What is the equation of the line tangent to the curve x^2 + y^3 = 9 at the point (-1, 2)? I am thrown off because it's no longer a circle. Does it matter?

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What is the equation of the line tangent to the curve x^2 + y^3 = 9 at the point (-1, 2)? I am thrown off because it's no longer a circle. Does it matter?

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Answer 1 · 2018-03-09T01:01:07

$y = \frac{x}{6} + \frac{13}{6}$ $y=x/6+13/6$

Explanation:

The function $x^{2} + y^{3} = 9$ $x^2+y^3=9$ ........... $[1]$ $[1]$ can be differentiated to give us the gradient [ slope] of the line required.

This can be done either implicitly or explicitly, it's easier done explicitly since we are given the $x and y$ $x and y$ values.

Differentiating both sides of $[1]$ $[1]$ implicitly,

$\frac{d}{d x} [x^{2} + y^{3}]$ $d/dx[x^2+y^3]$ = $[\frac{d}{d x} 9]$ $[d/dx 9]$ and thus,

$2 x + 3 y^{2} \frac{d y}{d x} = 0$ $2x+3y^2dy/dx=0$ , Rearranging, $\frac{d y}{d x} = \frac{- 2 x}{3 y^{2}}$ $dy/dx=[-2x]/[3y^2]$ and plugging in $x - 1, y = 2$ $x-1, y=2$

$\frac{d y}{d x} = \frac{1}{6}$ $dy/dx=1/6$ .

From the equation for a straight line... $[y - y 1] = \frac{1}{6} [[x - x 1]$ $[y-y1]=1/6[[x-x1]$ and so $[y - 2] = \frac{1}{6} [x - [- 1]]$ $[y-2]=1/6[x-[-1]]$ which gives us $y = \frac{1}{6} x + \frac{13}{6}$ $y=1/6x+13/6$ .

You can also use the chain rule to differentiate the function explicitly, ie, $\frac{d y}{d x} = \frac{d y}{d z} \cdot \frac{d z}{d x}$ $dy/dx=dy/dz*dz/dx$ where $z$ $z$ represents the contents of a bracket or similar function.

From $[1]$ $[1]$ ... $y = {[9 - x^{2}]}^{\frac{1}{3}}$ $y=[9-x^2]^[1/3$ , so $\frac{d y}{d x} = \frac{1}{3} {[9 - x^{2}]}^{- \frac{2}{3}}$ $dy/dx=1/3[9-x^2]^[-2/3$ $d / d x [9 - x^{2}]$ $d//dx[9-x^2]$

= $\frac{- 2 x}{3} {[{[\sqrt{9 - x^{2}}]}^{2}]}^{\frac{1}{3}} \cdot$ $[-2x]/3 [[sqrt[9-x^2]]^2]^[1/3]*$ , and substituting $x = - 1$ $x=-1$ , will give $\frac{d y}{d x} = \frac{1}{6}$ $dy/dx=1/6$ .

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What is the equation of the line tangent to the curve x^2 + y^3 = 9 at the point (-1, 2)? I am thrown off because it's no longer a circle. Does it matter?

1 Answer

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