Question

find the point where there is a horizontal tangent to the equation x^3+y^3-9xy=0 other than the origin.

Answer 1

These needs implicit differentiation doesn't it... D:<

Answer 2

ya

;(

Answer 3

not too bad

Answer 4

its also asking for the vertical tangent

Answer 5

ok here we go
\[3x^2+3y^2y'-9y-9xy'=0\]
\[3x^2-9y=9xy'-3y^2y'\]
\[3x^2-9y=(9x-3y^2)y'\]
\[y'=\frac{3x^2-9y}{9x-3y^2}\]
\[y'=\frac{x^2-3y}{3x-y^2}\]

Answer 6

check my algebra because it is late

Answer 7

ya thats right

Answer 8

so i need to find where that is = to 0 right?

Answer 9

oh lord ok

Answer 10

\[x=\pm\sqrt{3y}\]

Answer 11

so i guess you replace
\[x\] by
\[\sqrt{3y}\] in the original equation and solve. i would use a machine

Answer 12

would that give me the horiz tan?

Answer 13

yes

Answer 14

horizontal where the derivative is 0, this is going to be very annoying

Answer 15

\[(3y)^{\frac{3}{2}}+y^3-9y^{\frac{3}{2}}=0\]

Answer 16

is that right?

Answer 17

no it isn't let me fix it

Answer 18

ummm idk im bad at the fraction exponets lol ill draw what i got

Answer 19

\[(3y)^{\frac{3}{2}}+y^3-9\sqrt{3y}y=0\]

Answer 20

\[\sqrt{3y}^3+y^3-9\sqrt{3y}y=0\] is where we start yes?

Answer 21

this sucks