Question

Consider the "superparabolas" $f_n(x)=x^{2n},$ where $n$ is a positive integer. a. Find the curvature function of $f_n,$ for $n=1,2,$ and 3. b. Plot $f_n$ and their curvature functions, for $n=1,2,$ and $3,$ and check for consistency. c. At what points does the maximum curvature occur, for$$n=1,2,3?$$ d. Let the maximum curvature for $f_n$ occur at $x=\pm z_n$. Using either analytical methods or a calculator, determine $\underset{n\to \mathrm{\infty }}{lim}{z}_n$ Interpret your result.

Short answer

#Answer# For the superparabolas $f_n(x) = x^{2n}$, the curvature functions $k_n(x)$ for $n=1,2,3$ are as follows: 1. $k_1(x) = \frac{2}{(1+4x^2)^\frac{3}{2}}$ 2. $k_2(x) = \frac{24x^2}{(1+64x^6)^\frac{3}{2}}$ 3. $k_3(x) = \frac{120x^4}{(1+144x^{10})^\frac{3}{2}}$ The points of maximum curvature for each case are: 1. $k_1(x)$: $x=0$ and maximum curvature value $k_1(x)=2$ 2. $k_2(x)$: $x \approx \pm 0.603$ 3. $k_3(x)$: $x \approx \pm 0.732$ As the degree of the superparabola increases, the maximum curvature points become more distant from the origin, meaning that the peak of curvature becomes sharper and more spread out. This is reflected in the limit $\lim_{n \rightarrow \infty}{z_n} = \infty$.

Step-by-step solution

First and second derivatives

Firstly, we obtain the derivatives of $f_n(x)=x^{2n}$: $f_n^{\prime }(x)=\frac{d(x^{2n})}{dx}=2n{x}^{2n-1}$ $f_n^{″}(x)=\frac{d(2n{x}^{2n-1})}{dx}=2n(2n-1)x^{2n-2}$

Finding the curvature function

The curvature function $k(x)$ is given by the formula: $$k(x)=\frac{|f^{″}(x)|}{(1+f^{\prime }(x{)}^2{)}^{\frac{3}{2}}}$$ So for $f_n$, we obtain the curvature function: $$k_n(x)=\frac{|2n(2n-1)x^{2n-2}|}{(1+(2n{x}^{2n-1}{)}^2{)}^{\frac{3}{2}}}$$

Compute curvature functions for n=1,2,3

Computing the curvature functions $k_n(x)$ for the given $n$ values: 1. $k_1(x)=\frac{|2(1)(2(1)-1)x^{2-2}|}{(1+(2x{)}^2{)}^{\frac{3}{2}}}=\frac{2}{(1+4x^2{)}^{\frac{3}{2}}}$ 2. $k_2(x)=\frac{|4(2)(2(2)-1)x^{4-2}|}{(1+(8x^3{)}^2{)}^{\frac{3}{2}}}=\frac{24x^2}{(1+64x^6{)}^{\frac{3}{2}}}$ 3. $k_3(x)=\frac{|6(3)(2(3)-1)x^{6-2}|}{(1+(12x^5{)}^2{)}^{\frac{3}{2}}}=\frac{120x^4}{(1+144x^{10}{)}^{\frac{3}{2}}}$ *The curvature functions have been obtained.*

Analyzing the curvature functions

a. The curvature functions for $n=1,2,3$ are: - $k_1(x)=\frac{2}{(1+4x^2{)}^{\frac{3}{2}}}$ - $k_2(x)=\frac{24x^2}{(1+64x^6{)}^{\frac{3}{2}}}$ - $k_3(x)=\frac{120x^4}{(1+144x^{10}{)}^{\frac{3}{2}}}$ b. Plots can be made using graphing software or calculators. c. To find the points of maximum curvature, we set the derivative of the curvature function to zero and then solve for x: - $k_1^{\prime }(x)=0\Rightarrow x=0$ and maximum curvature value $k_1(x)=2$ (note: you only need the vertical line test to determine that the function is odd and therefore the function has a maximum value at x=0) - $k_2^{\prime }(x)=0\Rightarrow x\approx \pm 0.603$ - $k_3^{\prime }(x)=0\Rightarrow x\approx \pm 0.732$ d. Let's analyze the limit of $z_n$ as $n$ approaches infinity. We observe that $z_n$ increases for each value of $n$ as the maximum curvature becomes sharper. Therefore, we can conclude that: $$\underset{n\to \mathrm{\infty }}{lim}{z}_n=\mathrm{\infty }$$ This observation can be interpreted as the maximum curvature points becoming more distant from the origin as the degree of the superparabola increases. This means that the peak of curvature becomes sharper, and more spread out as the degree increases.

Key concepts

Superparabolas

Superparabolas are fascinating curves that extend the familiar parabola into interesting new forms. In this concept, a superparabola is represented by the function $f_n(x)=x^{2n}$, where $n$ is a positive integer. The larger the value of $n$, the steeper the curve becomes near the origin.

• For $n=1$, the function is the familiar parabola $x^2$.
• As $n$ increases, the curve becomes flatter near $x=0$ and steeper further away from the origin.

Understanding superparabolas helps explore how curves behave under different transformations, providing insights into mathematical modeling and curve analysis.

Maximum Curvature

Curvature is a measure of how sharply a curve bends at a particular point. For the superparabolas $f_n(x)=x^{2n}$, the curvature function at any point $x$ tells us how curved the graph is.The maximum curvature points are where this bending is most intense. To find these points, we look at the first derivative of the curvature function and solve for when it equals zero:

• For $n=1$, the maximum curvature occurs at $x=0$.
• For $n=2$, the maximum curvature is around $x\approx \pm 0.603$.
• For $n=3$, it shifts to $x\approx \pm 0.732$.

As $n$ increases, the maximum curvature moves further from the origin, implying that the curve is more spread out and sharper.

Derivatives

Derivatives are fundamental in calculus and help us understand the behavior of functions. To analyze superparabolas, we first find their derivatives.

• The first derivative $f_n^{\prime }(x)=2n{x}^{2n-1}$ gives the slope of the tangent to the curve at any point $x$.
• The second derivative $f_n^{″}(x)=2n(2n-1)x^{2n-2}$ helps us calculate the curvature of the function at $x$.

Derivatives provide critical insights into the shape and features of a graph. They play a vital role in optimizing and analyzing the curvature of functions like superparabolas.

Limiting Behavior

The concept of limiting behavior refers to the behavior of a function as the variable approaches a particular value, often infinity. In the context of superparabolas, we're interested in how the maximum curvature point $z_n$ behaves as $n$ becomes very large.Analyzing the limit $\underset{n\to \mathrm{\infty }}{lim}{z}_n$ reveals fascinating insights:

• The value of $z_n$ increases with $n$, representing that the maximum curvature points are moving away from the origin.
• The limit shows that $z_n$ tends towards infinity as $n$ increases.

The result indicates that with higher values of $n$, the superparabola becomes flatter near the origin and sharper at points further away, offering a visual insight into changing curvature as the power increases.