\begin{aligned} y^3+x^3y & = 10 \\ 3y^2\,\mathrm{d}y+x^3\,\mathrm{d}y+3x^2y\,\mathrm{d}x & = 0 \\ -\frac{3x^2y}{x^3+3y^2} = \frac{\mathrm{d}y}{\mathrm{d}x} & = y'(x,y) \\ \end{aligned}$

$\displaystyle{y'(1,2)=-\frac{3(1)^2(2)}{(1)^3+3(2)^2}=-\frac{6}{13}}$

$\begin{aligned} &\quad\enspace \frac{y-2}{x-1} = -\frac{6}{13} \\ & \Rightarrow \big\{ L:\quad 6x+13y-32 = 0\big\} \\ \end{aligned}$

This problem is not to be attempted.

M	T	W	T	F	S	S
			1	2	3	4
5	6	7	8	9	10	11
12	13	14	15	16	17	18
19	20	21	22	23	24	25
26	27	28	29	30	31