find tangent line x^3+y^3=y+21 at (3,-2) I got y=-27/11x +59/11 but it's saying its not right, and I cant see the error in my ways
oh hmm shouldnt it be 3x^2 +3y^2(dy/dx)=dy/dx and after solving that its the slope?
yep! :)
\[3y^2 y'-y'=-3x^2\]
\[y'(3x^2-1)=-3x^2\]
Now divide both sides by \[3x^2-1\]
\[3y^2-1\]
oops type-0
yes y doesn't usually change to x unless some kind of act of god happens
or myininaya screws up
I like doing it the quick way \[\frac{dy}{dx}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}\]
wait what you just typed to me is the slope? cuz thats dy/dx implicit d confuses the hell out of me
just use y' instead
dy/dx is the slope?
y'
y' always looks prettier to me
\[y'=\frac{dy}{dx}\]
cuz the equation I got up top is the tangent line with the slope as y'
wait what is your question? i'm confused
hahaha okay so how to do I get the slope of the tangent line?
\[slope=y'=\frac{dy}{dx}=\frac{-3x^2}{3y^2-1}\]
hmmm thats -27/11 and it isnt giving me the correct answer
the tangent line I posted is it with that slope so idk what i did wrong
\[y'|_{(3,-2)}=\frac{-3(3)^2}{3(-2)^2-1}=\frac{-27}{11}\] I don't think I understand your question yes that is the right slope
\[y=\frac{-27}{11}x+b\] What point do we know that is on this line (3,-2) plug this in to find b
okay so if that is my slope then y+2=-27/11(x-3) y= -27/11x +59/11 but it is saying that is incorrect
@Awstinf try typing it in standard form \[27x+11y=\frac{59}{11}\]
Sorry\[27x+11y=59\]
That is right lol
Ok well that just means you should subtract 2 both sides
\[y=\frac{-27}{11}(x-3)-2\] but you already did
so it just says y=?
yeah :\
So it does it say to put in slope intercept form?
yeah, but webwork is kind of screwy. Probably one of the worst systems I've ever dealt with. Wolfram should make something like it so we never have to deal with it ever again lol.
I'm afraid I don't know what to tell you Your answer is right
thank you for helping me! again!
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