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AFind y' if x^3 + y^3 = 6xy.
BFind the tangent to the folium of Descartes x3 + y3 = 6xy at the point (3, 3).
CAt what point in the first quadrant is the tangent line horizontal?

Verified Answer :

A)

\(\bf x^3+y^3=6xy\\\\ -----------------------------\\\\ 3x^2+3y^2\frac{dy}{dx}=6\left( 1\cdot y+x\frac{dy}{dx} \right)\\\\\\ 3\left( x^2+ y^2\frac{dy}{dx}\right)=6\left( 1\cdot y+x\frac{dy}{dx} \right) \\\\\\ x^2+ y^2\frac{dy}{dx}=2y+2x\frac{dy}{dx}\implies x^2+ y^2\frac{dy}{dx}-2y-2x\frac{dy}{dx}=0 \\\\\\ \cfrac{dy}{dx}(y^2-2x)=2y-x^2\implies \cfrac{dy}{dx}=\cfrac{2y-x^2}{y^2-2x}\)

B)

\(\bf \left. \cfrac{dy}{dx}=\cfrac{2y-x^2}{y^2-2x} \right|_{3,3}\implies \begin{cases} x=3\\ y=3 \end{cases}\implies -1 \\\\ -----------------------------\\\\ y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-3=-1(x-3)\\ \qquad \uparrow\\ \textit{point-slope form}\)

solve for "y", and that'd be the equation of the tangent at 3,3

C)   recall, the tangent is horizontal when the slope is 0, thus

\(\bf \cfrac{dy}{dx}=\cfrac{2y-x^2}{y^2-2x}\implies 0=\cfrac{2y-x^2}{y^2-2x}\implies 0=2y-x^2\implies \cfrac{x^2}{2}=\boxed{y} \\\\\\ now \qquad x^3+y^3=6xy\implies x^3+\left( \boxed{\cfrac{x^2}{2}} \right)^3=6x\left( \boxed{\cfrac{x^2}{2}} \right) \\\\\\ x^3+\cfrac{x^6}{8}=3x^3\implies \cfrac{8x^3+x^6}{8}=3x^3 \\\\\\ 8x^3+x^6-24x^3=0\implies x^6-16x^3=0 \\\\\\ x^3(x^2-16)=0\implies x=\{0,\pm 4\}\)

so.. x =0 is not exactly in the first quadrant,  x = -4 isn't either

so. the only choice is really x = 4... what's "y" when x = 4? well, just plug that into the \(\bf x^3+y^3=6xy\qquad or \qquad \cfrac{x^2}{2}=y\)

and solve for "y"

Using derivative concepts, it is found that:

  • a) \(y^{\prime} = \frac{2y - x^2}{y^2 - 2x}\)
  • b) -1
  • c) The tangent line is horizontal at points \((\sqrt{2y},y)\), considering y > 0.

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Question a:

To solve this, we use the implicit differentiation. Thus:

\(3x^2\frac{dx}{dx} + 3y^2\frac{dy}{dx} = 6x\frac{dy}{dx} + 6y\frac{dx}{dx}\)

\(3x^2 + 3y^2\frac{dy}{dx} = 6x\frac{dy}{dx} + 6y\)

\((3y^2 - 6x)\frac{dy}{dx} = 6y - 3x^2\)

\(\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}\)

\(\frac{dy}{dx} = \frac{3(2y - x^2)}{3(y^2 - 2x)}\)

\(y^{\prime} = \frac{2y - x^2}{y^2 - 2x}\)

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Question b:

This is \(y^{\prime}\) at \(x = 3, y = 3\). So

\(y^{\prime}(3,3) = \frac{2(3) - 3^2}{3^2 - 2(3)} = \frac{6 - 9}{9 - 6} = -\frac{3}{3} = -1\)

Thus -1 is the answer.

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Question c:

  • The tangent line is horizontal when \(y^{\prime} = 0\).
  • First quadrant means that both x and y are positive.

Thus:

\(\frac{2y - x^2}{y^2 - 2x} = 0\)

\(2y - x^2 = 0\)

\(x^2 = 2y\)

\(x = \sqrt{2y}\)

Thus:

The tangent line is horizontal at points \((\sqrt{2y},y)\), considering y > 0.

A similar problem is given at https://brainly.com/question/9543179

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