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Find the equation of the normal at the point $(2, 1)$ for the function $x^2 + y^3 - 2y = 3$

I'm still struggling a bit with the application of derivatives, but from what I understand I use the derivative to calculate the slope

\begin{align} \frac{dy}{dx}x^2 + y^3 - 2y &= \frac{dy}{dx} 3\\ (3y^2 - 2)\frac{dy}{dx} &= 0 - 2x\\ \frac{dy}{dx} &= \frac{-2x}{(3y^2 - 2)} \end{align}

I then plug the values of the point to get the slope $m = \frac{-2(2)}{(3(1)^2 - 2)} = \frac{-4}{1} = -4$

I then plug all of this into the equation $y - 1 = -4 (x + 2)$

Is this the correct approach?

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Everything looks good, except when you use $m = -4$ in the equation of the desired line: What you post in the end is the equation of the line tangent to the curve at $(2, 1)$.

The slope of the normal is the negative reciprocal of $-4$: $\dfrac 14$.

That gives us $$y - 1 = \frac 14(x - 2)$$

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  • $\begingroup$ Crumbs... that's right, I better remember that come exam time. $\endgroup$ – Leon Jun 10 '14 at 19:47
  • $\begingroup$ The slope should be $\frac{1}{4}$, and it should be $y-1=\frac{1}{4}(x-2)$... $\endgroup$ – stochasm Jun 10 '14 at 20:30
  • $\begingroup$ I caught that, just as you "spoke", @stochasm. Thanks! $\endgroup$ – Namaste Jun 10 '14 at 20:31
  • $\begingroup$ Yeah I was wondering, you edited it almost immediately after I clicked "Add Comment". I was like "wow...this dude is fast..." $\endgroup$ – stochasm Jun 10 '14 at 20:37

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