Problem in drawing a particular kind of curve

Question

Let be $u:\mathbb{R^+}\rightarrow\mathbb{R} $ a continous function such that u(x)

I consider the curve $\gamma(x)=(u(x),u'(x))$.

I want to draw this curve in the plane $t=u(x)$, $p=u'(x)$.

I have found a book with a picture of a general curve of this form and it shows that for a value of $t$ can be found two different values of p, one positive and one negative.

I don't understand this fact. I was convinced that the immagine of the curve was something like the graph of function: for a value of t I can find one and only one value of p.

Why can I find two values of p in correspondence of a t and no only one?

Thank you for the clarification!

Answer 1

Consider the function defined by $u(x) = x^2$.

From the parameter value $x=1$, we find that $t = u(x) = 1$ and $p = u'(x) = 2.$

But if $x=-1$, we find that $t = u(x) = 1$ and $p = u'(x) = -2.$

When we plot all values of $(u(x),u'(x))$ for $-1\leq x \leq 1$, we will therefore plot two points at the same value of $t,$ one at $t=1, p=2$ and one at $t=1, p=-2.$

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An even more interesting example is when we define $u(x) = \cos(x)$.

Then $u'(x) = -\sin(x)$, and the points $(t,p) = (u(x),u'(x)) = (\cos(x),-\sin(x))$ for $x \in \mathbb R$ are a unit circle in the $t,p$ plane.

Answer 2

Your curve represents an object moving on a straight line. The first component is the position and the second is the velocity. If I'm walking back and forth on a soccer field in a straight line, the first time I cross the midfield line my velocity will be positive. As I walk back in the other direction my velocity will be negative.