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Chapter 6: Problem 11

Find the arc length of the following curves on the given interval by integrating with respect to \(x\) $$y=\frac{1}{3} x^{3 / 2} \text { on }[0,60]$$

Short Answer

Expert verified
The arc length of the curve given by the function \(y = \frac{1}{3}x^{3/2}\) on the interval \([0, 60]\) is 168.

Step by step solution

01

Find the derivative of y with respect to x

To find the derivative of \(y\) with respect to \(x\), we can use the power rule: $$ \frac{d}{dx} x^n = nx^{n-1} $$ Applying the power rule to our function, we get $$ y' = \frac{d}{dx} \left(\frac{1}{3}x^{3/2}\right) = \frac{1}{2}x^{1/2} $$
02

Plug the derivative into the arc length formula

Now that we have the derivative \(y'\), we can plug it into the arc length formula: $$ L = \int_{0}^{60} \sqrt{1 + \left(\frac{1}{2}x^{1/2}\right)^2}\, dx $$ Simplify the expression inside the square root before proceeding: $$ L = \int_{0}^{60} \sqrt{1 + \frac{1}{4}x}\, dx $$
03

Compute the integral

To compute the integral, we can use substitution. Let \(u = 1 + \frac{1}{4}x\), so \(\frac{du}{dx} = \frac{1}{4}\). Thus, \(dx = 4\,du\). The limits of integration will also change accordingly: when \(x = 0\), \(u = 1\) and when \(x = 60\), \(u = 16\). Now, substitute in the integral: $$ L = \int_{1}^{16} \sqrt{u}\cdot 4\, du = 4 \int_{1}^{16} u^{1/2}\, du $$ Now, integrate with respect to u: $$ L = 4\left[\frac{2}{3}u^{3/2}\right]_{1}^{16} $$ Finally, evaluate the integral at the limits of integration: $$ L = 4\left[\frac{2}{3}(16)^{3/2} - \frac{2}{3}(1)^{3/2}\right] = 4\left[\frac{2}{3}(64) - \frac{2}{3}\right] = 4\left[\frac{2}{3}(63)\right] = 168 $$ Therefore, the arc length of the curve \(y = \frac{1}{3}x^{3/2}\) on the interval \([0, 60]\) is \(168\).

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Key Concepts

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

Integration
In mathematics, integration is a fundamental concept that represents the process of finding the whole from its parts. It's like reverse differentiation. When you integrate a function, you're essentially finding the area under its curve over a certain interval. This is particularly useful in calculating the arc length of a curve. In our exercise, we integrated the expression \( \sqrt{1 + (\frac{1}{2}x^{1/2})^2} \) over the interval \([0, 60]\). This involves transforming a complex expression into a simpler, aggregate form that represents the total arc length of the given curve

  • The aim is to sum an infinite number of infinitesimally small quantities.
  • Integration can be definite or indefinite. Here it is definite since it is over a specific interval.
  • It鈥檚 essentially finding an antiderivative.
By integrating step-by-step and simplifying the expression, we find the arc length, a crucial concept when dealing with curves.
Power Rule
The power rule is a basic principle in calculus that simplifies the process of taking derivatives. It states that if you have a function of form \( x^n \), its derivative is \( nx^{n-1} \). This rule makes it easier to derive polynomial functions efficiently. In the given problem, we used the power rule to find the derivative of \( y \), where \( y = \frac{1}{3}x^{3/2} \).

  • Subtract one from the exponent.
  • Multiply the term by the old exponent.
  • In our problem, this became \( y' = \frac{1}{2}x^{1/2} \).
The power rule is one of the first rules learned in calculus due to its simplicity and broad applicability for polynomial expressions, making it foundational for understanding derivatives deeply.
Derivative
A derivative represents the rate of change of a function with respect to a variable. Think of it as finding the speed of a moving object. Derivatives help determine the slope of the curve at a specific point and uncover critical features of functions, like maxima and minima.

  • In our exercise, the derivative helped us find the slope of the curve \( y = \frac{1}{3}x^{3/2} \).
  • It is necessary for calculating arc length as it factors into the arc length formula.
  • Derivatives can be computed using rules like the power rule, chain rule, product rule, etc.
The derivative is a core tool in calculus that allows us to understand dynamic and changing systems, positioning it as a fundamental concept in solving calculus problems.
Substitution Method
The substitution method is a technique used in integration to simplify complex integrals. It involves substituting a part of the integral, usually the inner function, with a new variable. This converts a tough integral into an easier one.

In our exercise, we chose \( u = 1 + \frac{1}{4}x \). This new variable \( u \) transformed the original integral into a more manageable form.

  • Step 1: Choose a substitution, \( u \).
  • Step 2: Express \( dx \) in terms of \( du \).
  • Step 3: Change limits of integration if needed.
  • Step 4: Rewrite the integral.
This method is particularly handy in solving integrals that are difficult to integrate using standard methods or when a direct antiderivative is hard to find, allowing for more straightforward calculations.

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. When using the shell method, the axis of the cylindrical shells is parallel to the axis of revolution. b. If a region is revolved about the \(y\) -axis, then the shell method must be used. c. If a region is revolved about the \(x\) -axis, then in principle, it is possible to use the disk/washer method and integrate with respect to \(x\) or the shell method and integrate with respect to \(y\)

At noon \((t=0),\) Alicia starts running along a long straight road at \(4 \mathrm{mi} / \mathrm{hr}\). Her velocity decreases according to the function \(v(t)=4 /(t+1),\) for \(t \geq 0 .\) At noon, Boris also starts running along the same road with a 2 -mi head start on Alicia; his velocity is given by \(u(t)=2 /(t+1),\) for \(t \geq 0 .\) Assume \(t\) is measured in hours. a. Find the position functions for Alicia and Boris, where \(s=0\) corresponds to Alicia's starting point. b. When, if ever, does Alicia overtake Boris?

Consider the region \(R\) in the first quadrant bounded by \(y=x^{1 / n}\) and \(y=x^{n},\) where \(n>1\) is a positive number. a. Find the volume \(V(n)\) of the solid generated when \(R\) is revolved about the \(x\) -axis. Express your answer in terms of \(n\). b. Evaluate \(\lim _{n \rightarrow \infty} V(n) .\) Interpret this limit geometrically.

Find the area of the region bounded by the curve \(x=\frac{1}{2 y}-\sqrt{\frac{1}{4 y^{2}}-1}\) and the line \(x=1\) in the first quadrant. (Hint: Express \(y\) in terms of \(x\).)

Consider the functions \(f(x)=a \sin 2 x\) and \(g(x)=(\sin x) / a,\) where \(a>0\) is a real number. a. Graph the two functions on the interval \([0, \pi / 2],\) for \(a=\frac{1}{2}, 1\) and 2. b. Show that the curves have an intersection point \(x^{*}\) (other than \(x=0)\) on \([0, \pi / 2]\) that satisfies \(\cos x^{*}=1 /\left(2 a^{2}\right),\) provided \(a>1 / \sqrt{2}\) c. Find the area of the region between the two curves on \(\left[0, x^{*}\right]\) when \(a=1\) d. Show that as \(a \rightarrow 1 / \sqrt{2}^{+}\). the area of the region between the two curves on \(\left[0, x^{*}\right]\) approaches zero.
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