- #1

- 13

- 0

## Homework Statement

The equations ##2x^3y+yx^2+t^2=0##, ##x+6+t-1=0## implicitly define a curve $$f(t) = \begin{pmatrix} x(t)\\y(t) \end{pmatrix}$$ that satisfies ##f(1)=\begin{pmatrix} -1\\1 \end{pmatrix}.##

Find the tangent line to the curve when ##t=1##.

## Homework Equations

The chain rule to find the derivative of f would give something like: $$\frac {df}{dt} = \frac {\partial f}{\partial x}\frac {\partial x}{\partial t} + \frac {\partial f}{\partial y}\frac {\partial y}{\partial t}.$$

And the equation of the tangent plane should be given by the formula: $$f(t_0)+\frac {df}{dt}(t_0) \cdot (t-t_0).$$

## The Attempt at a Solution

Differentiating both sides of each equation with respect to ##t## gives $$2x^3\frac {dy}{dt}+6x^2y\frac {dx}{dt}+2xy\frac {dx}{dt} +x^2\frac {dy}{dt} + 2t = 0$$$$\frac {dx}{dt}+\frac {dy}{dt}+1=0$$Solving for ##\frac {dx}{dt}## and ##\frac {dy}{dt}## when ##t=1## and ##f(1)=(-1,1)## gives $$\frac {dx}{dt} =-3/5$$$$\frac {dy}{dt}=-2/5.$$

Last edited: