Problem 21 - Advanced Engineering Mathematics

Question

Let \(C\) be a plane curve traced by \(\mathbf{r}(t)=f(t) \mathbf{i}+g(t) j\), where \(f\) and \(g\) have secood derivatives. Show that the curvature at a point is given by $$ \kappa=\begin{aligned} &\left|f^{\prime}(t) g^{\prime \prime}(t)-g^{\prime}(t) f^{\prime \prime}(t)\right| \\ &\left(\left[f^{\prime}(t)\right]^{2}+\left[g^{\prime}(t)\right]^{2}\right)^{3 / 2} \end{aligned} $$

Step-by-Step Solution

Answer

The curvature is given by \( \kappa=\frac{\left|f'(t)g''(t)-g'(t)f''(t)\right|}{\left((f'(t))^2+(g'(t))^2\right)^{3/2}} \).

Step 1: Definition of Curvature

Curvature \( \kappa \) is defined as the magnitude of the rate of change of the tangent vector \( \frac{d \mathbf{T}}{ds} \) with respect to arc length \( s \). For a plane curve parameterized by \( \mathbf{r}(t) = (f(t), g(t)) \), where \( s \) is the arc length, the curvature formula for a plane curve can be expressed using derivatives of \( \mathbf{r}(t) \).

Step 2: Compute Derivatives

First, compute the derivatives \( \mathbf{r}'(t) = (f'(t), g'(t)) \) and \( \mathbf{r}''(t) = (f''(t), g''(t)) \).

Step 3: Compute Tangent Vector

The tangent vector \( \mathbf{T}(t) \) is given by \( \frac{ \mathbf{r}'(t) }{ \| \mathbf{r}'(t) \| } \), where \( \| \mathbf{r}'(t) \| = \sqrt{(f'(t))^2 + (g'(t))^2} \).

Step 4: Differentiate Tangent Vector

Differentiate \( \mathbf{T}(t) \) with respect to \( t \). This involves using the quotient rule to differentiate \( \mathbf{T}(t) \).

Step 5: Express Derivative in Terms of Arc Length

Since \( \frac{d \mathbf{T}}{ds} = \left( \frac{d \mathbf{T}}{dt} \right) \div \left( \frac{ds}{dt} \right) \), where \( \frac{ds}{dt} = \| \mathbf{r}'(t) \| \), substitute to express \( \frac{d \mathbf{T}}{ds} \).

Step 6: Calculate Curvature

Curvature \( \kappa = \left| \mathbf{r}'(t) \times \mathbf{r}''(t) \right| \div \| \mathbf{r}'(t) \|^3 \). Compute \( \mathbf{r}'(t) \times \mathbf{r}''(t) = \det \begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ f'(t) & g'(t) & 0 \ f''(t) & g''(t) & 0 \end{pmatrix} = (f'(t)g''(t) - g'(t)f''(t))\mathbf{k} \) and simplify to \( \left| f'(t)g''(t) - g'(t)f''(t) \right| \).

Step 7: Simplify Denominator

The denominator becomes \( \left( \sqrt{(f'(t))^2 + (g'(t))^2} \right)^3 = \left[(f'(t))^2 + (g'(t))^2 \right]^{3/2} \).

Step 8: Combine for Final Formula

Substitute back to get \[ \kappa = \frac{\left| f'(t)g''(t) - g'(t)f''(t) \right|}{\left[(f'(t))^2 + (g'(t))^2 \right]^{3/2}} \]. This is the required curvature formula.

Key Concepts

Arc Length Parameterization

In calculus, arc length parameterization is a method used to simplify the description of a curve by adjusting the parameter so that its speed is constant. When a curve is described by functions of a particular variable, often time "t", the arc length \(s\) is typically not proportional to \(t\) in a simple linear way. Instead, the curve might be moving faster or slower at different points along \(t\). To handle this, we use the concept of arc length, which measures the distance along the curve from a fixed starting point.

Arc length parameterization ensures that as \(t\) increases, the curve is traced such that the parameter increases uniformly with respect to the actual distance covered along the curve. This method is particularly useful for advanced calculus concepts, such as computing curvature, where distances along the curve are crucial. To find the arc length \(s\) from \(t = a \) to \(t = b\), we use the integral:

\[ s = \int_{a}^{b} \sqrt{(f'(t))^2 + (g'(t))^2} \ dt \]

This formula calculates the total length of the curve from one point to another by summing up all the tiny segments along the path defined by the derivatives of \(f\) and \(g\).

Tangent Vector

Understanding tangent vectors is key to grasping how curves function in two-dimensional space. The tangent vector of a curve at a given point gives the direction in which the curve is heading. For a parametrized curve given by \( \mathbf{r}(t) = (f(t), g(t)) \), we define the tangent vector as the first derivative of \( \mathbf{r}(t) \):

\[ \mathbf{r}'(t) = (f'(t), g'(t)) \]

This vector does not have a constant length, which can lead to complications when analyzing curves. Hence, we normalize it to have a unit length by dividing by its magnitude, thus creating the unit tangent vector \( \mathbf{T}(t) \):

\[ \mathbf{T}(t) = \frac{ \mathbf{r}'(t) }{ \| \mathbf{r}'(t) \| } \]

Where \( \| \mathbf{r}'(t) \| = \sqrt{(f'(t))^2 + (g'(t))^2} \). This normalized vector maintains the direction of \( \mathbf{r}'(t) \) but with a standardized length of 1, making it very useful for mathematical operations like the computation of curvature.

Derivative

The derivative plays a vital role when exploring the behavior of smooth curves. In the context of plane curves described by functions \( f(t) \) and \( g(t) \), derivatives help us understand how the curve moves and changes direction. The first derivative \( \mathbf{r}'(t) = (f'(t), g'(t)) \) captures the instantaneous rate of change in the curve's position. Near any given point, this tells us how the curve is oriented or heading as \(t\) changes.

Taking derivatives also allows us to access higher-order behaviors of curves. The second derivative, \( \mathbf{r}''(t) = (f''(t), g''(t)) \), provides key insights into the curve's acceleration. Together, these derivatives can be used to establish the curvature of the curve, further investigating how much it bends at any given point.

Derivatives are not only for calculating velocities and accelerations but are also instrumental in determining factors like the tangent vector and curvature, necessary for a deeper understanding of the geometry of the curve.

Curvature Formula

Curvature is a geometric measure describing how sharply a curve bends at a particular point. It is a crucial concept when analyzing curves in mathematics and physics. For a plane curve defined by \( \mathbf{r}(t) = (f(t), g(t)) \), the formula for curvature \( \kappa \) is expressed as:

\[ \kappa = \frac{ \left| f'(t)g''(t) - g'(t)f''(t) \right|}{ \left[ (f'(t))^2 + (g'(t))^2 \right]^{3/2} } \]

To derive this formula, we consider the cross product of the derivatives, which gives the magnitude of the vector perpendicular to the plane of the curve. The numerator \( \left| f'(t)g''(t) - g'(t)f''(t) \right| \) represents this cross product magnitude and encapsulates how the curve's direction changes between points.

The denominator \( \left[ (f'(t))^2 + (g'(t))^2 \right]^{3/2} \) corresponds to cubic power of the speed's magnitude, normalizing the curvature with respect to the length of the curve. Understanding this formula provides insight into the natural "twist" within the curve, a critical component in both theoretical and applied contexts such as computer graphics and mechanical engineering.