Curvature of a Single Variable Real Valued Function Examples

Curvature of a Single Variable Real-Value Function Examples 1

Recall from the Curvature at a Point on a Single Variable Real Valued Function page that if we have a function $y = f(x)$ that is twice differentiable, then we can find the curvature of $f$ at any point $x \in D(f)$ with the following formula:

(1)

\begin{align} \kappa (x) = \frac{\mid f''(x) \mid}{(1 + (f'(x))^2)^{3/2}} \end{align}

We will now look at some examples of applying this formula.

Example 1

Find the curvature of the function $f(x) = 1 + x + x^2 + x^3 + x^4 + x^5$. Use this formula to calculate the curvature at $x = 0$ and $x = 1$.

We first find the first derivative, $f'(x) = 1 + 2x + 3x^2 + 4x^3 + 5x^4$, and then we find the second derivative $f''(x) = 2 + 6x + 12x^2 + 20x^3$.

Now applying the formula above, and we get that:

(2)

\begin{align} \kappa (x) = \frac{\mid 2 + 6x + 12x^2 + 20x^3 \mid}{\left ( 1 + (1 + 2x + 3x^2 + 4x^3 + 5x^4)^2\right )^{3/2}} \end{align}

If we plug in $x = 0$ we get $\kappa (0) = \frac{2}{2^{3/2}} = \frac{1}{\sqrt{2}}$.

Example 2

Compare the formulae for the curvature of the functions $f(x) = \sin x$ and $g(x) = \cos x$.

First let's calculate the curvature of $f$. We note that $f'(x) = \cos x$ and $f''(x) = -\sin x$, and so applying the formula for curvature, we get that:

(3)

\begin{align} \kappa_f (x) = \frac{\mid -\sin x \mid}{(1 + \cos ^2 x)^{3/2}} \end{align}

Now let's calculate the curvature of $g$. We note that $g'(x) = -\sin x$ and $g''(x) = -\cos x$, and so applying the formula for curvature, we get that:

(4)

\begin{align} \kappa_g (x) = \frac{\mid -\cos x \mid}{(1 + \sin^2 x)^{3/2}} \end{align}

If we compare the two formula, we notice that the two are almost identical. The only difference is the trigonometric functions.