Question

Use implicit differentiation to find an equation for the line normal to the curve y^2 + 2xy = x^2 - 2 at the point (1,-1)

Answer 1

Let's see your best shot at the derivative.

Answer 2

2y+2=2x ???

Answer 3

I get mixed up with using y'??

Answer 4

\(y^{2} \implies 2yy'\) There's a power rule and a chain rule in there!

\(2xy \implies 2xy' + 2y\) There's a product rule and a chain rule in there.

ALWAYS a Chain Rule.

Answer 5

ohh I see I see

Answer 6

:o

Answer 7

then after you just plug in the given coordinates

Answer 8

There is no such thing as "plug in".

1) Use the principle of substitution in the original equation to establish that the point (1,-1) is actually ON the curve.

2) Use the principle of substitution in the derivative to determine the slope of the TANGENT line at (1,-1)

3) Use the definition of a NORMAL line, compared to a TANGENT line, to find the slope of the NORMAL line.

4) Use the Point-Slope form to write down the equation of the Normal Line.

Answer 9

but when I substituted the coordinates I got -2y'+2y'=4

Answer 10

Hmm I'm confused what you're doing :(
Did you get this for your derivative?\[\Large 2yy'+2xy'+2y=2x\]

Answer 11

yes!

Answer 12

Did you solve for y'? What did you get?

Answer 13

I am not really sure how to approach that

Answer 14

Hmm it looks like you plugged in the coordinate pair before solving for y' explicitly. Which is actually fine, you just need to solve for y' now :)

Answer 15

Back up to the derivative that you took, it'll make a little bit more sense from there :x
Divide each side by 2,\[\Large yy'+xy'+y=x\]Subtract y to the other side,\[\Large yy'+xy'\quad=\quad x-y\]On the left side, factor a y' out of each term,\[\Large y'(y+x)\quad=\quad x-y\]Understand how that works? :o

Answer 16

Note: Please remember your algebra!!!! There is just not anything else going on, here. You already did all the calculus part of this problem. The rest of the problem should be second-nature algebra.