How do you find an equation of the line tangent to the curve at the given point $x^{3} + y^{3} = 36 x y$ $x^3+y^3=36xy$ given point (18,18)?

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How do you find an equation of the line tangent to the curve at the given point $x^{3} + y^{3} = 36 x y$ $x^3+y^3=36xy$ given point (18,18)?

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Answer 1 · 2016-10-21T03:58:39

Use implicit differentiation

$3 x^{2} + 3 y^{2} \frac{d y}{d x} = 36 (x (1) \frac{d y}{d x} + y (1))$ $3x^2+3y^2(dy)/dx=36(x(1)(dy)/dx+y(1))$

Simplify

$3 x^{2} + 3 y^{2} \frac{d y}{d x} = 36 (x \frac{d y}{d x} + y)$ $3x^2+3y^2(dy)/dx=36(x(dy)/dx+y)$

Distribute

$3 x^{2} + 3 y^{2} \frac{d y}{d x} = 36 x \frac{d y}{d x} + 36 y$ $3x^2+3y^2(dy)/dx=36x(dy)/dx+36y$

Isolate $\frac{d y}{d x}$ $dy/dx$

$3 y^{2} \frac{d y}{d x} - 36 x \frac{d y}{d x} = - 3 x^{2} + 36 y$ $3y^2(dy)/dx-36x(dy)/dx=-3x^2+36y$

Factor out $\frac{d y}{d x}$ $dy/dx$

$\frac{d y}{d x} (3 y^{2} - 36 x) = - 3 x^{2} + 36 y$ $(dy)/dx(3y^2-36x)=-3x^2+36y$

Divide to isolate $\frac{d y}{d x}$ $dy/dx$

$\frac{\frac{d y}{d x} 3 y^{2} - 36 x}{(3 y^{2} - 36 x)} = \frac{- 3 x^{2} + 36 y}{3 y^{2} - 36 x}$ $((dy)/dxcancel(3y^2-36x))/cancel((3y^2-36x))=(-3x^2+36y)/(3y^2-36x)$

Simplify

$\frac{d y}{d x} = \frac{- 3 x^{2} + 36 y}{3 y^{2} - 36 x}$ $(dy)/dx=(-3x^2+36y)/(3y^2-36x)$

Factor out a 3

$\frac{d y}{d x} = \frac{3 (- x^{2} + 12 y)}{3 (y^{2} - 12 x)}$ $(dy)/dx=(3(-x^2+12y))/(3(y^2-12x))$

Simplify

$\frac{d y}{d x} = \frac{3 (- x^{2} + 12 y)}{3 (y^{2} - 12 x)}$ $(dy)/dx=(cancel3(-x^2+12y))/(cancel3(y^2-12x))$

Simplify

$\frac{d y}{d x} = \frac{12 y - x^{2}}{y^{2} - 12 x}$ $(dy)/dx=(12y-x^2)/(y^2-12x)$

Substitute in the point (18,18) to find the slope

$\frac{d y}{d x} = \frac{12 (18) - 18^{2}}{18^{2} - 12 (18)}$ $(dy)/dx=(12(18)-18^2)/(18^2-12(18))$

$\frac{d y}{d x} = - 1$ $(dy)/dx=-1$

Now we have the slope(m), -1, and a point (18,18)

Use the point slope formula, $y = m (x - x_{1}) + y_{1}$ $y=m(x-x_1)+y_1$

$y = - 1 (x - 18) + 18$ $y=-1(x-18)+18$

Simplify

$y = - x + 18 + 18$ $y=-x+18+18$

$y = - x + 36$ $y=-x+36$

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How do you find an equation of the line tangent to the curve at the given point $x^{3} + y^{3} = 36 x y$ $x^3+y^3=36xy$ given point (18,18)?

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