### COMPUTING THE ARC LENGTH OF A DIFFERENTIABLE FUNCTION ON A CLOSED INTERVAL

The following problems involve the computation of arc length of differentiable functions on closed intervals. Let's first begin by finding a general formula for computing arc length. Consider a graph of a function of unknown length $L$ which can be represented as $y=f(x)$ for $a \le x \le b$ or $x=g(y)$ for $c \le y \le d$. We will derive the arc length formula using the differential of arc length, $ds$, a small change in arc length $s$, and write $ds$ in terms of $dx$, the differential of $x$, and $dy$, the differential of $y$ (See the graph below.).

Using the Pythagorean Theorem we will assume that $$(ds)^2 = (dx)^2 + (dy)^2$$ so that $$ds = \sqrt{ (dx)^2 + (dy)^2 }$$ It then follows that the total arc length $L$ from $x=a$ to $x=b$ is $$ARC = \displaystyle{ \int_{s=0}^{s=L} 1 \ ds } = \displaystyle{ \int_{x=a}^{x=b} \sqrt{ (dx)^2 + (dy)^2 } }$$ $$= \displaystyle{ \int_{a}^{b} \sqrt{ \Bigg(1 + {(dy)^2 \over (dx)^2}\Bigg) (dx)^2} }$$ $$= \displaystyle{ \int_{a}^{b} \sqrt{ 1 + \Big({dy \over dx}\Big)^2 } dx }$$ i.e., $$ARC = \displaystyle{ \int_{a}^{b} \sqrt{ 1 + \Big({dy \over dx}\Big)^2 } dx }$$ Similarly, it follows that the total arc length $L$ from $y=c$ to $y=d$ is $$ARC = \displaystyle{ \int_{s=0}^{s=L} 1 \ ds } = \displaystyle{ \int_{y=c}^{y=d} \sqrt{ (dx)^2 + (dy)^2 } }$$ $$= \displaystyle{ \int_{c}^{d} \sqrt{ \Bigg({(dx)^2 \over (dy)^2} + 1 \Bigg) (dy)^2} }$$ $$= \displaystyle{ \int_{c}^{d} \sqrt{ 1 + \Big({dx \over dy}\Big)^2 } dy }$$ i.e., $$ARC = \displaystyle{ \int_{c}^{d} \sqrt{ 1 + \Big({dx \over dy}\Big)^2 } dy }$$

Most of the following problems are average. But because the arc length formula includes a square root, most problems will require relatively intense and very careful algebraic simplification, including manipulation of fractions and creation of perfect squares.

Compute the area of the region enclosed by the graphs of the given equations. Use vertical cross-sections on Problems 1-16.

• PROBLEM 1 : Find the length of the graph for $y = 3x+2$ on the closed interval $-1 \le x \le 2$ .

• PROBLEM 2 : Find the length of the graph for $y = x^{3/2}$ on the closed interval $0 \le x \le 4$

• PROBLEM 3 : Find the length of the graph for $y = (3/2)x^{2/3}$ on the closed interval $0 \le x \le 1$ .

• PROBLEM 4 : Find the length of the graph for $y = 3 + (4/5)x^{5/4}$ on the closed interval $0 \le x \le 1$ .

• PROBLEM 5 : Find the length of the graph for $y= \displaystyle{ { x^4 \over 4 } + { 1 \over 8x^2 } }$ on the closed interval $2 \le x \le 4$ .

• PROBLEM 6 : Find the length of the graph for $y= \displaystyle{ { x^3 \over 6 } + { 1 \over 2x } }$ on the closed interval $1 \le x \le 3$ .

• PROBLEM 7 : Find the length of the graph for $y = (1/2)(e^x + e^{-x})$ on the closed interval $0 \le x \le \ln 2$ .

• PROBLEM 8 : Find the length of the graph for $y = (2/3)(x^2+1)^{3/2}$ on the closed interval $0 \le x \le 2$ .

• PROBLEM 9 : Find the length of the graph for $y = \displaystyle{ \ln(\cos x) }$ on the closed interval $0 \le x \le \pi/3$ .

• PROBLEM 10 : Find the length of the graph for $x = \displaystyle{ { y^4 \over 8 } + { 1 \over 4y^2 } }$ on the closed interval $1 \le y \le 2$ .

• PROBLEM 11 : Find the length of the graph for $x = \displaystyle{ { y^5 \over 5 } + { 1 \over 12y^3 } }$ on the closed interval $1/2 \le y \le 1$ .

• PROBLEM 12 : Find the length of the graph for $x = \ln y - \displaystyle{ y^2 \over 8 }$ on the closed interval $1 \le y \le 2$ .