Implicit differentiation: tangent to x^3 + y^3 = 3xy

J

jwpaine

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For the curve defined by the equation: \(\displaystyle \L x^3\,+y^3\,=\,3xy\), find an equation of the tangent at the point \(\displaystyle \L ( \frac{4}{3}\, ,\, \frac{2}{3} )\)

I do:

\(\displaystyle \L D(x^3\,+y^3)\,=\,D(3xy)\)

\(\displaystyle \L 3x^2\,+3y^2\cdot y'\,=\,0\)

\(\displaystyle \L y'\,=\, \frac{-3x^2}{3y^2}\)

Now can I simply plug in my ordered pair (x,y) into y' and then use that as my slope in the point-slope form for the equation of a line?

\(\displaystyle \L m = -\frac{\frac{4}{3}^2}{\frac{2}{3}^2} = -4\)

\(\displaystyle \L y - \frac{2}{3} = -4(x - \frac{4}{3})\)

Am I right?
 
Re: Implicit differentiation

jwpaine said:
For the curve defined by the equation: \(\displaystyle \L x^3\,+y^3\,=\,3xy\), find an equation of the tangent at the point \(\displaystyle \L ( \frac{4}{3}\, ,\, \frac{2}{3} )\)

I do:

\(\displaystyle \L D(x^3\,+y^3)\,=\,D(3xy)\)

\(\displaystyle \L 3x^2\,+3y^2\cdot y'\,=\,0\) <---RHS should be 3*(y + xy')

\(\displaystyle \L y'\,=\, \frac{-3x^2}{3y^2}\)

Now can I simply plug in my ordered pair (x,y) into y' and then use that as my slope in the point-slope form for the equation of a line?
Click to expand...
 
How do you get 3(y + xy') ?

John

Edit: nvm -

D(3xy) = 3 *D[xy] = 3[y + x * y'] because of the product rule and chain rule for differentiating y'

thanks!
 
Now if I solve for y'

\(\displaystyle \L 3x^2 + 3y^2\cdot y' = 3y + 3xy'\)

\(\displaystyle \L x^2 + y^2\cdot y' = y + xy'\)

\(\displaystyle \L y' = \frac{y + xy' - x^2}{y^2}\)

Now how would I find the slope of the tangent at my point (x,y) when I have a y' in my solved for y' ?
 
Hey JW. You shouldn't have a y' in your answer.

\(\displaystyle \L\\x^{3}+y^{3}-3xy=0\)

Differentiate, don't forget the product rule:

\(\displaystyle \L\\3x^{2}+\overbrace{3y^{2}y'}^{\text{derivative y^3}}-(\underbrace{3xy'+3y}_{\text{derivative 3xy}})=0\)

Solve for y':

\(\displaystyle \L\\y'=\frac{y-x^{2}}{y^{2}-x}\)

Now, sub in your x=4/3 and y=2/3 and you get a slop[e of 5/4 at that point. Now that you have the slope, use y=mx+b and find your line equation.
 
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