Question

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

Question: Prove that the curvature of the curve y=f(x) is given by the formula $$\kappa(x) = \frac{\left|f^{\prime\prime}(x)\right|}{\left(1+f^{\prime}(x)^{2}\right)^{3 / 2}}$$. Solution: We have proved the given curvature formula by following these steps: 1. Convert the given curve y=f(x) to the parametric representation x=t, y=f(t). 2. Find the first and second derivatives of x(t) and y(t) with respect to t. 3. Use the formula for curvature in terms of the parametric derivatives. 4. Simplify the expression and show that it is equal to the given curvature formula. The final proof is: $$\kappa(x) = \frac{\left|f^{\prime\prime}(x)\right|}{\left(1+f^{\prime}(x)^{2}\right)^{3 / 2}}$$

Step-by-step solution

Convert the curve to parametric representation

Rewrite the curve y=f(x) in parametric form as x(t)=t and y(t)=f(t).

Find the derivatives of x(t) and y(t)

Now we need to find the first and second derivatives of x(t) and y(t) with respect to t: First derivatives: $$\frac{dx}{dt} = \frac{d(t)}{dt} = 1$$ $$\frac{dy}{dt} = \frac{d[f(t)]}{dt} = f'(t)$$ Second derivatives: $$\frac{d^2x}{dt^2} = \frac{d(1)}{dt} = 0$$ $$\frac{d^2y}{dt^2} = \frac{d[f'(t)]}{dt} = f''(t)$$

Use the formula for curvature in terms of parametric derivatives

The formula for curvature in terms of parametric derivatives is given by: $$\kappa(t) = \frac{\left|\frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{dy}{dt}\frac{d^2x}{dt^2}\right|}{\left(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\right)^{3/2}}$$

Simplify the expression and show that it matches the given formula

Substitute the first and second derivatives of x(t) and y(t) into the curvature formula: $$\kappa(t) = \frac{\left|1\cdot f''(t) - f'(t)\cdot 0\right|}{\left((1)^2 + (f'(t))^2\right)^{3/2}}$$ This simplifies to: $$\kappa(t) = \frac{\left|f''(t)\right|}{\left(1 + (f'(t))^2\right)^{3/2}}$$ Since x=t, we can replace t with x in the above expression, and we obtain the given curvature formula: $$\kappa(x) = \frac{\left|f^{\prime\prime}(x)\right|}{\left(1+f^{\prime}(x)^{2}\right)^{3 / 2}}$$ Hence, we have proved that the curvature of the curve y=f(x) is given by the mentioned formula.

Key concepts

Parametric Representation

In the study of curves, the parametric representation is a powerful tool for describing paths in a coordinate system. Rather than expressing the curve as y in terms of x, or vice versa, each coordinate is described as a function of an independent parameter, typically denoted by t. So, for a curve described by the function f, we assign x(t)=t and y(t)=f(t). This method imbues us with the flexibility to describe a wider variety of curves, including those that cannot be represented as a function of x alone due to their shape, such as vertical lines or loops.

When we adopt a parametric representation, each point on the curve is a pair (x(t), y(t)), evolving as t changes. This approach is particularly useful when we need to calculate properties related to motion over time, such as velocity or, in this case, the curvature of the curve.

Differentiation

The cornerstone of calculus, differentiation, is the process of finding the rate at which a function changes at any point. When a curve is described parametrically, we find the first derivative of each coordinate with respect to the parameter t. The derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) represent the instantaneous rates of change of the coordinates as the parameter t changes, akin to how speed in physics is the derivative of position with respect to time.

This process gets us closer to understanding how the curve moves in space. For the x-coordinate, x(t)=t, its derivative with respect to t is 1, as the rate of change is constant. For the y-coordinate, y(t)=f(t), the derivative is \( f'(t) \), which will vary based on the function f.

Second Derivative

The second derivative of a function helps us comprehend the function's curvature, determining whether a point on the curve is concave up or down. In parametric terms, for the x-coordinate, x(t)=t, the second derivative \( \frac{d^2x}{dt^2} = 0 \), because the first derivative is constant and thus its rate of change is zero — a flat, uncurving line. On the other hand, for y(t)=f(t), the second derivative \( \frac{d^2y}{dt^2} = f''(t) \) gives us the rate of change of the slope provided by the first derivative \( f'(t) \), indicating how the curve bends over time. In the context of the curve's curvature, we interpret the second derivative as letting us see 'how much' and 'how quickly' the curve is bending at any point.

Curvature Formula

The curvature formula quantifies the extent to which a curve deviates from being straight. In parametric terms, it's defined as \( \kappa(t) = \frac{|\frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{dy}{dt}\frac{d^2x}{dt^2}|}{(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2)^{3/2}} \). This formula elegantly encapsulates the rate of change of the direction of the tangent to the curve relative to the length of the arc, resulting in a measure of 'how sharp' the curve is at a given point.

Applying the first and second derivatives calculated earlier collapses the formula to a simpler form because \frac{d^2x}{dt^2} equals zero. The absolute value ensures curvature is always a non-negative quantity, reflecting its nature as a measure of bending without direction. Eventually, we express the curvature in terms of the original function f, leading to the given curvature formula: \( \kappa(x) = \frac{|f''(x)|}{(1+f'(x)^2)^{3/2}} \). Here, we observe that the curvature not only depends on the second derivative, which tells us about the concavity of f, but is also moderated by the first derivative squared, which influences how the slope of the curve affects its curvature.