Finding the tangent line using implicit differentiation.

[Reckoning of Sand]

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 $$\frac {dx}{dt} =-3/5$$$$\frac {dy}{dt}=-2/5.$$

 

[ehild]

You need the equation of the tangent line. How can you write it in terms of x and y?

ehild

 

[Reckoning of Sand]

$f(x_0,y_0)+\nabla f(x,y)\cdot (x-x_0,y-y_0)?$

 

[ehild]

The curve is in the x,y plane and the tangent line is a straight line of form (y-yo)=a(x-xo). You need that tangent line. The slope of the line is a=dy/dx, you get it with implicit differentiation.

ehild

 

[Reckoning of Sand]

$f(t)$ defines $x$ and $y$ implicitly as functions $x=x(t)$ and $y=y(t)$. To get $\frac {dy}{dx}$, would I need to define $y$ implicitly as a function $y=y(x)$?

 

[ehild]

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 $f(1)=\begin{pmatrix} -1\\1 \end{pmatrix}.$

Are you sure that you copied the problem correctly?
X(1)≠-1, as -1+6+1-1 =5 .
Is not the second equation x+y+t-1=0?

ehild

 

[ehild]

$f(t)$ defines $x$ and $y$ implicitly as functions $x=x(t)$ and $y=y(t)$. To get $\frac {dy}{dx}$, would I need to define $y$ implicitly as a function $y=y(x)$?

No, you can get dy/dx as (dy/dt)/(dx/dt). That is implicit differentiation.

ehild

 

[Reckoning of Sand]

Then, $\frac {dy}{dx}=\frac {2}{3}$. And substituting the values for $y-y_0=a(x-x_0)$ gives $$y-1=\frac {2}{3}(x+1).$$

 

[ehild]

Looks good.

ehild