Show the Curves 2x^2+3y^2=5 and y^2=x^2 are orthogonal. I'm less interested in the answer and more interested in how to do it. Please help. Thank you

Question

Show the Curves 2x^2+3y^2=5 and y^2=x^2 are orthogonal.  I'm less interested in the answer and more interested in how to do it.  Please help.  Thank you

amistre64 · Answer

interesting .... im sure it has to do with their derivatives being orthogonal at all points (x,y)

amistre64 · Answer

there is also something i recall that an inner product equal to 0 defines orthogonal functions

anonymous · Answer

i wrote down in my notes that I first have to find the derivative of 2x^2+3y^2=5?  which ends up being: -2x/3y.  How is that?

amistre64 · Answer

implicit derivative

amistre64 · Answer

$$D_x[2x^2 + 3y^2 = 5]$$
$$D_x[2x^2] + D_x[3y^2] = D_x[5]$$
$$2D_x[x^2] + 3D_x[y^2] = 0$$
$$4xD_x[x] + 6yD_x[y] = 0$$
$$4x + 6y~y' = 0$$
$$y' = -\frac{4x}{6y}$$

anonymous · Answer

how did you go from 4x+6yy'=0 to y' = -4x/6y

anonymous · Answer

Orthogonal curves have orthogonal/perpendicular tangent lines at every point they intersect. Recall that for two perpendicular lines with slopes $m_1,m_2$ we have $m_1m_2=-1\Leftrightarrow m_2=-\frac1{m_1}\Leftrightarrow m_1=-\frac1{m_2}$. Now we just need to show that $y'_1y'_2=-1$.
First, we find $y'_1$ by implicit differentiation (w.r.t $x$):$$\frac{d}{dx}[2x^2+3y_1^2]=\frac{d}{dx}[5]\4x+6y_1y'_1=0\6y_1y'_1=-4x\y'_1=-\frac{4x}{6y_1}=-\frac{2x}{3y_1}$$We do something similar for $y'_2$:$$\frac{d}{dx}[y_2^2]=\frac{d}{dx}[x^2]\2y_2y'_2=2x\y'_2=\frac{2x}{2y_2}=\frac{x}{y_2}$$Now we need to show that for any time they intersect, i.e. $y_1=y_2$, that $y_1'y_2'=-1$... unfortunately:$$y_1'y_2'=-\frac{2x}{3y}\times\frac{x}y=-\frac{2x^2}{3y^2}$$Doesn't seem orthogonal to me.

anonymous · Answer

Ah I see that makes sense thanks for all the baby steps.  And what would be the ordered pair then?

anonymous · Answer

@Andysebb what ordered pair? where they intersect? Just substitute.$$y^2=x^2\3y^2=3x^2$$so$$2x^2+3y^2=5\2x^2+3x^2=5\5x^2=5\x^2=1$$so they will intersect at $(\pm1,\pm1)$.

anonymous · Answer

Notice that $y^2=x^2$ which means our earlier expression for $y_1'y_2'$ can be simplified to just $-\frac23$, which means they're not orthogonal.

amistre64 · Answer

4x+6yy'=0

solve for y' by the most basic of mathematical procedures .. subtract 4x from both sides

6yy' = -4x

then divide off the 6y

y' = -4x/6y

and reduce ....