Question

[Calculus] The slope of the tangent is -1 at the point (0,1) on x^3 - 6xy - ky^3 = a , where k and a are constants. The values of the constants a and k are what?

Answer 1

Set f' (0,1) = -1

Answer 2

@Chlorophyll what would that look like?

Answer 3

Take dy/ dx -> y'

Answer 4

ok so 3x^2- 6x(y')+y(6)-2y^2 (y')?

Answer 5

hmmm... i think it might be:
\[3x ^{2} - 6xy \prime + 6y - 3ky ^{2} y \prime = 0\]

Answer 6

oh yeah i meant 3 not 2

Answer 7

Yes, almost correct, just fix:
-6 y

Answer 8

so the k stays in there?

Answer 9

Of course (ax)' = a

Answer 10

ok, what's next?

Answer 11

Factorize to isolate y' = ...

Answer 12

Okay, can you do that? :) pleaseeeeeeeee!

Answer 13

y'=-3x^2-6y/6x-3ky

Answer 14

*3ky^2

Answer 15

y' = ( x² - 2y) / ( -2x + ky²)

Answer 16

you simplified, right?

Answer 17

Yes, just make it neat!
Now plug value y' = -1 at the point ( 0, 1) to obtain k!

Answer 18

k=2?

Answer 19

Yup :)
Now plug k and the point back => a

Answer 20

Do we plug k and the point back into the original equation to get a? :O

Answer 21

That's only place you have a, right ;)

Answer 22

Questions, comments?

Answer 23

okay so a would equal -2? :)

Answer 24

Beautifully done :)

Answer 25

Beautifully explained by my fellow Chloroform mate :)
Thank you! :)