How Reparametrization Changes Speed

One Geometric Path, Different Traversal Rates

A parametrized curve describes points in the plane as a parameter varies. The set of points traced out is its geometric trace; the parameter also determines how quickly, and in which direction, those points are visited. A single geometric path can therefore have several parametrizations with different speeds.

For example, traversing a line segment in one second and traversing it in half a second produces the same trace, but not the same speed. Changing the parameter lets us describe this difference precisely. The key is to track both the original curve and how quickly the new parameter moves through the original parameter values.

Definition and Speed Rule

Let $\mathbf r(t)=(x(t),y(t))$ be a differentiable parametrization, with $t$ in an interval $I$. Its velocity is $\mathbf r'(t)=(x'(t),y'(t))$, and its speed is the magnitude of that velocity:

v(t)=\|\mathbf r'(t)\| =\sqrt{(x'(t))^2+(y'(t))^2}.

Now let $\phi$ be a differentiable function from an interval $J$ into $I$. Replacing $t$ by $t=\phi(u)$ gives a new parametrization $\widetilde{\mathbf r}(u)=\mathbf r(\phi(u))$. The variable $u$ is the new parameter; $\phi(u)$ tells us which original parameter value is being used at $u$.

Definition: The parametrization $\widetilde{\mathbf r}(u)=\mathbf r(\phi(u))$ is obtained by a change of parameter $\phi$. Its velocity and speed are $$ \widetilde{\mathbf r}'(u)=\mathbf r'(\phi(u))\phi'(u), \qquad \widetilde v(u)=\|\mathbf r'(\phi(u))\|\,|\phi'(u)| =v(\phi(u))|\phi'(u)|. $$

The factor $v(\phi(u))$ is the original speed at the corresponding point on the curve. The factor $|\phi'(u)|$ measures how quickly the original parameter value changes as $u$ changes. Since speed is nonnegative, the absolute value is essential.

Why the Speed Rule Holds

The rule follows directly from the chain rule for each coordinate. Write $\mathbf r(t)=(x(t),y(t))$. Then the new coordinates are $x(\phi(u))$ and $y(\phi(u))$, so

\frac{d}{du}x(\phi(u))=x'(\phi(u))\phi'(u), \qquad \frac{d}{du}y(\phi(u))=y'(\phi(u))\phi'(u).

Thus the new velocity vector is $\phi'(u)$ times the original velocity vector evaluated at $t=\phi(u)$. Taking its magnitude gives

\begin{aligned} \widetilde v(u) &=\sqrt{\bigl(x'(\phi(u))\phi'(u)\bigr)^2+ \bigl(y'(\phi(u))\phi'(u)\bigr)^2}\\ &=|\phi'(u)|\sqrt{\bigl(x'(\phi(u))\bigr)^2+ \bigl(y'(\phi(u))\bigr)^2}\\ &=v(\phi(u))|\phi'(u)|. \end{aligned}

This also explains the difference between velocity and speed. The sign of $\phi'(u)$ affects the direction of the velocity vector, while speed uses its magnitude and so includes $|\phi'(u)|$.

Worked Example: A Linear Change of Parameter

Worked Example: Doubling the Traversal Rate

Let $\mathbf r(t)=(3t,4t)$ for $0\le t\le1$, and use the change of parameter $t=\phi(u)=2u$ for $0\le u\le\frac12$. Find the new parametrization and compare speeds.

First, substitute $2u$ for $t$:

\widetilde{\mathbf r}(u)=\mathbf r(2u)=(6u,8u).

The original velocity is $\mathbf r'(t)=(3,4)$, so its speed is

v(t)=\sqrt{3^2+4^2}=5.

The new velocity and speed are

\widetilde{\mathbf r}'(u)=(6,8), \qquad \widetilde v(u)=\sqrt{6^2+8^2}=10.

The formula gives the same result: $\phi'(u)=2$, so $v(\phi(u))|\phi'(u)|=5\cdot2=10$. Both parametrizations trace the line segment from $(0,0)$ to $(3,4)$: the new one starts at $(0,0)$ and ends at $\widetilde{\mathbf r}(\frac12)=(3,4)$. The speed has doubled while the trace is unchanged.

Check: differentiating $(6u,8u)$ gives $(6,8)$, whose magnitude is $10$, as calculated. The endpoints also agree with those of the original segment.

Linear Changes and Direction of Traversal

A linear change of parameter has the form $\phi(u)=au+b$, where $a$ and $b$ are constants. Its derivative is the constant $a$, so the new speed is $|a|$ times the original speed at the corresponding point. If $a>0$, the original parameter increases as $u$ increases; if $a<0$, it decreases. The latter reverses the direction of traversal when the change covers the same interval.

Worked Example: Reversing the Direction Without Changing the Trace

Consider $\mathbf r(t)=(t,t^2)$ for $0\le t\le1$. Set $t=\phi(u)=1-u$ for $0\le u\le1$. Find the new parametrization, its speed, and the direction of traversal.

Substitution gives

\widetilde{\mathbf r}(u) =\mathbf r(1-u) =(1-u,(1-u)^2).

The original velocity and speed are

\mathbf r'(t)=(1,2t), \qquad v(t)=\sqrt{1+4t^2}.

Since $\phi'(u)=-1$, the new velocity and speed are

\widetilde{\mathbf r}'(u)=(-1,-2(1-u)), \qquad \widetilde v(u)=\sqrt{1+4(1-u)^2}.

The speed formula confirms this: $v(\phi(u))|\phi'(u)|=\sqrt{1+4(1-u)^2}\cdot1$. At $u=0$, the new curve is at $(1,1)$; at $u=1$, it is at $(0,0)$. The original curve visits those endpoints in the opposite order, so the trace is the same but the direction is reversed.

Check: differentiating each coordinate of $(1-u,(1-u)^2)$ gives $-1$ and $-2(1-u)$. Their squared magnitudes sum to $1+4(1-u)^2$, as required.

Nonlinear Changes Can Vary the Speed

If $\phi'(u)$ varies, the speed factor can vary along the path. At a corresponding point where the original speed is positive, a factor $|\phi'(u)|>1$ makes the new speed greater than the original speed there; a factor $0<|\phi'(u)|<1$ makes it smaller. If $\phi'(u)=0$, the new speed is zero at that value of $u$. These comparisons require positive original speed: if the original speed is zero, multiplying it by any finite factor still gives zero.

Worked Example: A Nonlinear Change Speeds Up and Slows Down

Let $\mathbf r(t)=(t,t^2)$ for $0\le t\le1$, and take $t=\phi(u)=u^2$ for $0\le u\le1$. Find the new parametrization and compare speeds along the interval.

The new parametrization is

\widetilde{\mathbf r}(u)=\mathbf r(u^2)=(u^2,u^4).

Its velocity and speed, computed directly, are

\widetilde{\mathbf r}'(u)=(2u,4u^3), \qquad \widetilde v(u)=\sqrt{4u^2+16u^6} =2u\sqrt{1+4u^4},

where $u\ge0$ has been used to write $\sqrt{4u^2}=2u$. For the original curve, $v(t)=\sqrt{1+4t^2}$. Also $\phi'(u)=2u$, so the speed rule gives

v(u^2)|\phi'(u)| =\sqrt{1+4u^4}\,(2u) =2u\sqrt{1+4u^4}.

For $0<u<\frac12$, the factor $2u$ is between zero and one, and the original speed is positive, so the new speed is lower at the corresponding point. For $\frac12<u\le1$, the factor is greater than one, so the new speed is higher. At $u=0$, the factor is zero and the new speed is zero. The trace is preserved because $u^2$ increases from $0$ to $1$ and covers the original parameter interval.

Check: $\widetilde{\mathbf r}(0)=(0,0)=\mathbf r(0)$, and $\widetilde{\mathbf r}(1)=(1,1)=\mathbf r(1)$. The two speed calculations agree for every $u$ in the interval.

Trace, Orientation, and Pauses

The trace of $\mathbf r$ on $I$ is the set $\{\mathbf r(t):t\in I\}$. The new curve has trace $\{\mathbf r(\phi(u)):u\in J\}$, which is the part of the original trace corresponding to parameter values in $\phi(J)$. Consequently, the entire original trace is preserved if $\phi(J)=I$. A continuous, strictly increasing change onto $I$ preserves the order of traversal; a continuous, strictly decreasing change onto $I$ reverses it. If a change is not one-to-one, it may cause points to be revisited, even when it still covers the same trace.

A zero value of $\phi'(u)$ means the new parametrization has zero velocity at that instant. It does not, by itself, imply that the trace has changed or that the curve has reached an endpoint. The next example shows how the parameter change can pause at an interior point.

Worked Example: A Change of Parameter Creates a Pause

Let $\mathbf r(t)=(t,0)$ for $-1\le t\le1$, and set $t=\phi(u)=u^3$ for $-1\le u\le1$. Find the new speed and identify the pause.

The new parametrization is $\widetilde{\mathbf r}(u)=(u^3,0)$, so

\widetilde{\mathbf r}'(u)=(3u^2,0), \qquad \widetilde v(u)=3u^2.

The original velocity is $(1,0)$, giving $v(t)=1$. Since $\phi'(u)=3u^2\ge0$, the rule gives $v(u^3)|\phi'(u)|=1\cdot3u^2=3u^2$, in agreement with direct differentiation. At $u=0$, the speed is zero and the point is $(0,0)$; for $u\ne0$, the speed is positive. The change $u^3$ increases from $-1$ to $1$ and covers the whole original parameter interval, so the segment is traversed in the same direction, with an instantaneous stop at its midpoint.

Check: the endpoints are $\widetilde{\mathbf r}(-1)=(-1,0)$ and $\widetilde{\mathbf r}(1)=(1,0)$, as for the original segment. Differentiating $(u^3,0)$ verifies the stated velocity and speed.

A Common Mistake: Ignoring the Absolute Value

Suppose $\phi'(u)=-1$. A tempting but incorrect statement is that the new speed equals $-v(\phi(u))$. That expression is negative, but speed cannot be negative. For the reversing change $t=1-u$ in the earlier example, the incorrect calculation would give $-\sqrt{1+4(1-u)^2}$. The correct calculation is

\widetilde v(u)=v(1-u)|-1| =\sqrt{1+4(1-u)^2}.

The negative sign belongs to the velocity’s direction, not to the speed. Another related error is to say that a factor greater than one always makes the new speed greater. The comparison only holds when the original speed at that point is positive. For instance, if $\mathbf r(t)=(t^2,0)$ and $\phi(u)=2u$, then at $u=0$ the original speed $v(0)$ is zero and $|\phi'(0)|=2$, but the new speed is still $0\cdot2=0$, not greater than zero.

Why Reparametrization Matters

Reparametrization separates the shape of a path from the way it is traversed. The same trace may be covered at constant speed, at a changing speed, in the reverse direction, or with a pause, depending on the chosen parameter change. This distinction is useful whenever a problem asks about motion or traversal rate rather than only about the set of points on the curve.

The practical procedure is short: substitute $t=\phi(u)$ to obtain the new curve, differentiate using the chain rule, and take the magnitude to find speed. Then check the range and monotonicity of $\phi$ to determine which part of the original trace is covered and whether traversal direction changes. Keep the two questions separate: the speed formula describes the rate at each corresponding point, while the range and behavior of $\phi$ describe how the curve is traced.

Key takeaway: Under a differentiable change $t=\phi(u)$, the new speed is the original speed at $t=\phi(u)$, multiplied by $|\phi'(u)|$. The trace is preserved when the new parameter covers the full original parameter interval; the sign of $\phi'$ determines whether traversal direction is maintained or reversed when $\phi$ is monotone.

Check Your Understanding

Use the speed rule and the behavior of the parameter change to answer these questions.

If the original speed at a point is $7$ and $|\phi'(u)|=3$, what is the new speed at the corresponding parameter value?
For $t=1-u$ on $0\le u\le1$, does the direction of traversal stay the same or reverse? What is the speed factor?
If $v(\phi(u))=0$ and $|\phi'(u)|=5$, what is the new speed? Explain why a factor greater than one does not make it greater.
For $t=u^2$ on $0\le u\le1$, does the change cover the full interval $0\le t\le1$? At what value of $u$ is the speed factor zero?
Why is the absolute value of $\phi'(u)$ used in the speed formula rather than $\phi'(u)$ itself?