Find the curvature of the curve y = integral from 0 to x

Find the curvature of the curve $y = \int_0^x (1+t)\ln(1+t) \, dt$ at the point $(0,0)$.

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

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

To find the curvature at $(0,0)$, we need to find $y'(0)$ and $y''(0)$.

Step 1: Verify that the curve passes through $(0,0)$

At $x = 0$:

$$y(0) = \int_0^0 (1+t)\ln(1+t) \, dt = 0$$

So the curve does indeed pass through the origin. ✓

Step 2: Find $y'(x)$ using the Fundamental Theorem of Calculus

By the Fundamental Theorem of Calculus:

$$y'(x) = \frac{d}{dx}\left(\int_0^x (1+t)\ln(1+t) \, dt\right) = (1+x)\ln(1+x)$$

At $x = 0$:

$$y'(0) = (1+0)\ln(1+0) = 1 \cdot \ln(1) = 1 \cdot 0 = 0$$

Step 3: Find $y''(x)$

We need to differentiate $y'(x) = (1+x)\ln(1+x)$ using the product rule:

$$y''(x) = \frac{d}{dx}[(1+x)\ln(1+x)]$$

Let $u = 1+x$ and $v = \ln(1+x)$. Then:

$$u' = 1 \quad \text{and} \quad v' = \frac{1}{1+x}$$

Using the product rule:

$$y''(x) = u'v + uv' = 1 \cdot \ln(1+x) + (1+x) \cdot \frac{1}{1+x}$$

$$y''(x) = \ln(1+x) + 1$$

At $x = 0$:

$$y''(0) = \ln(1+0) + 1 = \ln(1) + 1 = 0 + 1 = 1$$

Step 4: Calculate the curvature at $(0,0)$

Using the curvature formula:

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

Substituting $y'(0) = 0$ and $y''(0) = 1$:

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

Final Answer

$$\boxed{\kappa = 1}$$

The curvature of the curve $y = \int_0^x (1+t)\ln(1+t) \, dt$ at the point $(0,0)$ is $\kappa = 1$.

Geometric interpretation

What does curvature mean?

The curvature $\kappa$ measures how quickly the curve is changing direction at a point. A larger curvature means the curve is "bending" more sharply.

• For a straight line: $\kappa = 0$ (no bending)
• For a circle of radius $r$: $\kappa = \frac{1}{r}$ (constant curvature)

Interpretation of $\kappa = 1$:

A curvature of $\kappa = 1$ means that at the point $(0,0)$, the curve has the same curvature as a circle of radius 1.

The circle that best approximates the curve at $(0,0)$ is called the osculating circle (or "kissing circle"), and it has radius:

$$R = \frac{1}{\kappa} = \frac{1}{1} = 1$$

Behavior near the origin:

Since $y'(0) = 0$, the curve has a horizontal tangent at the origin.

Since $y''(0) = 1 > 0$, the curve is concave upward near the origin.

The curve can be approximated near the origin by its Taylor expansion:

$$y \approx y(0) + y'(0)x + \frac{y''(0)}{2}x^2 = 0 + 0 \cdot x + \frac{1}{2}x^2 = \frac{x^2}{2}$$

This parabola $y = \frac{x^2}{2}$ has curvature at the origin:

$$\kappa_{\text{parabola}} = \frac{|1|}{(1+0)^{3/2}} = 1$$

which confirms our result!

Additional observations

The osculating circle:

The osculating circle at $(0,0)$ has:

• Center at $(0, 1)$ (since the tangent is horizontal and the curve is concave up)
• Radius $R = 1$
• Equation: $x^2 + (y-1)^2 = 1$

Near the origin, the curve $y = \int_0^x (1+t)\ln(1+t) \, dt$ closely follows this circle.

Why the Fundamental Theorem of Calculus is useful:

Without the FTC, we would need to:

1. Find the antiderivative of $(1+t)\ln(1+t)$ explicitly
2. Evaluate it at the limits
3. Differentiate the result twice

Instead, the FTC allows us to immediately write:

• $y'(x) = (1+x)\ln(1+x)$ (one differentiation becomes zero differentiations!)
• $y''(x) = \ln(1+x) + 1$ (two differentiations becomes one differentiation!)

This is much simpler!

Summary

1. Using the Fundamental Theorem of Calculus: $y'(x) = (1+x)\ln(1+x)$
2. Differentiating once more: $y''(x) = \ln(1+x) + 1$
3. At the origin: $y'(0) = 0$ and $y''(0) = 1$
4. The curvature is: $\kappa = \frac{|y''(0)|}{(1+(y'(0))^2)^{3/2}} = \frac{1}{1} = 1$
5. This means the curve bends like a circle of radius 1 at the origin.

Key lesson: The Fundamental Theorem of Calculus is incredibly powerful for analyzing curves defined by integrals. It allows us to find derivatives without explicitly evaluating the integral, making problems involving curvature, tangent lines, and other geometric properties much more tractable.