Question

For a smooth curve given by the parametric equations \(x=f(t)\) and \(y=g(t)\), prove that the curvature is given by \(K=\frac{\left|f^{\prime}(t) g^{\prime \prime}(t)-g^{\prime}(t) f^{\prime \prime}(t)\right|}{\left\\{\left[f^{\prime}(t)\right]^{2}+\left[g^{\prime}(t)\right]^{2}\right\\}^{3 / 2}}\)

Short answer

The derivation of the curvature formula for a curve given by the parametric equations \(x=f(t)\), \(y=g(t)\) uses the concept that curvature is the rate of change of direction per unit change in length. By differentiating the normalized tangent vector and dividing by the magnitude of the velocity vector, the required formula is obtained: \( K=\frac{\left|f^{\prime}(t) g^{\prime\prime}(t)-g^{\prime}(t) f^{\prime\prime}(t)\right|}{\left( (f^{\prime}(t))^{2}+(g^{\prime}(t))^{2}\right)^{3 / 2}}\) .

Step-by-step solution

Computing the Derivatives

First, find the first derivatives \(f'(t)\) and \(g'(t)\) as well as the second derivatives \(f''(t)\) and \(g''(t)\) for the given functions \(f(t)\) and \(g(t)\). These entities will be utilized in the formula for curvature.

Find the Speed

Compute for the speed \(v(t)\) which is the magnitude of the velocity vector (derivative of the position vector). Speed is represented as \(\sqrt{(f'(t))^2+(g'(t))^2}\).

Recalling the Definition of Curvature

By definition, curvature is the rate of change of the direction of the tangent vector, divided by the change in position, or in a sense, the speed. Therefore, derive the curvature formula applying this idea.

Formulating the Curvature Formula

In light of the actual formula, differentiate the unit tangent vector and take its magnitude divided by \(v(t)\), which gives the curvature \(K(t)\). Since the unit tangent vector is \(\frac{1}{v(t)} \times (f'(t),g'(t))\), its derivative is the difference of derivatives of \(f'(t)\) and \(g'(t)\) with time. Taking the magnitude of the derivative and dividing it by \(v(t)\), we get \(K(t)= \frac{\left|f^{\prime}(t) g^{\prime\prime}(t)-g^{\prime}(t) f^{\prime\prime}(t)\right|}{\left( (f^{\prime}(t))^{2}+(g^{\prime}(t))^{2}\right)^{3 / 2}}\) since we need to raise \(v(t)\) to power 3/2 for proper units.

Key concepts

Parametric Equations

Parametric equations are a powerful way to describe a curve by specifying the coordinates of points on the curve as functions of a single parameter. In essence, they allow us to trace the shape of a curve by varying just one variable.

Consider two functions, one for each coordinate axis: the function f(t) for the x-coordinate and g(t) for the y-coordinate. As the parameter t, which can be thought of as time, varies, the point (f(t), g(t)) moves along the curve, drawing the shape as it changes values.

• In the context of the exercise, x=f(t) and y=g(t) delineate the curve in the Cartesian plane.
• Parametric equations are useful because they can represent a wide variety of curves, including those that cannot be described by a simple y=f(x) equation, such as vertical lines or loops.

Derivatives

The concept of derivatives is at the core of calculus and serves as a tool to analyze varying quantities. If you conceptualize a curve as a path traced by a moving point, the derivative represents the rate at which the point's coordinates are changing at any given moment.

Mathematically, if you have a function h(t), its derivative h'(t) signifies the rate of change of the function's value with respect to the variable t.

• When applied to parametric equations, the derivatives f'(t) and g'(t) give us the instantaneous rate of change of each coordinate with respect to the parameter t.
• The second derivatives, f''(t) and g''(t), indicate how the rate of change itself is changing, allowing us to understand the curve’s behavior in terms of acceleration.
• These derivatives are essential for computing the curvature of a curve as they reflect both speed and direction.

Velocity Vector

In physics and mathematics, the velocity vector provides us with not just the speed of a moving object, but also its direction of movement. Specifically, for a curve defined by parametric equations, the velocity vector is the derivative of the position vector with respect to time.

The components of the velocity vector are simply the derivatives of the respective position coordinates: if the position vector is given by (x(t), y(t)), then the velocity vector is given by (f'(t), g'((t)).

• The magnitude of the velocity vector, or speed, is crucial as it combines the rates of change in both the x and y directions. In our exercise, it's denoted as v(t) = \( \sqrt{(f'(t))^2+(g'(t))^2} \).
• This speed is then used to normalize the velocity vector to find the unit tangent vector, which is used in turn to calculate the curvature.

Unit Tangent Vector

The unit tangent vector is a vector that is tangent to the curve and has a magnitude of one. It's fundamental in understanding the direction that a curve is headed at any given point. You get the unit tangent vector by dividing the velocity vector by its magnitude, effectively normalizing it.

In the step-by-step solution, the unit tangent vector is simply the velocity vector scaled down to have a magnitude of one: T(t) = \(1/v(t)\) * (f'(t), g'(t)).

• The curvature of the curve at any point is then determined by how much this unit tangent vector changes as you move along the curve, which is calculated by taking the derivative of the unit tangent vector.
• This leads to the ultimate equation for curvature given in the exercise, which significantly relies on both the first and second derivatives of the parametric equations.