Speed Along a Parametrized Curve

Motion Along a Parametrized Curve

A parametrization describes a point in the plane by giving its coordinates as functions of a parameter $t$: $x=x(t)$ and $y=y(t)$. When $t$ represents time, these functions describe motion. Their derivatives tell us how quickly each coordinate changes, but neither derivative alone tells us the total rate at which the point moves.

For example, a point can move quickly in both the horizontal and vertical directions at once. Its total speed combines those two coordinate rates. The resulting quantity is the instantaneous speed along the parametrized curve.

Definition: Suppose $x(t)$ and $y(t)$ are differentiable at $t$. The velocity vector is $\mathbf{v}(t)=(x'(t),y'(t))$. The instantaneous speed is the magnitude of this vector: $$ v(t)=\sqrt{(x'(t))^2+(y'(t))^2}. $$ Speed is nonnegative. It measures how quickly the point moves, not the direction in which it moves.

The coordinate derivatives $x'(t)$ and $y'(t)$ are the horizontal and vertical components of velocity. Squaring them, adding, and taking a square root gives the length of the velocity vector. Direction is carried by the vector itself; speed retains only its magnitude.

Calculating Speed from Coordinate Derivatives

To find speed at a specified parameter value, differentiate both coordinate functions, evaluate both derivatives at that value, and use the formula. Keep the parameter distinct from the coordinates: the derivatives describe the rate of coordinate change with respect to $t$.

Differentiate each coordinate.
Find $x'(t)$ and $y'(t)$.

Evaluate at the required parameter.
Substitute the specified value of $t$ into both derivatives.

Compute the magnitude.
Square the two derivative values, add them, and take the nonnegative square root.

Worked Example: Speed with Physical Units

A particle’s coordinates, measured in metres, are $x(t)=(3\text{ m/s}^2)t^2$ and $y(t)=(4\text{ m/s})t$, where $t$ is measured in seconds. Find its speed at $t=1\text{ s}$.

First differentiate with respect to time:

\frac{dx}{dt}=(6\text{ m/s}^2)t, \qquad \frac{dy}{dt}=4\text{ m/s}.

At $t=1\text{ s}$, the velocity components are $6\text{ m/s}$ and $4\text{ m/s}$. Therefore,

v(1) =\sqrt{(6\text{ m/s})^2+(4\text{ m/s})^2} =\sqrt{36+16}\text{ m/s} =\sqrt{52}\text{ m/s} =2\sqrt{13}\text{ m/s}.

To two decimal places, the speed is approximately $7.21\text{ m/s}$. As a check, the velocity vector at this time is $(6,4)\text{ m/s}$, whose magnitude is $\sqrt{6^2+4^2}=2\sqrt{13}\text{ m/s}$, as calculated.

Worked Example: Speed That Is Constant

Let $x(t)=5\cos(2t)$ and $y(t)=5\sin(2t)$, with coordinates measured in centimetres and $t$ in seconds. Find the speed at $t=\frac{\pi}{4}\text{ s}$, and determine whether the formula gives the same speed at other times.

Differentiate both coordinates:

x'(t)=-10\sin(2t)\text{ cm/s}, \qquad y'(t)=10\cos(2t)\text{ cm/s}.

At $t=\frac{\pi}{4}\text{ s}$, $2t=\frac{\pi}{2}$, so $\sin(2t)=1$ and $\cos(2t)=0$. Thus the velocity components are $-10\text{ cm/s}$ and $0\text{ cm/s}$, giving

v\left(\frac{\pi}{4}\right) =\sqrt{(-10)^2+0^2}\text{ cm/s} =10\text{ cm/s}.

For any $t$, the speed is

v(t)=\sqrt{100\sin^2(2t)+100\cos^2(2t)} =\sqrt{100\bigl(\sin^2(2t)+\cos^2(2t)\bigr)} =10\text{ cm/s},

using $\sin^2(2t)+\cos^2(2t)=1$. This also checks the value at $t=\frac{\pi}{4}\text{ s}$: substitution gives $10\text{ cm/s}$. The velocity direction changes, but its magnitude stays constant.

Why the Formula Measures Instantaneous Speed

Speed is a rate of distance traveled with respect to time. This is not always the same as the rate of change of the distance between the starting and ending positions: a path may bend or loop between those positions. To connect the definition of speed to the coordinate-derivative formula, consider a short parameter interval around $t$.

Write $\mathbf{r}(u)=(x(u),y(u))$, and let $L$ be the length of the path traveled as the parameter runs from $t$ to $t+h$. Assume $x$ and $y$ have continuous derivatives on an interval around $t$. The path length is at least the straight-line distance between its endpoints. Also, it is at most the integral of the magnitude of the velocity over the interval. The latter bound follows by splitting the interval into pieces: the distance between the endpoints of each piece is at most the integral of the velocity’s magnitude over that piece, and adding these bounds gives the result.

Why the speed formula works: For a curve with continuous coordinate derivatives near $t$, the distance traveled over a short interval, divided by the length of that interval, approaches the magnitude of the velocity at $t$. Thus instantaneous speed is $\sqrt{(x'(t))^2+(y'(t))^2}$.

Here is the squeeze argument in detail. The endpoint displacement gives a lower bound for $L$, while the velocity integral gives an upper bound. The inequalities apply whether $h$ is positive or negative, using the interval between the two parameter values:

\left\|\mathbf{r}(t+h)-\mathbf{r}(t)\right\| \leq L \leq \int_{\min(t,t+h)}^{\max(t,t+h)} \left\|\mathbf{r}'(u)\right\|\,du.

Divide by $|h|$. As $h$ approaches zero, the left-hand expression divided by $|h|$ approaches $\|\mathbf{r}'(t)\|$, by the definition of the derivative. The right-hand expression divided by $|h|$ also approaches $\|\mathbf{r}'(t)\|$: it is an average of the continuous function $\|\mathbf{r}'(u)\|$ over an interval shrinking to $t$. Therefore the middle expression has the same limit. Since $\mathbf{r}'(t)=(x'(t),y'(t))$, its magnitude is precisely the stated speed formula.

This argument explains the role of differentiability and continuity: the velocity must describe the motion on arbitrarily short intervals around the parameter value. At a parameter where a coordinate derivative does not exist, this formula does not define instantaneous speed there.

Distance Traveled Is Not Endpoint Displacement

For a nonzero parameter change $h$, the magnitude of the endpoint displacement divided by $|h|$ is a useful quantity, but it is not generally the average speed. It uses only the two endpoint positions and ignores what happens between them. Average speed over the interval is the length of the path traveled divided by the elapsed time (or, more generally, by the absolute parameter change when $t$ is the time parameter).

A full turn around a circle makes the distinction clear. For $x(t)=\cos t$ and $y(t)=\sin t$, starting at $t=0$ and ending at $t=2\pi$, the endpoint positions are both $(1,0)$. The endpoint displacement is zero, so its magnitude divided by $2\pi$ is zero. But the point travels one full circumference, which is $2\pi$ units. Its average speed over this interval is therefore $\frac{2\pi}{2\pi}=1$ unit per unit of time. The instantaneous-speed formula agrees: $x'(t)=-\sin t$, $y'(t)=\cos t$, so $v(t)=\sqrt{\sin^2t+\cos^2t}=1$.

Worked Example: Zero Instantaneous Speed

Consider $x(t)=t^3$ and $y(t)=t^2$. Find the speed at $t=0$ and $t=1$.

Differentiate the coordinates:

x'(t)=3t^2, \qquad y'(t)=2t.

At $t=0$, both components are zero, so

v(0)=\sqrt{(3\cdot 0^2)^2+(2\cdot 0)^2} =\sqrt{0+0} =0.

At $t=1$, the components are $3$ and $2$, giving

v(1)=\sqrt{3^2+2^2} =\sqrt{9+4} =\sqrt{13}.

To check, the velocity vectors at the two values are $\mathbf{v}(0)=(0,0)$ and $\mathbf{v}(1)=(3,2)$. Their magnitudes are $0$ and $\sqrt{13}$, respectively, matching the computed speeds. A zero speed at an instant means the velocity vector is zero there; it does not mean that the coordinate functions or the curve are constant everywhere.

Speed, Direction, and Regularity

Speed and velocity answer different questions. The velocity vector records both rates and direction. Speed is its nonnegative magnitude. For instance, reversing the direction of a velocity vector changes its components’ signs but not its magnitude.

A parameter value is regular when the derivative pair $(x'(t),y'(t))$ is not $(0,0)$. At a regular value, the speed is positive. If both derivatives vanish, the speed is zero, and the instantaneous velocity vector gives no nonzero direction of motion at that value. This observation identifies a stationary parameter value, but by itself it does not classify the shape of the curve there.

Coordinate derivatives	Speed	Interpretation
At least one of $x'(t)$, $y'(t)$ is nonzero	Positive	The point has a nonzero instantaneous velocity.
Both $x'(t)$ and $y'(t)$ are zero	Zero	The instantaneous velocity vector vanishes.

Common Mistake: Using Displacement as Average Speed

A tempting calculation is to take the distance between the positions at the ends of an interval and divide by the time elapsed, then call the result average speed. That calculation gives the magnitude of average velocity, or endpoint-displacement rate. It can underestimate the distance traveled, and for a closed loop it can be zero even when the point has been moving throughout the interval.

Correction: Average speed over an interval is path length divided by the elapsed time. Endpoint displacement divided by elapsed time measures average velocity as a vector; taking its magnitude gives the endpoint-displacement rate, not necessarily average speed.

For an interval that shrinks toward a parameter value on a smooth curve, the distinction disappears in the limit: the local path-length rate approaches the magnitude of the velocity. That is why instantaneous speed can be calculated from the coordinate derivatives even though a finite-interval endpoint-displacement rate need not equal average speed.

Check Your Understanding

Use the definition and the distinction between traveled distance and displacement to answer these questions.

If $x'(t)=3$ and $y'(t)=4$, what is the instantaneous speed at that parameter value?
Can the speed be negative? Explain using the formula.
What does it mean for the speed to be zero at a parameter value?
Why can endpoint displacement divided by elapsed time fail to equal average speed?
For $x(t)=\cos t$, $y(t)=\sin t$, calculate the instantaneous speed from the coordinate derivatives.

Coordinate derivatives	Speed	Interpretation
At least one of \(x'(t)\), \(y'(t)\) is nonzero	Positive	The point has a nonzero instantaneous velocity.
Both \(x'(t)\) and \(y'(t)\) are zero	Zero	The instantaneous velocity vector vanishes.