Answered: y = Vx for 0

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

Determine the arc length of the curve. Set up the integral to find the lengthand then integrate in your calculator

Expert Solution

Step 1

the given function is:

$y = \sqrt{x}$ for $0 \leq x \leq 3$

we have to determine the length of the curve.

as we know that the length L of the curve y=f(x) from x=a to x=b is given by:

$L = \int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$

Step 2

as $y = \sqrt{x}$

differentiating both the sides with respect to x, we get

$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{d (\sqrt{x})}{d x} \\ \frac{d y}{d x} & = & \frac{d (x^{\frac{1}{2}})}{d x} \\ \frac{d y}{d x} & = & \frac{1}{2} x^{\frac{1}{2} - 1} \\ \frac{d y}{d x} & = & \frac{1}{2} x^{\frac{- 1}{2}} \\ \frac{d y}{d x} & = & \frac{1}{2 \sqrt{x}} \end{array}$

Step 3

as we have $0 \leq x \leq 3$ .

therefore the upper limit of x is 3 and the lower limit of x is 0.

therefore the length L of the curve $y = \sqrt{x}$ from x=0 to x=3 is :

$\begin{array}{rcl} L & = & \int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x \\ = & \int_{0}^{3} \sqrt{1 + {(\frac{1}{2 \sqrt{x}})}^{2}} d x \\ = & \int_{0}^{3} \sqrt{1 + \frac{1}{4 x}} d x \end{array}$

let the integral $\int_{0}^{3} \sqrt{1 + \frac{1}{4 x}} d x$ be I.

Step 4

therefore,

$I = \int_{0}^{3} \sqrt{1 + \frac{1}{4 x}} d x$

let 4x=u

therefore,

4dx=du

$d x = \frac{d u}{4}$

now when x=0, u=4(0)=0

when x=3, u=4(3)=12

Step 5

therefore now substitute these values in the integral I.

therefore,

$I = \int_{0}^{3} \sqrt{1 + \frac{1}{4 x}} d x = \int_{0}^{12} \sqrt{1 + \frac{1}{u}} \frac{d u}{4} = \frac{1}{4} \int_{0}^{12} \sqrt{\frac{u + 1}{u}} d u = \frac{1}{4} \int_{0}^{12} \frac{\sqrt{u + 1}}{\sqrt{u}} d u$

now let the integral $\int \frac{\sqrt{u + 1}}{\sqrt{u}} d u$ be $I_{1}$

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