When $x$-component of velocity is extremum in motion in a plane?

While revising the topic, "Motion in a plane" from "Concepts of Physics by HC. Verma", I had the following question in my mind.

Question:~ When x- component of velocity of a particle is extremum provided that it travels at uniform velocity along x-direction in X-Y plane or in other words $$a_x$$(x-component of acceleration is $$0$$), i.e $$\overrightarrow a = a_y \hat j$$?

In order to answer this question, I've attempted as follows,

My answer:~ $$\overrightarrow v = v_x \hat i + v_y \hat j$$ ...(Velocity vector) $$\overrightarrow v = \frac{dx}{dt} \hat i + \frac{dy}{dt} \hat j$$ If we represent $$|\overrightarrow x|$$ as magnitude of $$\overrightarrow x$$, then

$$|\overrightarrow v| = v = \sqrt[]{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}$$ $$vdt= \sqrt[]{dx^2 + dy^2}$$ $$vdt = dx\left(\sqrt[]{1+y'^{2}}\right)$$ $$v = v_x \left ( \sqrt[]{1+y'^{2}}\right)$$ Here, $$v_x = \frac{v}{\sqrt[]{1+y'^{2}}}$$ is $$x$$-component of velocity. Now we need to make $$v_x$$ extremum with respect to x coordinate since we have information that particle travels at uniform velocity along x-direction, i.e $$\frac{\partial v}{\partial x} = 0$$. We set, $$\frac{d v_x}{d x} = 0$$ to make $$v_x$$ extremum. We then have, $$\frac{y"y'}{1 + y'^2} = 0$$

From, here three possibilities arise.

1. $$y" = 0$$
2. $$y' = 0$$
3. $$y' = Undefined$$

My main query is as follows.

If we consider a projectile motion, which is a nice example of motion in the plane with constant acceleration, we know that at maximum height, $$x$$- component of velocity is maximum. In that case $$\overrightarrow a = -g \hat j$$. Hence, in this case, $$y' = 0$$ is a valid case of $$x$$-component of velocity to be maximum, and $$y" = 0$$ represents the case for which $$x$$-component is minimum at boundaries. But, then what $$y' = undefined$$ represents. Or more specifically, what is an example that includes $$y' = undefined$$ be the condition for x-component of velocity to be minimum?

• If the x component of velocity is constant its maximum and minimum are equal to that constant value. This is independent of any motion in the y direction. The values of y, y' and y'' are not defined by the value of x; they must be defined separately. Jul 28 '21 at 10:03

Using the word extremum for the x-component of velocity is misleading in this context, since you have stated that the x-component of velocity is constant. The statement that $$v_x$$ is constant immediately implies that any derivative of $$v_x$$ is zero. It also implies that $$v_y$$ and $$y'$$ are proportional, as $$v_y=y'v_x$$.
When you differentiate both sides of $$v = v_x \left ( \sqrt[]{1+y'^{2}}\right)$$ with respect to $$x$$ you should get $$v'=v_x \frac{y'y''}{\sqrt{1+y'^2}}$$ While you did not state explicitly that you are considering the situation where the speed is extremal, this is what you have done by putting $$v'=0$$. The following are the conditions when the speed is extremal.
If $$y'=0$$ then $$v_y=y'v_x=0$$; any change in $$v_y$$ will increase the speed.
If $$y''=0$$ then $$v_y$$ is a maximum or a minimum; any change in $$v_y$$ will change the speed.
In the limiting case where $$\frac{1}{\sqrt{1+y'^2}}\to0$$, we would have $$v_y\to\infty$$, but in this case $$\frac{y'}{\sqrt{1+y'^2}}\to1$$ and so we would get $$v'/y''\to 1$$. This would be a situation where the speed increases to infinity, but at a rate (acceleration) that decreases to zero. Note that we do not need to worry about the possibility of an imaginary solution because $$\sqrt{1+y'^2}\ge 1$$ for all real $$y'$$.