/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 The formula $$ \kappa(x)=\fr... [FREE SOLUTION] | 91Ó°ÊÓ

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The formula $$ \kappa(x)=\frac{\left|f^{\prime \prime}(x)\right|}{\left[1+\left(f^{\prime}(x)\right)^{2}\right]^{3 / 2}} $$ derived in Exercise \(5,\) expresses the curvature \(\kappa(x)\) of a twice- differentiable plane curve \(y=f(x)\) as a function of \(x\) . Find the curvature function of each of the curves. Then graph \(f(x)\) together with \(\kappa(x)\) over the given interval. You will find some surprises. $$ y=x^{2}, \quad-2 \leq x \leq 2 $$

Short Answer

Expert verified
The curvature function is \( \kappa(x) = \frac{2}{(1 + 4x^2)^{3/2}} \).

Step by step solution

01

Understand the Formula

The given formula for curvature is \( \kappa(x) = \frac{|f''(x)|}{[1 + (f'(x))^2]^{3/2}} \). This involves the first and second derivatives of the function \( f(x) \), which in this case is \( y = x^2 \). Our task is to find \( f'(x) \) and \( f''(x) \) to plug into this formula.
02

Calculate the First Derivative

Differentiate \( y = x^2 \) with respect to \( x \). This gives us \( f'(x) = 2x \). This derivative represents the slope of the tangent line to the curve at any point \( x \).
03

Calculate the Second Derivative

Differentiate \( f'(x) = 2x \) to find \( f''(x) \). The result is \( f''(x) = 2 \). This second derivative tells us about the concavity of the function and is used in the curvature formula.
04

Substitute Derivatives into the Curvature Formula

Now substitute \( f'(x) = 2x \) and \( f''(x) = 2 \) into the curvature formula. The curvature formula now becomes: \( \kappa(x) = \frac{|2|}{[1 + (2x)^2]^{3/2}} = \frac{2}{(1 + 4x^2)^{3/2}} \).
05

Evaluate Curvature over the Interval

The formula for curvature is now \( \kappa(x) = \frac{2}{(1 + 4x^2)^{3/2}} \). For the interval \(-2 \leq x \leq 2\), we will compute this expression at various points to understand how the curvature behaves along the curve.
06

Graph the Functions

Graph \( y = x^2 \) and the curvature \( \kappa(x) = \frac{2}{(1 + 4x^2)^{3/2}} \) over the interval \(-2 \leq x \leq 2\) using graphing software or manually plotting points. Note where the curvature is largest and smallest.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a key concept in calculus, involving the process of finding a derivative. A derivative measures how a function changes as its input changes. It's like calculating the rate of change or the slope at a specific point on a curve. In our task, we're working with the function \( y = x^2 \). This is a simple polynomial function, making it straightforward to differentiate.

To find the first derivative, use the power rule, where you bring down the exponent as a coefficient and reduce the exponent by one. Applying this to \( y = x^2 \), the first derivative is \( f'(x) = 2x \). This tells us the slope of the tangent at any point \( x \).

The second derivative, \( f''(x) \), provides information about the concavity or how the curve is bending. Differentiating \( f'(x) = 2x \) gives \( f''(x) = 2 \). This second derivative is a constant, suggesting a uniform concavity, which we'll see is important in understanding curvature. Differentiation lays the foundation for advanced concepts, such as curvature.
Curvature Formula
The curvature formula measures how much a curve deviates from a straight line at a particular point. For a function \( y = f(x) \), the curvature \( \kappa(x) \) is calculated using a specific formula that requires the first and second derivatives of the function. It is expressed as:
\[\kappa(x)=\frac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}\]
This formula allows us to understand the 'bendiness' of the function \( f(x) \) at different points along the x-axis.

Substituting the derivatives we derived from \( y = x^2 \), where \( f'(x) = 2x \) and \( f''(x) = 2 \), into the formula, we arrive at:
\[\kappa(x) = \frac{2}{(1 + 4x^2)^{3/2}}\]
Curvature helps us to articulate how sharply a curve is bending at any given point, crucial for fields like physics and engineering. Observing this formula, you might notice how adding \((f'(x))^2\) in the denominator emphasizes significant changes when \( x \) is far from zero, highlighting rapid fluctuations in curvature.
Graphing Functions
Graphing functions like \( y = x^2 \) alongside their curvature function \( \kappa(x) = \frac{2}{(1 + 4x^2)^{3/2}} \) offers visual insight into how they behave over a specified interval. For our case, from \( -2 \) to \( 2 \). By graphing, you can actually see how the theoretical calculations hold true.

When graphing \( y = x^2 \), you observe a smooth parabola that has a minimum point at \( x = 0 \). In contrast, the curvature function helps to visually represent how the steepness of the parabolic curve changes. Areas where the parabola is less steep correspond to lower values of \( \kappa(x) \), and where it becomes steeper, \( \kappa(x) \) increases.

Graphs are powerful tools for confirming analytical calculations and providing an intuitive sense of how mathematical relationships manifest. When sketching the graph of a function and its curvature, pay attention to the peaks and troughs, as these can tell you a lot about where and how the curvature varies along the curve. Whether using graphing software or graph paper, sketching functions and their properties is a practical way to deepen your grasp of mathematics.

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Most popular questions from this chapter

Length is independent of parametrization To illustrate that the length of a smooth space curve does not depend on the parametrization you use to compute it, calculate the length of one turn of the helix in Example 1 with the following parametrizations. a. \(\mathbf{r}(t)=(\cos 4 t) \mathbf{i}+(\sin 4 t) \mathbf{j}+4 t \mathbf{k}, \quad 0 \leq t \leq \pi / 2\) \(\mathbf{b} . \mathbf{r}(t)=[\cos (t / 2)] \mathbf{i}+[\sin (t / 2)] \mathbf{j}+(t / 2) \mathbf{k}, \quad 0 \leq t \leq 4 \pi\) \(\mathbf{c} \cdot \mathbf{r}(t)=(\cos t) \mathbf{i}-(\sin t) \mathbf{j}-t \mathbf{k}, \quad-2 \pi \leq t \leq 0\)

As mentioned in the text, the tangent line to a smooth curve \(\mathbf{r}(t)=f(t) \mathbf{i}+g(t) \mathbf{j}+h(t) \mathbf{k}\) at \(t=t_{0}\) is the line that passes through the point \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right)\) parallel to \(\mathbf{v}\left(t_{0}\right),\) the curve's velocity vector at \(t_{0}\) Find parametric equations for the line that is tangent to the given curve at the given parameter value \(t=t_{0}\) . \begin{equation} \mathbf{r}(t)=t^{2} \mathbf{i}+(2 t-1) \mathbf{j}+t^{3} \mathbf{k}, \quad t_{0}=2 \end{equation}

Find the velocity and acceleration vectors in terms of \(\mathbf{u}_{r}\) and \(\mathbf{u}_{\theta} .\) \begin{equation}r=a \sin 2 \theta \quad \text { and } \quad \frac{d \theta}{d t}=2 t\end{equation}

The speedometer on your car reads a steady 35 mph. Could you be accelerating? Explain.

Range and height versus speed a. Show that doubling a projectile's initial speed at a given launch angle multiplies its range by 4 . b. By about what percentage should you increase the initial speed to double the height and range?

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