Implicit Differentiation: Show two curves to be orthogonal

[Jakotheshadows]

Show that 2x^2 +y^2 = 3 and x = y^2 are orthogonal.

Work:

d/dx(2x^2 + y^2) = d/dx(3) and d/dx(x) = d/dx(y^2)
4x + (dy/dx)*2y = 0 and 1 = 2y*(dy/dx)
(dy/dx)*2y = -4x and 1/2y = dy/dx (Looks like I killed the x :? )
(dy/dx) = -2x/y
These derivatives don't appear to be opposite reciprocals. Please tell me if my differentiation is wrong.

I investigated further to see where the two curves might intersect:
(I'm assuming this works with implicit functions)
2x^2 + y^2 - 3 = y^2 - x
2x^2 + x - 3 = 0 --> (-6) = (-2)*3
(2x^2 - 2x) + (3x - 3) = 0
2x(x -1) + 3(x -1) = 0
(2x+3)(x-1)=0
2x+3 = 0
x = -3/2
x-1 = 0
x = 1

Substituting x values into the derivatives:
1/2y and -2/y (for x = 1) doesn't look like an opposite reciprocal..
1/2y and [-2(-3/2)]/y = 3/y definitely not an opposite reciprocal..

 

[BigGlenntheHeavy]

\(\displaystyle 2x^{2}+y^{2} = 3\ and\ x = y^{2}\)

\(\displaystyle y^{2} = 3-2x^{2}\ and\ y^{2} = x\)

\(\displaystyle x= 3-2x^{2}, 2x^{2}+x-3=0, (2x+3)(x-1)=0,\ when\ x=1,y={+or-1},\ when\ x=-\frac{3}{2},y \ is\ undefined\)

\(\displaystyle y' = \frac{-2x}{y}, y' = \frac{1}{2y}\)

\(\displaystyle (1,1) \ y' =-2, y'=\frac{1}{2}.\ (-2)(\frac{1}{2} )= -1\)

\(\displaystyle (1,-1) \ y'=2, y'=\frac{-1}{2}.\ (2)(\frac{-1}{2})=-1\)

which implies the tangent lines at these two points are perpendicular.

 

[Jakotheshadows]

\(\displaystyle 2x^{2}+y^{2} = 3\ and\ x = y^{2}\)

\(\displaystyle (1,1) \ y' =-2, y'=\frac{1}{2}.\ (-2)(\frac{1}{2} )= -1\)

\(\displaystyle (1,-1) \ y'=2, y'=\frac{-1}{2}.\ (2)(\frac{-1}{2})=-1\)

which implies the tangent lines at these two points are perpendicular.

Ah! Thank you. I need to work on interpreting my own work. I pretty much did the same things you did, except for thinking to figure out which of the values I found for x made sense for y, finding y for those values of x that do, and substituting those into the derivatives I found.

 

[BigGlenntheHeavy]

You're welcome, glad to be of help.

 

[mliuall]

You could also multiply the two derivatives and substitute y² for x, and then you end up with -1.