Math Problem Statement

At what point does the curve have maximum curvature? y = 6e^x

Solution

To determine the point at which the curve y=6exy = 6e^x has maximum curvature, we need to compute the curvature formula and find its maximum value.

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

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

Step 1: Find the first and second derivatives of y=6exy = 6e^x.

  1. The first derivative f(x)f'(x) is: f(x)=ddx(6ex)=6exf'(x) = \frac{d}{dx}(6e^x) = 6e^x

  2. The second derivative f(x)f''(x) is: f(x)=ddx(6ex)=6exf''(x) = \frac{d}{dx}(6e^x) = 6e^x

Step 2: Plug the derivatives into the curvature formula.

Now that we have f(x)=6exf'(x) = 6e^x and f(x)=6exf''(x) = 6e^x, we substitute them into the curvature formula:

κ(x)=6ex(1+(6ex)2)3/2\kappa(x) = \frac{|6e^x|}{\left( 1 + (6e^x)^2 \right)^{3/2}}

This simplifies to:

κ(x)=6ex(1+36e2x)3/2\kappa(x) = \frac{6e^x}{\left( 1 + 36e^{2x} \right)^{3/2}}

Step 3: Maximize the curvature.

To find the point where the curvature is maximized, we take the derivative of κ(x)\kappa(x) with respect to xx, set it equal to zero, and solve for xx.

Let's compute this derivative.

κ(x)=ddx(6ex(1+36e2x)3/2)\kappa'(x) = \frac{d}{dx} \left( \frac{6e^x}{\left( 1 + 36e^{2x} \right)^{3/2}} \right)

Using the quotient rule and simplifying might give us the critical points where the maximum occurs.

Let me now compute this derivative and find where it equals zero.The real critical point occurs at approximately x2.138x \approx -2.138. This is where the curve y=6exy = 6e^x has maximum curvature.

Would you like more details on how this result was derived, or have any other questions?

Here are 5 related questions you might find interesting:

  1. What is the curvature of the curve at x=2.138x = -2.138?
  2. How does curvature behave as xx \to \infty for exponential functions like 6ex6e^x?
  3. Can the point of maximum curvature be determined without derivatives for some simpler functions?
  4. How does curvature change for a polynomial function compared to an exponential one?
  5. What physical interpretations can be linked to the concept of curvature?

Tip: Curvature is particularly useful in geometry and physics to understand how sharply a curve bends at a particular point!

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Math Problem Analysis

Mathematical Concepts

Calculus
Curvature
Exponential Functions
Derivative

Formulas

Curvature formula: κ(x) = |f''(x)| / (1 + (f'(x))^2)^(3/2)
Derivative of exponential: d/dx(e^x) = e^x

Theorems

Curvature theorem for curves
Quotient rule in differentiation

Suitable Grade Level

Grades 11-12, Early College