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Finding the tangent line using implicit differentiation.

  • #1

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:

Answers and Replies

  • #2
ehild
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You need the equation of the tangent line. How can you write it in terms of x and y?

ehild
 
  • #3
##f(x_0,y_0)+\nabla f(x,y)\cdot (x-x_0,y-y_0)?##
 
  • #4
ehild
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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
 
  • #5
##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)##?
 
  • #6
ehild
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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
 
  • #7
ehild
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##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
 
  • #8
Then, ##\frac {dy}{dx}=\frac {2}{3}##. And substituting the values for ##y-y_0=a(x-x_0)## gives $$y-1=\frac {2}{3}(x+1).$$
 
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  • #9
ehild
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Looks good.

ehild
 

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