Question

Determine an equation of the tangent line and normal line at (2,2) on the curve (x^2+y^2-2x)^2 = 2(x^2+y^2)

Answer 1

My guess on how to solve this Q: Find the derivative, take the derivative and plug point into derivative and that gives you slope of tangent line. Since x's and y's are all mixed up on the same side, I would solve using implicit differentiation.

My problem is that I don't know how to take the derivative of such a complicated function. I also don't know what they mean by normal line.

Answer 2

Ok yes you have your steps thought out correctly.
Let's see if we can take the derivative of this bad boy.\[\Large (x^2+y^2-2x)^2\quad=\quad 2(x^2+y^2)\]

Answer 3

\[\Large \color{royalblue}{\left[(x^2+y^2-2x)^2\right]'}\quad=\quad \color{royalblue}{\left[2(x^2+y^2)\right]'}\]We start by applying the power rule on the left side,\[\Large 2(x^2+y^2-2x)\color{royalblue}{(x^2+y^2-2x)'}\quad=\quad \color{royalblue}{\left[2(x^2+y^2)\right]'}\]
That extra blue term is showing up due to the chain rule.
Chain rule tells us to multiply by the derivative of the inside.
(I'm using blue to show where we still need to take a derivative, I hope that's clear.)

Answer 4

Taking the derivative of the blue part on the left will give us:\[\Large 2(x^2+y^2-2x)\color{orangered}{(2x+2yy'-2)}\quad=\quad \color{royalblue}{\left[2(x^2+y^2)\right]'}\]
Understand where that y' is coming from?

Answer 5

Mmk well I gotta get to bed.. -_-

I'll just make a note here for you:
`The slope of a normal line is perpendicular to that of the tangent line`

So after you find your tangent slope, finding the normal slope is as easy as taking it's negative reciprocal.\[\LARGE m_{norm}\quad=\quad -\frac{1}{m_{\tan}}\]

Answer 6

The slope is a negative reciprocal of the other*
the lines are perpendicular*

blah I said that poorly the first time :) whatev.

Answer 7

eeek! Thank-you so much. I got it. Just checked this board. g'night xoxo