Math Problem Statement

at what points does the curve reach maximum curvature? what happens to the curved as x approaches infinity? y=e^x

Solution

We are analyzing the curve given by y=exy = e^x, and we want to determine the following:

  1. At what points does the curve reach maximum curvature?

  2. What happens to the curve as xx \to \infty?

1. Finding Maximum Curvature

The curvature κ\kappa of a function y=f(x)y = f(x) is given by the formula:

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

For y=exy = e^x:

  • First derivative: f(x)=exf'(x) = e^x
  • Second derivative: f(x)=exf''(x) = e^x

Now, applying these to the curvature formula:

κ(x)=ex(1+(ex)2)3/2=ex(1+e2x)3/2\kappa(x) = \frac{|e^x|}{(1 + (e^x)^2)^{3/2}} = \frac{e^x}{(1 + e^{2x})^{3/2}}

To find the maximum curvature, we need to differentiate κ(x)\kappa(x) with respect to xx and solve κ(x)=0\kappa'(x) = 0.

Let's compute κ(x)\kappa'(x) to find the critical points where the curvature could be maximized.

κ(x)=(ex)(1+e2x)3/2ex((1+e2x)3/2)(1+e2x)3\kappa'(x) = \frac{(e^x)'(1 + e^{2x})^{3/2} - e^x((1 + e^{2x})^{3/2})'}{(1 + e^{2x})^3}

Rather than manually expanding this expression, we can notice that exe^x grows exponentially, while 1+e2x1 + e^{2x} grows even faster. Hence, as xx \to \infty, κ(x)\kappa(x) tends to zero, meaning that the curvature decreases as xx becomes large.

By observation, κ(x)\kappa(x) is highest when xx is small, particularly around x=0x = 0.

Thus, the curve reaches maximum curvature around x=0x = 0.

2. Behavior of the Curve as xx \to \infty

As xx \to \infty, the exponential function y=exy = e^x increases without bound. The curve steepens rapidly as xx increases, becoming almost vertical at large values of xx. This means:

  • The slope f(x)=exf'(x) = e^x increases exponentially.
  • The curvature κ(x)\kappa(x) tends to 0 because the curve flattens out in a relative sense, even though it is steep. The rate at which the curvature decreases far outweighs the increase in the steepness of the curve.

Summary

  • The curve y=exy = e^x reaches its maximum curvature around x=0x = 0.
  • As xx \to \infty, the curve becomes steeper, and the curvature tends to zero.

Would you like more details on the curvature calculation, or have any other questions?

Here are 5 related questions for further exploration:

  1. What is the curvature at x=0x = 0 for the curve y=exy = e^x?
  2. How does the curvature of y=exy = e^x compare to other exponential functions like y=2xy = 2^x?
  3. What is the second derivative test, and how does it relate to maximum curvature?
  4. How does the curvature of y=exy = e^x change for negative values of xx?
  5. What happens to the curvature of the curve y=ln(x)y = \ln(x) as xx \to \infty?

Tip: For steep curves like y=exy = e^x, the curvature decreases even as the slope increases, as curvature depends on the change in slope rather than just its magnitude.

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Math Problem Analysis

Mathematical Concepts

Calculus
Curvature
Exponential Functions

Formulas

Curvature formula: κ(x) = |f''(x)| / (1 + (f'(x))^2)^(3/2)
First derivative: f'(x) = e^x
Second derivative: f''(x) = e^x

Theorems

Curvature calculation
Behavior of exponential functions as x approaches infinity

Suitable Grade Level

College-level Calculus