Curvature of Parametric Plane Curves

Curvature from a Parametrization

A plane curve can be described by coordinate functions $x(t)$ and $y(t)$, with parameter $t$ ranging over an interval. Its position vector is $\mathbf{r}(t)=(x(t),y(t))$. The parameter need not measure distance along the curve, and its rate of change need not be constant. The curvature formula for a parametrized curve accounts for that varying speed.

Curvature measures how quickly the direction of travel changes per unit distance along the curve. It is therefore nonnegative: the direction may turn clockwise or counterclockwise, but curvature records the magnitude of the turning. This distinction matters when the same geometric curve is traversed in opposite directions.

Definition: Let $x(t)$ and $y(t)$ be twice continuously differentiable on an interval, and suppose the parametrization is regular there, meaning $x'(t)^2+y'(t)^2>0$. The curvature at parameter value $t$ is $$ \kappa(t)=\frac{|x'(t)y''(t)-y'(t)x''(t)|}{\bigl(x'(t)^2+y'(t)^2\bigr)^{3/2}}. $$

The regularity condition ensures that the velocity is nonzero, so the curve has a well-defined unit tangent. If both first derivatives vanish at a parameter value, this formula has a zero denominator and does not give the curvature there. A separate analysis may be needed to decide whether the geometric curve has a meaningful tangent or curvature at such a point.

In the numerator, $x'y''-y'x''$ compares the velocity and acceleration. Its absolute value removes the sign associated with the direction of turning. In the denominator, $x'^2+y'^2$ is the square of the speed, so its power $3/2$ is the speed cubed.

Why the Formula Works

The unit tangent is the velocity divided by its magnitude. Write the speed as $v(t)=\sqrt{x'(t)^2+y'(t)^2}$, which is positive wherever the parametrization is regular. Then $\mathbf{T}(t)=(x'(t)/v(t),y'(t)/v(t))$. Curvature is the magnitude of the unit tangent’s change per unit distance: $\kappa=|\mathbf{T}'(t)|/v(t)$.

To connect this definition to the coordinate formula, let $\mathbf{N}(t)=(-y'(t)/v(t),x'(t)/v(t))$, the unit vector obtained by rotating $\mathbf{T}$ through a right angle. Since $\mathbf{T}\cdot\mathbf{T}=1$, differentiating gives $\mathbf{T}'\cdot\mathbf{T}=0$. Thus $\mathbf{T}'$ is perpendicular to $\mathbf{T}$, so its magnitude is the absolute value of its component in the $\mathbf{N}$ direction.

\begin{aligned} \mathbf{T}'\cdot\mathbf{N} &=\left(\frac{x''}{v}-\frac{x'v'}{v^2}\right)\left(-\frac{y'}{v}\right) +\left(\frac{y''}{v}-\frac{y'v'}{v^2}\right)\left(\frac{x'}{v}\right)\\ &=\frac{x'y''-y'x''}{v^2}. \end{aligned}

The terms involving $v'$ cancel. Because $\mathbf{T}'$ is perpendicular to $\mathbf{T}$, this gives $|\mathbf{T}'|=|x'y''-y'x''|/v^2$. Dividing by the speed $v$, as required for turning per unit distance, yields the stated formula because $v^3=(x'^2+y'^2)^{3/2}$.

Key result: For a regular twice-differentiable plane parametrization, calculate the determinant-like numerator $x'y''-y'x''$, take its absolute value, and divide by the speed cubed. The absolute value makes curvature independent of whether the curve turns clockwise or counterclockwise.

Worked Examples

Worked Example: A Parametrized Parabola

Find the curvature of $\mathbf{r}(t)=(t,t^2)$ at $t=1$, and check it using the unit tangent.

Here $x(t)=t$ and $y(t)=t^2$, so $x'=1$, $y'=2t$, $x''=0$, and $y''=2$. The numerator is $ |x'y''-y'x''|=|1(2)-(2t)(0)|=2$. The squared speed is $x'^2+y'^2=1+4t^2$. Therefore $\kappa(t)=2/(1+4t^2)^{3/2}$, and at $t=1$, $\kappa(1)=2/5^{3/2}=2/(5\sqrt{5})$.

For a check, the unit tangent is $\mathbf{T}(t)=(1,2t)/\sqrt{1+4t^2}$. Differentiation gives $\mathbf{T}'(t)=(-4t,2)/(1+4t^2)^{3/2}$. Thus $\mathbf{T}'(1)=(-4,2)/(5\sqrt{5})$, whose magnitude is $\sqrt{16+4}/(5\sqrt{5})=2/5$. The speed at $t=1$ is $\sqrt{5}$, so $|\mathbf{T}'(1)|/v(1)=(2/5)/\sqrt{5}=2/(5\sqrt{5})$, agreeing with the formula.

Worked Example: Curvature of an Ellipse

For the ellipse $\mathbf{r}(t)=(3\cos t,2\sin t)$, find the curvature at $t=0$.

The derivatives are $x'=-3\sin t$, $y'=2\cos t$, $x''=-3\cos t$, and $y''=-2\sin t$. The numerator simplifies to \[ |x'y''-y'x''| =|6\sin^2t+6\cos^2t| =6. \] The squared speed is $9\sin^2t+4\cos^2t$. At $t=0$, it is $4$, so the speed cubed is $2^3=8$. Hence $\kappa(0)=6/8=3/4$.

To check by differentiating the unit tangent, at $t=0$ the velocity is $(0,2)$, the speed is $2$, and the acceleration is $(-3,0)$. The speed derivative is $v'=\mathbf{r}'\cdot\mathbf{r}''/v=((0,2)\cdot(-3,0))/2=0$. Since $\mathbf{T}'=\mathbf{r}''/v-\mathbf{r}'v'/v^2$, this gives $\mathbf{T}'(0)=(-3/2,0)$, with magnitude $3/2$. Dividing by the speed $2$ gives curvature $3/4$, as before.

What Reversing Direction Changes

Reversing a parametrization changes the direction in which the curve is traversed. For example, if $\mathbf{r}(t)=(x(t),y(t))$, the reversed parametrization $\widetilde{\mathbf{r}}(u)=\mathbf{r}(-u)$ travels along the same path in the opposite direction. Its first derivative is the negative of the original velocity at the corresponding point, while its second derivative has the same sign as the original acceleration.

Consequently, the numerator $x'y''-y'x''$ changes sign under reversal, but its absolute value does not. The speed is also unchanged. Curvature therefore remains the same, although the signed direction of turning is reversed.

Worked Example: Reversing a Parabola’s Direction

Reverse the parametrization $\mathbf{r}(t)=(t,t^2)$, then calculate the curvature at the point $(1,1)$.

The reversed parametrization is $\widetilde{\mathbf{r}}(u)=(-u,u^2)$. The point $(1,1)$ occurs when $u=-1$. For this parametrization, $x'=-1$, $y'=2u$, $x''=0$, and $y''=2$. The absolute numerator is $|(-1)(2)-(2u)(0)|=2$. At $u=-1$, the squared speed is $1+4=5$, so $\widetilde{\kappa}(-1)=2/5^{3/2}=2/(5\sqrt{5})$.

For a direct check, at $u=-1$ the velocity is $(-1,-2)$, the speed is $\sqrt{5}$, and the acceleration is $(0,2)$. The speed derivative is $v'=((-1,-2)\cdot(0,2))/\sqrt{5}=-4/\sqrt{5}$. Therefore \[ \mathbf{T}'(-1) =\frac{(0,2)}{\sqrt{5}}-\frac{(-1,-2)(-4/\sqrt{5})}{5} =\frac{(-4,2)}{5\sqrt{5}}. \] Its magnitude is $\sqrt{20}/(5\sqrt{5})=2/5$. Dividing by the speed $\sqrt{5}$ gives $2/(5\sqrt{5})$, confirming the result. The curvature is unchanged even though the direction of traversal has reversed.

The same check works for a constant rescaling of the parameter. If $t=au+b$, where $a\ne 0$, then each first derivative gains a factor $a$, and each second derivative gains a factor $a^2$. The numerator gains a factor $a^3$, while the speed cubed gains a factor $|a|^3$. Taking the numerator’s absolute value cancels that scaling. When $a<0$, the traversal direction is reversed; when $a>0$, it is preserved. In either case, the curvature at corresponding points is unchanged.

A Common Sign Mistake

A frequent error is to calculate $x'y''-y'x''$ and report its value without taking the absolute value. For the reversed parabola above, that signed numerator at $u=-1$ is $(-1)(2)-(2u)(0)=-2$. Using it directly would give $\kappa=-2/5^{3/2}$, a negative answer.

That result is not curvature: curvature is a magnitude and cannot be negative. The signed numerator can record the direction of turning, but the curvature formula uses its absolute value. The correction is $|-2|/5^{3/2}=2/(5\sqrt{5})$. A second useful check is that the denominator is the speed cubed, not the speed squared; omitting one factor of speed generally produces an incorrect value.

How to Apply the Formula Reliably

Check regularity.
Confirm that $x'^2+y'^2>0$ at the parameter value of interest.

Differentiate both coordinates twice.
Record $x'$, $y'$, $x''$, and $y''$ before substituting the parameter value.

Compute the numerator and speed.
Evaluate $|x'y''-y'x''|$ and $x'^2+y'^2$ carefully.

Form the ratio and check it.
Divide by $(x'^2+y'^2)^{3/2}$, then, when useful, compare with $|\mathbf{T}'|/v$.

Parametric curvature is useful because a curve need not be expressible conveniently as a single graph $y=f(x)$. A parametrization can describe loops, ellipses, and paths whose horizontal coordinate reverses direction. The formula handles these cases directly, provided the velocity does not vanish at the point being considered.

Check Your Understanding

Use the definitions and formula in this tutorial to answer the following questions.

What condition on $x'(t)$ and $y'(t)$ is required for the parametrization to be regular?
For $\mathbf{r}(t)=(x(t),y(t))$, which expression gives the squared speed?
Why does the curvature formula use an absolute value in its numerator?
When a parametrization is reversed, what happens to the sign of $x'y''-y'x''$, and what happens to curvature?
For $\mathbf{r}(t)=(t,3t^2)$, write the formula for $\kappa(t)$ and evaluate it at $t=0$.