derivatives and optical illusions

[Guest]

I have this problen where I need to find (dy/dx) for the indicated values of x and y.

Hyperbolas: xy=c where x=1 y=4 c=4

i took the der. of both sides and got xy' + yx'=c'
but if I substitute the expression in for c', stick in the values and then solve for y' I'm getting 0/0. And is there anyway to check if your answers for derivatives are right?

 

[stapel]

I think "c" is supposed to be a constant. So you've got:

. . . . .xy = c

. . . . .y = c/x

. . . . .y = cx^-1

Then:

. . . . .dy/dx = c(-1)x^-2

. . . . .dy/dx = -c/x²

Plugging in the initial x- and y-values, we get:

. . . . .(1)(4) = c = 4

So:

. . . . .y = 4/x

...and:

. . . . .dy/dx = -4/x²

At x = 1, we get:

. . . . .dy/dx = -4/(1)² = -4

Unless I'm missing something (and I could be), this is all they're looking for. I have no idea how any of this might relate to optical illusions (mentioned in your subject line).

Eliz.

 

[soroban]

Hello, Assunta!

I have this problen where I need to find (dy/dx) for the indicated values of x and y.

Hyperbolas: xy = c, where x = 1, y = 4, c = 4.

i took the der. of both sides and got: xy' + yx' = c' . . . . no!

The derivative of x is 1 . . . x' = 1
The derivative of a constant is zero . . . c' = 0

You have: .xy' + y .= .0 .---> .y' .= .-y/x