Problem 21 - Advanced Engineering Mathematics

Question

Let \(C\) be a plane curve traced by \(r(t)=f(t) \dot{i}+g(t) \dot{j}\), where \(f\) and \(g\) have second 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

Curvature formula is derived using cross product and speed.

Step 1: Understanding the Problem

We need to show that the curvature \( \kappa \) of a plane curve defined by \( r(t) = f(t) \dot{i} + g(t) \dot{j} \) is given by a specific formula. Curvature measures how sharply a curve bends at a given point.

Step 2: Derivative of the Position Vector

Compute the first derivative of the position vector \( r(t) = f(t) \dot{i} + g(t) \dot{j} \). The result is \( r'(t) = f'(t) \dot{i} + g'(t) \dot{j} \).

Step 3: Calculate Speed

The speed is obtained by computing the magnitude of the velocity vector \( r'(t) \), which is \( \| r'(t) \| = \sqrt{[f'(t)]^2 + [g'(t)]^2} \).

Step 4: Derivative of the Velocity Vector

Compute the second derivative \( r''(t) \) of the position vector, which gives \( r''(t) = f''(t) \dot{i} + g''(t) \dot{j} \).

Step 5: Cross Product for Curvature

Find the cross product of \( r'(t) \) and \( r''(t) \). Since the vectors are in the plane, this cross product is represented by the determinant \( f'(t)g''(t) - g'(t)f''(t) \).

Step 6: Curvature Formula

The curvature \( \kappa \) is given by the formula \( \kappa = \frac{|f'(t)g''(t) - g'(t)f''(t)|}{([f'(t)]^2 + [g'(t)]^2)^{3/2}} \), using the cross product and speed calculated in previous steps.

Step 7: Conclusion of Derivation

The derived formula for curvature matches the given formula, confirming the correctness of the solution approach.

Key Concepts

Derivative of Position Vector

To determine how a vector changes as it moves along a curve, we start with the position vector, denoted as \( r(t) = f(t) \mathbf{i} + g(t) \mathbf{j} \). This vector describes the curve \( C \) in the plane. To understand the path's behavior, we compute the first derivative of this vector with respect to time, \( t \). This derivative gives us the velocity vector, \( r'(t) = f'(t) \mathbf{i} + g'(t) \mathbf{j} \).
The velocity vector indicates the direction and rate of change of position, acting as a tangent direction to the curve at each point. The components \( f'(t) \) and \( g'(t) \) specifically represent how much the position changes with time in the \( x \)- and \( y \)-directions, respectively. It's crucial to differentiate accurately to gain insights into the curve's dynamics.

Velocity and Speed in Curved Motion

Velocity is a vector quantity derived from the first derivative of the position vector, as described earlier. It captures both the direction and magnitude of motion of a point on the curve. The magnitude of the velocity vector is referred to as "speed". This is a scalar quantity that tells us how fast a point moves along the curve, regardless of direction.

• The formula for speed, derived from the velocity vector \( r'(t) \), is given by \( \| r'(t) \| = \sqrt{[f'(t)]^2 + [g'(t)]^2} \).
• The magnitude expression shows that speed is always non-negative and takes into account both horizontal and vertical changes.

When the speed is zero, the point on the curve is momentarily stationary. Analyzing both velocity and speed helps in understanding how a point moves on a plane curve over time.

Cross Product in Two Dimensions

In two-dimensional vector analysis, the cross product isn't as intuitive as in three dimensions, but it's still a valuable tool for computing the curvature of a plane curve. The cross product in this context often uses the determinants to reflect orthogonal aspects between vectors.

• When vectors \( r'(t) \) and \( r''(t) \) are considered, their cross product results in a scalar represented by \( f'(t)g''(t) - g'(t)f''(t) \).
• This scalar value helps capture the effect of how the change in one direction interacts with the change in the orthogonal direction.

The importance of this determinant lies in its ability to account for how the curvature forms, linking the behavior of the curve's changes in the plane directly to its twisting or bending characteristics.

Curvature Formula Derivation

The formula for curvature \( \kappa \) highlights how the curve bends at any given point. The curvature is derived using both the velocity vector and its derivative, alongside their cross product. Specifically, curvature quantifies how a curve deviates from a straight line.

• The expression derived for curvature is \( \kappa = \frac{|f'(t)g''(t) - g'(t)f''(t)|}{([f'(t)]^2 + [g'(t)]^2)^{3/2}} \).
• In the formula, the numerator, which is the determinant of the cross product, captures rotational change aspects.
• The denominator involves the speed raised to the power of \( \frac{3}{2} \), ensuring it accounts for both linear and quadratic terms across the curve.

This powerful formula fundamentally tells us how sharply a curve bends: the higher the curvature, the more pronounced the curve's deviation from a straight path at point \( t \). Understanding the derivation of curvature involves appreciating the interplay of multiple calculus concepts.