Curvature at a Point on a Curve Examples 3

# Curvature at a Point on a Curve Examples 3

Recall that if $\vec{r}(t) = (x(t), y(t), z(t))$ is a vector-valued function that traces the smooth curve $C$, then the curvature of $C$ at the point $\vec{r}(t)$ can be given by either of the formulas:

• $\kappa (s) = \biggr \| \frac{d\hat{T}(s)}{ds} \biggr \| = \biggr \| \frac{d\hat{T}(s) / dt}{ds / dt} \biggr \|$ if we find an arc length parameterization of $\vec{r}(t)$ as $\vec{r}(s)$.
• $\kappa (t) = \frac{\| \hat{T'}(t) \|}{\| \vec{r'}(t) \|}$ if $\hat{T'}(t)$ is not too difficult to calculate.
• $\kappa (t) = \frac{\| \vec{r'}(t) \times \vec{r''}(t) \|}{\| \vec{r'}(t) \|^3}$.

We will now look at some more examples of finding the curvature of a space curve.

## Example 1

Find the curvature of the curve given by the vector-valued function $\vec{r}(t) = (\sin t \cos t, \sin^2 t, \cos t)$.

We have that the derivative of $\vec{r}(t)$ is given by:

(1)
\begin{align} \quad \vec{r'}(t) = (\cos^2 t - \sin^2 t, 2\sin t \cos t, -\sin t) \end{align}

The second derivative of $\vec{r}(t)$ is given by:

(2)
\begin{align} \quad \quad \vec{r''}(t) = (-2 \cos t \sin t - 2\sin t \cos t, 2\cos^2 t - 2 \sin^2 t, - \cos t) = (-4 \sin t \cos t, 2(\cos^2 t - \sin^2 t), -\cos t \end{align}

The cross product of $\vec{r'}(t)$ and $\vec{r''}(t)$ is:

(3)
\begin{align} \quad \quad \vec{r'}(t) \times \vec{r''}(t) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ \cos^2 t - \sin^2 t & 2\sin t \cos t & -\sin t\\ -4 \sin t \cos t & 2(\cos^2 t - \sin^2 t) & - \cos t \end{vmatrix} \\ \quad \quad = [-2 \sin t \cos^2 t + 2 \sin t(\cos^2 t - \sin^2 t)] \vec{i} + [\cos t (\cos^2 t - \sin^2 t) + \sin t(4 \sin t \cos t)] \vec{j} + [ 2 (\cos^2 t - \sin^2 t)^2 + 8 \sin^2 t \cos^2 t ] \vec{k} \\ \quad \quad = (-2 \sin^3 t ) \vec{i} + (\cos^3 t + 3 \sin^2 t \cos t) \vec{j} + (2 \cos^4 t + \sin^4 t + 6 \sin^2 t \cos^2 t) \vec{k} \end{align}

The norm of $\vec{r'}(t) \times \vec{r''}(t)$ is:

(4)
\begin{align} \quad \| \vec{r'}(t) \times \vec{r''}(t) \| = \sqrt{4 \sin^6 t + (\cos^3 t + 3 \sin^2 t \cos t)^2 + (2 \cos^4 t + \sin^4 t + 6 \sin^2 t \cos^2 t)^2 } \end{align}

Lastly, the norm of $\vec{r'}(t)$ is:

(5)
\begin{align} \quad \| \vec{r'}(t) \| = \sqrt{(\cos^2 t - \sin^2 t)^2 +4 \sin^2 t \cos^2 t + \sin^2 t} \end{align}

Therefore the curvature $\kappa$ is given by:

(6)
\begin{align} \quad \kappa (t) = \frac{\| \vec{r'}(t) \times \vec{r''}(t) \|}{\| \vec{r'}(t) \|^3} = \frac{\sqrt{4 \sin^6 t + (\cos^3 t + 3 \sin^2 t \cos t)^2 + (2 \cos^4 t + \sin^4 t + 6 \sin^2 t \cos^2 t)^2 }}{\left ( \sqrt{(\cos^2 t - \sin^2 t)^2 +4 \sin^2 t \cos^2 t + \sin^2 t}\right )^3 } \end{align}

## Example 2

Prove that if $y = f(x)$ is a twice differentiable real-valued function that traces the curve $C$ on the $xy$-plane, then the curvature at any point $x \in D(f)$ is given by $\frac{ \mid f''(x) \mid}{(1 + (f'(x))^2)^{3/2}}$. What does this formula say about the curvature of straight lines of the form $y = mx + b$?

We will formally prove this statement on the Curvature at a Point on a Single Variable Real Valued Function.

We can easily parameterize a curve $C$ on the $xy$-plane as:

(7)
\begin{align} \quad \vec{r}(t) = (x, f(x), 0) \end{align}

If we differentiate $\vec{r}(t)$ we get:

(8)
\begin{align} \quad \vec{r'}(t) = (1, f'(x), 0) \end{align}

If we differentiate $\vec{r}(t)$ again, we get that:

(9)
\begin{align} \quad \vec{r''}(t) = (0, f''(x), 0) \end{align}

Therefore we compute the cross product $\vec{r'}(t) \times \vec{r''}(t)$ as:

(10)
\begin{align} \quad \vec{r'}(t) \times \vec{r''}(t) = \begin{bmatrix} \vec{i} & \vec{j} & \vec{k}\\ 1 & f'(x) & 0\\ 0 & f''(x) & 0 \end{bmatrix} = f''(x) \end{align}

Therefore we have that $\| \vec{r'}(t) \times \vec{r''}(t) \| = \sqrt{(f''(x))^2} = \mid f''(x) \mid$.

Now we compute the norm of $\vec{r'}(t)$ as:

(11)
\begin{align} \quad \| \vec{r'}(t) \| = \sqrt{1^2 + (f'(x))^2} = \sqrt{1 + (f'(x))^2} \end{align}

Plugging what we have just obtained into the curvature formula, we get that:

(12)
\begin{align} \quad \kappa (x) = \frac{\mid f''(x) \mid}{\left (\sqrt{1 + (f'(x))^2} \right )^3} \end{align}