Prove the orthogonality of these curves

Question

I am having trouble proving the orthogonality of the following curves: $$\begin{cases} x = \frac{1}{2}(v_1^2 - v_2^2), v_1=\text{const} \\ y = v_1v_2, v_2=\text{const} \end{cases}$$

Here's everything I know (and tried):

I've been told that in order to check the orthogonality of the given curves, for this problem we need to see if using the formula for scalar multiplication of gradients $(\text{grad}x, \text{grad}y)$ we get $0$. However, I am having trouble actually setting up the gradients even before scalar multiplication. Any ideas on how to derive the gradients here?

$$∇f=(∂x/∂f, ∂y/∂f)$$ How exactly do we find $∂x/∂f$ and $∂y/∂f$ for the system with constants?

Answer 1

As it's written, the problem is very confusing.

Here is my first interpretation: The curves are obtained by setting $x$ and $y$ constant. Then differentiate with respect to $v_1$ implicitly:

\begin{align*} 0 &= v_1 - v_2\,\dfrac{dv_2}{dv_1} \\ 0 &= v_2 + v_1\,\dfrac{dv_2}{dv_1}. \end{align*}

So you have slope $\dfrac{dv_2}{dv_1}$ equal to $\dfrac{v_1}{v_2}$ for the first curve and $-\dfrac{v_2}{v_1}$ for the second curve. They are indeed negative reciprocals.

Here is what is probably the intended interpretation: The $v_1$ and $v_2$ constant should not be adjacent to $x$ and $y$ at all, but describe how we obtain the curves from $x$ and $y$ as functions of $v_1$ and $v_2$.

We have $x=\frac12v_1^2-\frac12v_2^2$ and $y=v_1v_2$. We obtain the first curve by setting $v_1$ constant and varying $v_2$; so $$\frac{dx}{dv_2} = -v_2, \quad \frac{dy}{dv_2} = v_1.$$ For the second curve we set $v_2$ constant and vary $v_1$; so $$\frac{dx}{dv_1} = v_1, \quad \frac{dy}{dv_1} = v_2.$$ The first curve has slope $$\frac{dy}{dx} = \frac{\frac{dy}{dv_2}}{\frac{dx}{dv_2}} = \frac{v_1}{-v_2},$$ while the second has slope $$\frac{dy}{dx} = \frac{\frac{dy}{dv_1}}{\frac{dx}{dv_1}} = \frac{v_2}{v_1}.$$ These slopes are indeed negative reciprocals.

Answer 2

Your two (families of) curves seem to be $$C_r:c(s)=((r^2-s^2)/2,rs)$$ and $$\Gamma_s:\gamma(r)=((r^2-s^2)/2,rs).$$ At some point $((r^2-s^2)/2,rs)\in C_r\cap\Gamma_s$, the tangent to $C_r$ is directed by $c'(s)=(-s,r)$ and the tangent to $\Gamma_s$ is directed by $\gamma'(r)=(r,s)$.

These two vectors are orthogonal.