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(I know the title isn’t clear. I didn’t know how to best describe my question in the title)

So, we have a position-time relation : $\vec r$ = $2t\hat i$ + $4t^2\hat j$

It shows that the particle is initially at $x$ = $0$, $y$ = $0$

Its corresponding velocity vector is : $\vec v$ = $2\hat i$ + $8t\hat j$.

It means its $v$ along $x-axis$ is constant throughout the motion of the particle.

Its trajectory equation is $y$ = $x^2$, which is a parabola symmetric about $y-axis$ (although we’d only use the part of the curve on the right side of $+ve$ $y-axis$)

From its $y$ vs $x$ curve, I tried to obtain its $v_y$ vs $v_x$ relation, and I got : $v_y$ = $2xv_x$

Now if I put any value of $x$ (particle’s position) and its velocity along $x-axis$ at that value of $x$, I correctly get the value of $v_y$. For example, the particle was at $x$ = $2$ at $t$ = $1$, and its $v_x$ at $t$ = $1$ was $2$ units. Substituting these values I get $v_y$ = $8$ units, which is true, as we can verify from the velocity-time relation that its $v_y$ was $8$ units at $t$ = $1$.

I tried other values of $x$ & $v_x$, or $x$ & $v_y$, and I always seemed to get correct values of $v_y$, and $v_x$. Except for when I tried calculating $v_x$ when the particle was at $x=0$ at $t$ = $0$.

Since $v_x$ = $\frac{v_y}{2x}$, if I put $x$ = $0$ and $v_y$ = $0$, I end up with an indeterminate form $\frac{0}{0}$. ($v_y$ was $0$ at $x$ = $0$)

Shouldn’t the $v_y$ vs $v_x$ relation show that $v_x$ is equal to $2$ units at $x$ = $0$? Actually $v_x$ is constant, and equal to $2$ units at any instant, according to the velocity-time relation.

Am I doing something wrong?

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    $\begingroup$ It is possible to find the limit of an indeterminate form : $\underset{t\to 0}{\mathop{\lim }}\,\frac{8t}{4t}=2$ $\endgroup$
    – Vincent Fraticelli
    Commented Feb 9, 2019 at 16:02
  • $\begingroup$ I haven't studied limits yet. Have studied a bit of calculus (derivates, integrals, etc) that is required at the beginning of our physics syllabus. I know a little bit about limits, as in instantaneous velocity, instantaneous acceleration etc, but haven't studied it yet. How did that $\frac{8t}{4t}$ come out to be equal to 2? $\endgroup$
    – 4d_
    Commented Feb 9, 2019 at 16:10
  • $\begingroup$ You are speaking of the limit of $\frac{{{v}_{y}}}{2x}=\frac{8t}{4t}$ ? $\endgroup$
    – Vincent Fraticelli
    Commented Feb 9, 2019 at 16:13
  • $\begingroup$ Yeah, that's what I want to understand. Why is it equal to 2? $\endgroup$
    – 4d_
    Commented Feb 9, 2019 at 16:14
  • $\begingroup$ I think it's very intuitive, even if you have not studied the limits. If $t$ is different from 2, as small as it is, the ratio is 2. So "at the limit" the ratio is 2 ! $\endgroup$
    – Vincent Fraticelli
    Commented Feb 9, 2019 at 16:22

2 Answers 2

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That point you mention, at $t=0$, as being a problem, arises from the relation $Ux=Uy/2x$ which can not be inferred unless x#$0$. Since the tranjectory equation you start up is $y=x^2$, then it is always (for any t) true that $Uy=2xUx$, but in order to get Ux you have to divide both sides by $2x$. That division is permitted only if x#$0$. So, if you consider the relation that is true for any t, that is the relation $Uy=2xUx$, then at $t=0$ it is $x=0$ and even if $Ux$ # $0$ the last relation yields $Uy=0$ which it should be true for $t=0$.

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You are not doing anything wrong. But you are doing something deliberately stupid.

When you get a result 0/0 that says "You can't get the answer this way!" But it leaves open the possibility that the answer can be obtained another way.

It's like starting with an equation like $ x=3$. Multiply both sides by $y$ and get $xy=3y$. Divide both sides by $y$ and get $x=3y/y$. If you mischievously set $y=0$ you get $0/0$ for $x$. But that doesn't invalidate the rules from other values.

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  • $\begingroup$ Thank you, Sir. Yeah I blindly assumed that I could directly put zero for $x$ and for $v_y$ in $\frac{v_y}{2x}$, or for $t$ in $\frac{8t}{4t}$. Apparently, I have to use limits here which I haven’t studied yet (still at school). Which is why I blindly put zero in the fractions. Now I understand it better, but I will understand it fully only when I study limits, I guess $\endgroup$
    – 4d_
    Commented Feb 10, 2019 at 3:15

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