Question

what is the slope of the tangent line to the curve (x^3 + y)^2 -2a^2xy = b^2 at the point P(0,1) when a=(2)^(1/2) and b = 1

Answer 1

@ganeshie8 @Robert136 @nincompoop

Answer 2

I'm getting DNE but its wrong

Answer 3

For this problem can you input the values for a and b, then find the derivative? Or does the inputting of values happen after differentiating..

Answer 4

It looks really ugly if you differentiate it along with a and b :\

Answer 5

I would use implicit differentiation to find the derivative.

Answer 6

i really have no idea , iv imputed the values for a and be and then solved for y and derived that , but that seems to be wrong

Answer 7

Yeah it is hahahahaha

Answer 8

Because that's basically stating the tangent line is found at a point before even being found.

Answer 9

you only need to find f'(x,y)

Answer 10

but before u start it says when a=(2)^(1/2) and b = 1 , a and b seems like small values and you can sub them why it mention them in ur question ? seems like there is a formula in ur book or something

Answer 11

\[\vec n = <2(x^3+y)3x^2 - 2a^2y, 2(x^3+y)-2a^2x>\]
\[= <6x^2(x^3+y) - 2a^2y, 2(x^3+y)-2a^2x>\]

x = 0

\[= <-2a^2y,2>\]

\[y = 1, a = \sqrt{2}\]

\[\vec n = <-4,2,>\]

Answer 12

don't solve for y
just find the derivative of both sides w.r.t. x
a and b are constants
you can input those values in later or not
once you have differentiated both sides
you can input the values for a and b if you haven't already
and then also input 0 for x and 1 for y
then solve for y' to find the slope.

Answer 13

see thats what i'm talking about :D like @irishboy123