Find the equation of the tangent and normal lines to the curve2x3 + 2y3

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Ronsmart @Ronsmart

September 2022 1 0 Report

Find the equation of the tangent and normal lines to the curve
2x3 + 2y3 - 9xy = 0 the point (1, 2).

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jeblang

$\mathbb{SOLUTION:}$

$\mathsf{2x^3+2y^3-9xy=0}$

To determine the slope of the tangent line, get the first derivative of the curve and evaluate it at the given x and y coordinates of the point.

$>Using implicit differentiation:<img src=$

$\mathsf{6x^2+6y^2\dfrac{dy}{dx}-9\left(x\dfrac{dy}{dx}+y\right)=0}$

Isolate dy/dx

$\mathsf{6x^2+6y^2\dfrac{dy}{dx}-9x\dfrac{dy}{dx}-9y=0}$

$\mathsf{9x\dfrac{dy}{dx}-6y^2\dfrac{dy}{dx}=6x^2-9y}$

$\mathsf{(9x-6y^2)\dfrac{dy}{dx}=6x^2-9y}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{6x^2-9y}{9x-6y^2}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{2x^2-3y}{3x-2y^2}} \longleftarrow \textsf{evaluate at x = 1, y = 2}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{2(1)-3(2)}{3(1)-2(2)^2}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{4}{5}}$

$>The slope of the tangent is m₁ = 4/5. The tangent line is also passing through point (1, 2).Using point-slope form of a line:<img src=$

$\mathsf{m=m_1=\dfrac{4}{5}}$

$\mathsf{(x_1,y_1)=(1,2)}$

$><img src=$

$\mathsf{y-4=\dfrac{4}{5}x-\dfrac{4}{5}}$

$\mathsf{y=\dfrac{4}{5}x-\dfrac{4}{5}+2}$

$\mathsf{y=\dfrac{4}{5}x+\dfrac{6}{5}} \Longleftarrow \textsf{equation of the tangent line}$

$>The slope of the normal is line is equal to the negative reciprocal of the slope of the tangent line. m₂ = -(1/m₁) = -(5/4). The normal line is also passing through point (1, 2).Using point-slope form of a line:<img src=$

$\mathsf{m=m_2=-\dfrac{5}{4}}$

$\mathsf{(x_1,y_1)=(1,2)}$

$><img src=$

$\mathsf{y-2=-\dfrac{5}{4}x+\dfrac{5}{4}}$

$\mathsf{y=-\dfrac{5}{4}x+\dfrac{5}{4}+2}$

$\mathsf{y=-\dfrac{5}{4}x+\dfrac{13}{4}} \Longleftarrow \textsf{equation of the normal line}$

$><img src=$

The equation of the tangent line is:

$\boxed{\mathsf{y=\dfrac{4}{5}x+\dfrac{6}{5}}}$

The equation of the normal line is:

$\boxed{\mathsf{y=-\dfrac{5}{4}x+\dfrac{13}{4}}}$

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Find the equation of the tangent and normal lines to the functiony = x2 - 6x + 4...

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Find the equation of the tangent and normal lines to the functiony = x2 - 6x + 4...

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