Question

Special formula: Curvature for \(y=f(x)\) Assume that \(f\) is twice differentiable. Prove that the curve \(y=f(x)\) has curvature $$ \kappa(x)=\frac{\left|f^{\prime \prime}(x)\right|}{\left(1+f^{\prime}(x)^{2}\right)^{3 / 2}} $$ (Hint: Use the parametric description \(x=t, y=f(t) .\) )

Short answer

In this problem, we were asked to prove the curvature formula for a curve of the form \(y = f(x)\). We used the parametric description of the curve given by \(\boldsymbol{r}(t) = (t, f(t))\), computed the first and second derivatives, and then calculated the curvature using the given formula for parametric curves. We finally simplified the expression to obtain the desired formula: $$ \kappa(x) = \frac{\left|f^{\prime\prime}(x)\right|}{\left(1+f^{\prime}(x)^{2}\right)^{3 / 2}} $$ This is the formula for the curvature of a curve described by a function \(y = f(x)\), as required.

Step-by-step solution

Compute the derivatives of the parametric curve

Rewrite the curve as \(\boldsymbol{r}(t) = (t, f(t))\). Then, we find the derivatives of the parametric description: First derivative: $$ \boldsymbol{r}'(t) = \frac{d(t, f(t))}{dt} = \left(1, f'(t)\right) $$ Second derivative: $$ \boldsymbol{r}''(t) = \frac{d^2(t, f(t))}{dt^2} = \left(0, f''(t)\right) $$

Calculate the cross product of the derivatives

Now, we compute the cross product of the two derivatives using \(|\boldsymbol{r}'(t)\times \boldsymbol{r}''(t)| = \|\boldsymbol{r}'(t)\|\|\boldsymbol{r}''(t)\|\sin{\theta}\): $$ |\boldsymbol{r}'(t) \times \boldsymbol{r}''(t)| = \left|\begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ 1 & f'(t) & 0 \\ 0 & f''(t) & 0 \end{vmatrix}\right| = \left|- f'(t) f''(t) \boldsymbol{k}\right| = |f'(t)f''(t)| $$

Calculate the magnitude of the first derivative

Now we compute the magnitude of the first derivative, \(\|\boldsymbol{r}'(t)\|\): $$ \|\boldsymbol{r}'(t)\| = \sqrt{(1)^2 + (f'(t))^2} = \sqrt{1 + (f'(t))^2} $$

Compute the curvature and simplify

Using the curvature formula for a parametric curve, plug in the results obtained in the previous steps: $$ \kappa(t) = \frac{\left|\boldsymbol{r}'(t) \times \boldsymbol{r}''(t)\right|}{\left\|\boldsymbol{r}'(t)\right\|^{3}} = \frac{|f'(t)f''(t)|}{(\sqrt{1 + (f'(t))^2})^3} $$ Since \(t=x\), the curvature can be expressed as \(\kappa(x)\): $$ \kappa(x) = \frac{|f'(x)f''(x)|}{(\sqrt{1 + (f'(x))^2})^3} $$ Finally, apply \(|f'(x)f''(x)| = |f''(x)||f'(x)| = |f''(x)||{f^{\prime}(x)}|^2\) : $$ \kappa(x) = \frac{\left|f^{\prime\prime}(x)\right|}{\left(1+f^{\prime}(x)^{2}\right)^{3 / 2}} $$ We have derived the curvature formula for a curve of the form \(y = f(x)\).

Key concepts

Parametric Curves

Parametric curves provide a powerful way to represent mathematical curves by expressing both the x and y coordinates as functions of a third parameter, typically denoted as t. This representation allows for more flexibility compared to the traditional y as a function of x method. By setting

• $x = g(t)$
• $y = h(t)$

we can describe a curve in terms of these functions $g(t)$ and $h(t)$. This method is especially useful for curves that loop or intersect themselves, which can be challenging to describe using $y=f(x)$.
Parametric curves are not limited by the restrictions of vertical line intersections or changes in direction, making them ideal for modeling complex geometric shapes and paths such as ellipses, circles, and cycloids.
In this exercise, we parameterize the curve using $x = t$ and $y = f(t)$, to align with the hints provided in the exercise.

Derivatives

Understanding derivatives is crucial when dealing with parametric curves, as they help us explore the curve's behavior. Derivatives provide information about rates of change and the slope at every point along the curve.
For a parametric curve $(x = g(t), y = h(t))$, the first derivative $oldsymbol{r}'(t)$ measures the velocity vector or how the curve changes at each moment. In the given function $y = f(t)$, the vector becomes $oldsymbol{r}'(t) = (1, f'(t))$. Here, the derivative of $t$ with respect to $t$ is 1, and the derivative of $f(t)$ with respect to $t$ is $f'(t)$.

• The second derivative $oldsymbol{r}''(t)$ follows similarly, providing the acceleration vector or how the rate of change itself changes over time. This is given by $oldsymbol{r}''(t) = (0, f''(t))$.

These derivatives are pivotal for understanding the curve's concavity and thus play a significant role in determining curvature.

Cross Product

The cross product is a mathematical operation used often in vector calculus to find a vector perpendicular to two given vectors. In the context of parametric curves, it is particularly helpful for deriving curvature formulas.
When calculating the curvature, the cross product \(oldsymbol{r}'(t) \times oldsymbol{r}''(t)\) gives a vector orthogonal to the tangent and normal vectors of the curve. For

• \(\boldsymbol{r}'(t) = (1, f'(t))\)
• \(\boldsymbol{r}''(t) = (0, f''(t))\)

the cross product becomes \(|\boldsymbol{r}'(t) \times \boldsymbol{r}''(t)| = |f'(t)f''(t)|\). This scalar value simplifies the computation by eliminating the \(\boldsymbol{k}\) component, reducing it to a meaningful magnitude for curvature.
The cross product is essential here because it links the geometry of the curve (how sharply it turns) with the algebraic expression of the parametric curve's derivatives.

Curvature Formula

Curvature quantifies how sharply a curve bends at a point. In determining curvature for a function \(y = f(x)\), we employ a parametric approach and compare it with traditional curvature measures.
The core formula for curvature involves both first and second derivatives. It is given by

• \(\kappa(x) = \frac{|f''(x)|}{(1+f'(x)^2)^{3/2}}\)

This expression arises from the parametric derivatives and the cross product we previously discussed.
The numerator, \(|f''(x)|\), emphasizes the impact of concavity or convexity within the curve, while the denominator \((1+f'(x)^2)^{3/2}\) attributes weight to the overall shape and slope. The formula balances these elements to provide an accurate representation of a curve's bending behavior.
Understanding this curvature formula allows us to quantify geometric features of curves in applications such as physics, engineering, and computer graphics, where precise geometric modeling is critical.