Find the maximum curvature of the curve y = integral

Find the maximum curvature of the curve $y = \int_0^x \sinh t \, dt$.

The curvature $\kappa$ of a curve $y = f(x)$ is given by:

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

To find the maximum curvature, we will:

1. Find $y'(x)$ and $y''(x)$ using the Fundamental Theorem of Calculus
2. Express the curvature $\kappa(x)$ as a function of $x$
3. Find where $\kappa(x)$ is maximized

Step 1: Find $y'(x)$ using the Fundamental Theorem of Calculus

By the Fundamental Theorem of Calculus:

$$y'(x) = \frac{d}{dx}\left(\int_0^x \sinh t \, dt\right) = \sinh x$$

Step 2: Find $y''(x)$

$$y''(x) = \frac{d}{dx}(\sinh x) = \cosh x$$

Step 3: Express the curvature

The curvature is:

$$\kappa(x) = \frac{|y''(x)|}{(1 + (y'(x))^2)^{3/2}} = \frac{|\cosh x|}{(1 + \sinh^2 x)^{3/2}}$$

Since $\cosh x > 0$ for all $x$, we have $|\cosh x| = \cosh x$:

$$\kappa(x) = \frac{\cosh x}{(1 + \sinh^2 x)^{3/2}}$$

Step 4: Simplify using the hyperbolic identity

Recall the fundamental hyperbolic identity:

$$\cosh^2 x - \sinh^2 x = 1$$

Therefore:

$$1 + \sinh^2 x = \cosh^2 x$$

Substituting this into the curvature formula:

$$\kappa(x) = \frac{\cosh x}{(\cosh^2 x)^{3/2}} = \frac{\cosh x}{\cosh^3 x} = \frac{1}{\cosh^2 x}$$

Step 5: Find the maximum of $\kappa(x)$

We need to maximize:

$$\kappa(x) = \frac{1}{\cosh^2 x} = \operatorname{sech}^2 x$$

Since $\cosh x \geq 1$ for all $x$, with equality only at $x = 0$, we have:

$$\cosh^2 x \geq 1 \quad \text{with equality at } x = 0$$

Therefore:

$$\frac{1}{\cosh^2 x} \leq 1 \quad \text{with equality at } x = 0$$

This means $\kappa(x)$ achieves its maximum at $x = 0$.

Step 6: Calculate the maximum curvature

At $x = 0$:

$$\kappa(0) = \frac{1}{\cosh^2 0} = \frac{1}{1^2} = 1$$

Also, at $x = 0$:

$$y(0) = \int_0^0 \sinh t \, dt = 0$$

So the maximum curvature occurs at the origin $(0, 0)$.

Final Answer

$$\boxed{\kappa_{\max} = 1}$$

The maximum curvature is $\kappa_{\max} = 1$, occurring at the point $(0, 0)$.

Verification

Let's verify our calculation by checking the curvature formula directly at $x = 0$:

$$\kappa(0) = \frac{|\cosh 0|}{(1 + \sinh^2 0)^{3/2}} = \frac{1}{(1 + 0)^{3/2}} = \frac{1}{1} = 1$$

This confirms our answer. ✓

Behavior of curvature as $x \to \pm\infty$

As $x \to \pm\infty$, we have $\cosh x \to \infty$, so:

$$\kappa(x) = \frac{1}{\cosh^2 x} \to 0$$

This means the curve becomes straighter (less curved) as we move away from the origin in either direction.

Geometric interpretation

The curve:

Since $y = \int_0^x \sinh t \, dt$, we can evaluate this integral explicitly:

$$y = \int_0^x \sinh t \, dt = \left[\cosh t\right]_0^x = \cosh x - \cosh 0 = \cosh x - 1$$

So the curve is actually:

$$y = \cosh x - 1$$

This is the catenary (the shape of a hanging chain) shifted down by 1 unit.

Properties at the origin:

At $(0, 0)$:

• $y'(0) = \sinh 0 = 0$ (horizontal tangent)
• $y''(0) = \cosh 0 = 1 > 0$ (concave upward)
• $\kappa(0) = 1$ (maximum curvature)

The curve has a horizontal tangent at the origin and curves most sharply there.

The osculating circle:

At the origin, the osculating circle has:

• Radius: $R = \frac{1}{\kappa} = \frac{1}{1} = 1$
• Center: $(0, 1)$ (above the origin, since the curve is concave up)
• Equation: $x^2 + (y-1)^2 = 1$

Comparison with other points:

Let's check the curvature at a few other points:

At $x = 1$:

$$\kappa(1) = \frac{1}{\cosh^2 1} = \frac{1}{(1.543)^2} \approx 0.420$$

At $x = 2$:

$$\kappa(2) = \frac{1}{\cosh^2 2} = \frac{1}{(3.762)^2} \approx 0.071$$

The curvature decreases rapidly as we move away from the origin.

Graph of curvature

The curvature function $\kappa(x) = \operatorname{sech}^2 x = \frac{1}{\cosh^2 x}$ has the following properties:

• Symmetric about $x = 0$: $\kappa(-x) = \kappa(x)$
• Maximum at $x = 0$: $\kappa(0) = 1$
• Monotonically decreasing for $x > 0$ and monotonically increasing for $x < 0$
• Approaches 0 as $x \to \pm\infty$
• Always positive: $\kappa(x) > 0$ for all $x$

Alternative approach: Finding critical points

We could also find the maximum by taking the derivative of $\kappa(x)$ and setting it to zero.

$$\kappa(x) = \frac{1}{\cosh^2 x} = \cosh^{-2} x$$

$$\frac{d\kappa}{dx} = -2\cosh^{-3} x \cdot \sinh x = \frac{-2\sinh x}{\cosh^3 x}$$

Setting $\frac{d\kappa}{dx} = 0$:

$$-2\sinh x = 0 \implies \sinh x = 0 \implies x = 0$$

To verify this is a maximum, check the second derivative or note that $\kappa(x)$ decreases on both sides of $x = 0$.

Summary

1. Using the Fundamental Theorem of Calculus: $y' = \sinh x$ and $y'' = \cosh x$
2. The curvature simplifies to: $\kappa(x) = \frac{1}{\cosh^2 x}$
3. Since $\cosh x$ is minimized at $x = 0$ with $\cosh 0 = 1$, the curvature is maximized at $x = 0$
4. The maximum curvature is $\kappa_{\max} = 1$ at the origin $(0, 0)$
5. The curve $y = \cosh x - 1$ is a shifted catenary with maximum bending at the origin

Key lesson: When finding maximum curvature, it's often helpful to simplify the curvature formula using trigonometric or hyperbolic identities. In this case, the identity $1 + \sinh^2 x = \cosh^2 x$ was crucial for obtaining the simple form $\kappa(x) = \frac{1}{\cosh^2 x}$, making it easy to identify the maximum.