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Math V21B Section 10.2

Introduction

Please try this problem on your own before looking at my solution. There may be other ways to solve this; I make no claim that this method is the quickest, prettiest, or most efficient.

Problem 40

The original problem is to find the arc length of the parametric curve $$x(t) = t + {t^{1/2}};\;\;\;\;\;y(t) = t - {t^{1/2}};\;\;\;\;\;0 \leqslant t \leqslant 1$$

The differential arc length, in parametric form, is $$ds = \sqrt {{{\left[ {x'\left( t \right)} \right]}^2} + {{\left[ {y'\left( t \right)} \right]}^2}} \;dt = \sqrt {{{\left( {1 + \frac{1}{2}{t^{ - 1/2}}} \right)}^2} + {{\left( {1 - \frac{1}{2}{t^{ - 1/2}}} \right)}^2}} \;dt$$ Squaring the parentheses and combining like terms, $$ds = \sqrt {\left( {1 + 2 \cdot \frac{1}{2}{t^{ - 1/2}} + \frac{1}{4}{t^{ - 1}}} \right) + \left( {1 - 2 \cdot \frac{1}{2}{t^{ - 1/2}} + \frac{1}{4}{t^{ - 1}}} \right)} \;dt = \sqrt {\left( {1 + {t^{ - 1/2}} + \frac{1}{4}{t^{ - 1}}} \right) + \left( {1 - {t^{ - 1/2}} + \frac{1}{4}{t^{ - 1}}} \right)} \;dt$$ $$ds = \sqrt {2 + \frac{1}{2}{t^{ - 1}}} \;dt$$

The integrated (total) arc length is therefore given by $$s = \int {ds} = \int_0^1 {\sqrt {2 + \frac{1}{2}{t^{ - 1}}} \;dt} $$ which is a potentially intimidating integral. In fact, to solve it will require invoking many of the least popular techniques of integration from Chapter 7 of the textbook. So if you have read this far, beware: things are about to get interesting! We will be using $u$-substitution, trigonometric substitution, improper integrals, and integration by tables to complete this integration.

After some trial and error, I discovered that a helpful $u$-substitution would be $$u = {t^{ - 1/2}}$$ from which we may derive (by repeatedly squaring both sides of the equation) the useful relationships $${u^2} = {t^{ - 1}}\;\;\;\text{and}\;\;\;{u^4} = {t^{ - 2}}$$ and (by taking the differential of the first equation in the line above and carrying out some manipulations to isolate $dt$) $$2u\;du = - {t^{ - 2}}\;dt \to 2u\;du = - {u^4}\;dt{\text{ (from the second equation above)}} \to \frac{{ - 2u\;du}}{{{u^4}}} = dt \to dt = - 2{u^{ - 3}}\;du$$

Plugging these substitutions into the integral for the arc length $s$ gives $$s = \int_?^? {\sqrt {2 + \frac{1}{2}{u^2}} \cdot \left( { - 2{u^{ - 3}}\;du} \right)} = - 2 \cdot \int_?^? {\sqrt {2 + \frac{1}{2}{u^2}} \cdot \left( {{u^{ - 3}}\;du} \right)} $$ To find the values of the question marks in the limits of integration, we plug in $t=0$ for the lower limit and $t=1$ for the upper limit. The upper limit is fairly straightforward: $$u(1) = {(1)^{ - 1/2}} = \frac{1}{{{1^{1/2}}}} = \frac{1}{{\sqrt 1 }} = \frac{1}{1} = 1$$ The lower limit is peculiar: $$u(0) = {(0)^{ - 1/2}} = \frac{1}{{\sqrt 0 }} = {\text{undefined}}$$ which makes the problem appear unsolvable.

Instead of evaluating $u$ at $t=0$, let us try to salvage the situation by evaluating the limit of integration as the value of $u$ as $t$ approaches zero, rather than trying to set $t$ exactly equal to zero. As the interval of integration is on the positive side of $t=0$, let us compute the right-sided limit only (i.e., we are using $t$ values very close to zero, but only on the positive side of zero). We may thus obtain the value of $u$ by defining it as $$u(0) = \mathop {\lim }\limits_{t \to {0^ + }} {t^{ - 1/2}} = \mathop {\lim }\limits_{t \to {0^ + }} \frac{1}{{{t^{1/2}}}} = \mathop {\lim }\limits_{t \to {0^ + }} \frac{1}{{\sqrt t }} = + \infty {\text{ (positive because }}t{\text{ is always positive)}}$$ So now, for the arc length, we have the improper integral (which we are fairly confident will converge) $$s = - 2 \cdot \int_\infty ^1 {\sqrt {2 + \frac{1}{2}{u^2}} \cdot \left( {{u^{ - 3}}\;du} \right)} = 2 \cdot \int_1^\infty {\sqrt {2 + \frac{1}{2}{u^2}} \cdot \left( {{u^{ - 3}}\;du} \right)} $$ where the latter integral is obtained by exchanging the limits of integration and reversing the sign of the constant in front to compensate. Factoring a (1/2) out of the radicand (the expression inside the square root) and moving the negative powers of $u$ to the denominator as a positive exponent gives $$s = 2 \cdot \int_1^\infty {\frac{{\sqrt {\frac{1}{2}\left( {4 + {u^2}} \right)} }}{{{u^3}}}\;du} = 2 \cdot \int_1^\infty {\frac{{\sqrt {\frac{1}{2}} \sqrt {\left( {4 + {u^2}} \right)} }}{{{u^3}}}\;du} = 2 \cdot \int_1^\infty {\frac{{\frac{1}{{\sqrt 2 }}\sqrt {\left( {4 + {u^2}} \right)} }}{{{u^3}}}\;du} = \frac{2}{{\sqrt 2 }} \cdot \int_1^\infty {\frac{{\sqrt {\left( {4 + {u^2}} \right)} }}{{{u^3}}}\;du} $$

The radicand is now of the form ${a^2} + {u^2}$, where $4 = {a^2}$ or $a=2$. Employ the trigonometric substitutions $u = 2\tan \theta \to du = 2{\sec ^2}\theta \;d\theta$ to obtain the next expression for arc length $s$: $$s = \sqrt 2 \int_?^? {\frac{{\sqrt {4 + 4{{\tan }^2}\theta } }}{{{{\left( {2\tan \theta } \right)}^3}}} \cdot \left( {2{{\sec }^2}\theta \;d\theta } \right)} = \sqrt 2 \int_?^? {\frac{{\sqrt {4\left( {1 + {{\tan }^2}\theta } \right)} }}{{8{{\tan }^3}\theta }} \cdot 2{{\sec }^2}\theta \;d\theta } = \frac{{2\sqrt 2 }}{8}\int_?^? {\frac{{\sqrt 4 \sqrt {1 + {{\tan }^2}\theta } }}{{{{\tan }^3}\theta }}{{\sec }^2}\theta \;d\theta } $$ To replace the question marks in the limits of integration, we first need to isolate $\theta$ from the trigonometric substitution. From $u = 2\tan \theta$ we obtain $$\frac{u}{2} = \tan \theta \to \theta = {\tan ^{ - 1}}\left( {\frac{u}{2}} \right)$$ From the lower limit $u=1$ we obtain $\theta = {\tan ^{ - 1}}\left( {\frac{1}{2}} \right)$, and from the upper limit $u=\infty$ we obtain $\theta = \frac{\pi }{2}$. (The latter is true because $$\mathop {\lim }\limits_{u \to \infty } {\tan ^{ - 1}}u = \frac{\pi }{2}$$ which you may verify by examining and interpreting the upper horizontal asymptote of the graph of the inverse-tangent function.)

Continuing to simplify the arc-length integral with the new limits yields $$s = \frac{{2\sqrt 2 \sqrt 4 }}{8}\int_{{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)}^{\frac{\pi }{2}} {\frac{{\sqrt {{{\sec }^2}\theta } }}{{{{\tan }^3}\theta }}{{\sec }^2}\theta \;d\theta } = \frac{{4\sqrt 2 }}{8}\int_{{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)}^{\frac{\pi }{2}} {\frac{{\sec \theta }}{{{{\tan }^3}\theta }}{{\sec }^2}\theta \;d\theta } = \frac{{\sqrt 2 }}{2}\int_{{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)}^{\frac{\pi }{2}} {\frac{{{{\sec }^3}\theta }}{{{{\tan }^3}\theta }}\;d\theta } $$ $$s = \frac{{\sqrt 2 }}{2}\int_{{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)}^{\frac{\pi }{2}} {\frac{{\left( {1/{{\cos }^3}\theta } \right)}}{{\left( {{{\sin }^3}\theta /{{\cos }^3}\theta } \right)}}\;d\theta } = \frac{{\sqrt 2 }}{2}\int_{{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)}^{\frac{\pi }{2}} {\frac{1}{{{{\sin }^3}\theta }}\;d\theta } = \frac{{\sqrt 2 }}{2}\int_{{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)}^{\frac{\pi }{2}} {{{\csc }^3}\theta \;d\theta } $$

The process of integrating ${{\csc }^3}\theta$ closely follows the example for integrating ${{\sec }^3}\theta$ provided in Chapter 7 of the textbook. The antiderivative has a parallel structure to the antiderivative of ${{\sec }^3}\theta$; rather than deriving it, let us take advantage of the integral table in the back of the textbook. Integral #72 in that table gives the antiderivative; replacing $u$ in the table with $\theta$ yields $$s = \frac{{\sqrt 2 }}{2}\left[ { - \frac{1}{2}\csc \theta \cot \theta + \frac{1}{2}\ln \left| {\csc \theta - \cot \theta } \right|} \right]_{{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)}^{\frac{\pi }{2}} = \frac{{\sqrt 2 }}{4}\left[ { - \csc \theta \cot \theta + \ln \left| {\csc \theta - \cot \theta } \right|} \right]_{{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)}^{\frac{\pi }{2}}$$ To evaluate the lower limit $$\theta={{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)$$ it is necessary to draw a corresponding right triangle; taking the tangent of both sides of the above equation instructs us to draw a triangle for which $\tan \theta = \frac{1}{2}$, as shown:

The hypotenuse, $\sqrt 5$, is obtained by applying the Pythagorean theorem to the legs (1 and 2). From the triangle, we see that $$\csc \left( {{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)} \right) = \csc \theta = \sqrt 5 $$ and $$\cot \left( {{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)} \right) = \cot \theta = 2$$ Also note that, for the upper limit, $$\csc \left( {\frac{\pi }{2}} \right) = \frac{1}{{\sin \left( {\pi /2} \right)}} = 1\;\;\;\text{and}\;\;\;\cot\left( {\frac{\pi }{2}} \right) = \frac{{\cos \left( {\pi /2} \right)}}{{\sin \left( {\pi /2} \right)}} = \frac{0}{1} = 0$$

Evaluating the limits yields, at long last: $$s = \frac{{\sqrt 2 }}{4}\left\{ {\left[ { - \csc \frac{\pi }{2}\cot \frac{\pi }{2} + \ln \left| {\csc \frac{\pi }{2} - \cot \frac{\pi }{2}} \right|} \right] - \left[ { - \csc \left( {{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)} \right)\cot \left( {{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)} \right) + \ln \left| {\csc \left( {{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)} \right) - \cot \left( {{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)} \right)} \right|} \right]} \right\}$$ $$s = \frac{{\sqrt 2 }}{4}\left\{ {\left[ { - \left( 1 \right)\left( 0 \right) + \ln \left| {\left( 1 \right) - \left( 0 \right)} \right|} \right] - \left[ { - \left( {\sqrt 5 } \right)\left( 2 \right) + \ln \left| {\sqrt 5 - 2} \right|} \right]} \right\} = \frac{{\sqrt 2 }}{4}\left[ {0 + \ln 1 + 2\sqrt 5 - \ln \left( {\sqrt 5 - 2} \right)} \right]$$ $$\boxed{\displaystyle{s = \frac{{\sqrt 2 }}{4}\left[ {2\sqrt 5 - \ln \left( {\sqrt 5 - 2} \right)} \right]}}$$