Question

Find the point on C where the curve 2x^2 - 3xy - y^3 = 1 has a horizontal tangent line?

I found dy/dx = (4x-3y) / (3x+3y^2). Then I set 4x-3y = 0 and got y = 4/3x.
After substituting y = 4/3x into the original equation, I got 2x^2 - 4x^2 - (4/3x)^3 = 1, which when simplified, is -2x^2 - (4/3x)^3 = 1.

How do I solve for x in this case?

Answer 1

you're using implicit derivative?

Answer 2

yes

Answer 3

k

Answer 4

you plugged your y = 4/3x into all the y in the original equation properly?

Answer 5

i think so. i checked three times. do you think i might have made a mistake in the process?

Answer 6

let me see

Answer 7

2x^2 - 3xy - y^3 = 1
2x^2 - 3x(4/3x) - (4/3x)^3 = 1
2x^2 - 12x/3x - (64/27x^3) = 1
-64/(27 x^3)+2 x^2-4 = 1

Answer 8

i see where you made 3x (4/3x) = 12x/3x. but how do i find x after this?

Answer 9

we have both x^3 and x^2 in the equation...

Answer 10

something's amiss

Answer 11

show me the implicit derivation just to make sure we're not solving for the wrong x

Answer 12

you know what, i actually have an idea! i'm solving it right now. i'll show you the work when it's done. it might take a few minutes.

Answer 13

k

Answer 14

ah... that didn't work out. i'll show you my work from finding the slope to substituting y = 4/3x into the original equation.

Answer 15

the x is not in the denominator

Answer 16

y = 4x/3

Answer 17

that changes the whole thing

Answer 18

sorry my computer froze

Answer 19

i'm following your work right now..

Answer 20

2x^2 - 3xy - y^3 = 1
2x^2 - 3x(4x/3) - (4x/3)^3 = 1
2x^2 - (12x^2)/3 - 64x^3/9 = 1
2x^2 - 4x^2 - (64x^3)/9 = 1


I think it is easy from here on

Answer 21

make all denominators 9?

Answer 22

or remove the denominator by multiplying 9 on both sides of the equation

Answer 23

yes i did that

Answer 24

i got -18x^2 - 64x^3 = 9 and -2x^2 (9+32x) = 9

Answer 25

and now i'm solving for x by doing -2x^2 = 9 and 9+32x=9

Answer 26

x^2 = -9/2 ... i don't know what x will be when i put a radical sign over -9/2. for 9+32x = 9 i got x = 0...

Answer 27

why?

Answer 28

what do you mean... why?

Answer 29

you factored it out ?

Answer 30

ah.... you know what! i give up.... :'(

Answer 31

i dont think i can solve this problem :'((

Answer 32

the factoring is the most tedious part of the problem

Answer 33

what about you?

Answer 34

x = -3/32-3/32 (-1)^(2/3) (3/(259-16 sqrt(262)))^(1/3)+1/32 3^(2/3) (16sqrt(262)-259)^(1/3)
or
x = -3/32+3/32 (-3/(259-16 sqrt(262)))^(1/3)-1/32 (-1)^(1/3) 3^(2/3) (16 sqrt(262)-259)^(1/3)
or
x = -3/32-3/32 (3/(259-16 sqrt(262)))^(1/3)+1/32 (-3)^(2/3) (16 sqrt(262)-259)^(1/3)

Answer 35

oh my gosh... what is that... haha

Answer 36

I don't even want to know

Answer 37

lol..... :)

Answer 38

thanks for your help! sooooo much!! i probably took too much of your precious time..

Answer 39

i thnk it's time for us to move on with this problem..

Answer 40

I'll handle this later

Answer 41

haha, you don't have to :) you already did a lot for me :)

Answer 42

thanks again! i'm closing the question.

Answer 43

k