# 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?

Mar 9, 2018

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

#### Explanation:

The function ${x}^{2} + {y}^{3} = 9$...........$\left[1\right]$ 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 $x \mathmr{and} y$ values.

Differentiating both sides of $\left[1\right]$ implicitly,

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{6}$.

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

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

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

=$\frac{- 2 x}{3} {\left[{\left[\sqrt{9 - {x}^{2}}\right]}^{2}\right]}^{\frac{1}{3}} \cdot$, and substituting $x = - 1$ , will give$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{6}$.