Question

Consider the spiral curves given parametrically by: (see picture and answer question)

Answer 1

a. The arc length is given by the integral

$L(r) = \displaystyle \int_3^\infty \sqrt{x'(t)^2 + y'(t)^2} \, dt \\\\ ~~~~~~~~ = \int_3^\infty \sqrt{\left(\frac{t\cos(t) - r\sin(t)}{t^{r+1}}\right)^2 + \left(-\frac{t\sin(t) + r\cos(t)}{t^{r+1}}\right)^2} \, dt \\\\ ~~~~~~~~ = \int_3^\infty \sqrt{\frac{(t^2+r^2)\cos^2(t) + (t^2+r^2)\sin^2(t)}{\left(t^{r+1}\right)^2}} \, dt \\\\ ~~~~~~~~ = \boxed{\int_3^\infty \frac{\sqrt{t^2+r^2}}{t^{r+1}} \, dt}$

b. The integrand roughly behaves like

$\dfrac t{t^{r+1}} = \dfrac1{t^r}$

so the arc length integral will converge for $\boxed{r>1}$.

c. When $r=3$, the integral becomes

$L(3) = \displaystyle \int_3^\infty \frac{\sqrt{t^2+9}}{t^4} \, dt$

Pull out a factor of $t^2$ from under the square root, bearing in mind that $\sqrt{x^2} = |x|$ for all real $x$.

$L(3) = \displaystyle \int_3^\infty \frac{\sqrt{t^2} \sqrt{1+\frac9{t^2}}}{t^4} \, dt \\\\ ~~~~~~~~ = \int_3^\infty \frac{|t| \sqrt{1+\frac9{t^2}}}{t^4} \, dt \\\\ ~~~~~~~~ = \int_3^\infty \frac{t \sqrt{1+\frac9{t^2}}}{t^4} \, dt \\\\ ~~~~~~~~ = \int_3^\infty \frac{\sqrt{1+\frac9{t^2}}}{t^3} \, dt$

since for $3\le t<\infty$, we have $|t|=t$.

Now substitute

$s=1+\dfrac9{t^2} \text{ and } ds = -\dfrac{18}{t^3} \, dt$

Then the integral evaluates to

$L(3) = \displaystyle -\frac1{18} \int_2^1 \sqrt{s} \, ds \\\\ ~~~~~~~~ = \frac1{18} \int_1^2 s^{1/2} \, ds \\\\ ~~~~~~~~ = \frac1{27} s^{3/2} \bigg|_1^2 \\\\ ~~~~~~~~ = \frac{2^{3/2} - 1^{3/2}}{27} = \boxed{\frac{2\sqrt2-1}{27}}$

Answer 2

a) The improper integral in simplified form is equal to $L = \int\limits^{\infty}_{3} {\frac{\sqrt{t^{2}+r^{2}}}{t^{r + 1}} } \, dt$.

b) r > 1 for a spiral with finite length.

c) The length of the spiral when r = 3 is (1 - 2√2) / 9 units.

How to characterize and analyze a group of parametric functions

a) The arc length formula for 2-dimension parametric functions is defined below:

L = ∫ √[(dx / dt)² + (dy / dt)²] dt, for [α, β]         (1)

If we know that $\dot x (t) = \frac{t \cdot \cos t - r \cdot \sin t}{t^{r+1}}$, $\dot y(t) = \frac{t\cdot \sin t + r\cdot \cos t}{t^{r + 1}}$, α = 0 and β → + ∞ then their arc length formula is:

$L = \int\limits^{\infty}_{3} {\sqrt{\left(\frac{t\cdot \cos t - r\cdot \sin t}{t^{r + 1}}\right)^{2}+\left(\frac{t\cdot \sin t + r\cdot \cos t}{t^{r+1}}\right)^{2}} } \, dt$

By algebraic handling and trigonometric formulae (cos ² t + sin² t = 1):

$L = \int\limits^{\infty}_{3} {\frac{\sqrt{t^{2}+r^{2}}}{t^{r + 1}} } \, dt$      (2)

The improper integral in simplified form is equal to $L = \int\limits^{\infty}_{3} {\frac{\sqrt{t^{2}+r^{2}}}{t^{r + 1}} } \, dt$.

b) By ratio comparison criterion, we notice that √(t² + r²) is similar to √t² = t and $\frac{\sqrt{t^{2}+r^{2}}}{t^{r + 1}}$ is similar to $\frac{t}{t^{r +1}} = \frac{1}{t^{r}}$.

The integral found in part a) has a finite length if and only the governing grade of the denominator is greater that the governing grade of the numerator. and according to the ratio comparson criterion, the absolute value of the ratio is greater than 0 and less than 1. Therefore, r > 1 for a spiral with finite length.

c) Now we proceed to integrate the function:

L = ∫ [√(t² + 9) / t⁴] dt, for [3, + ∞].

L = ∫ [t · √(1 + 9 / t²) / t⁴] dt, for [3, + ∞].

By using the algebraic substitutions: u = 1 + 9 / t², du = - (18 / t³) dt → - (1 / 18) du.

L = ∫ √u du, for [3, + ∞].

L = - (1 / 9) · √(u³), for [3, + ∞].

L = - (1 / 9) · [√(1 + 9 / t²)³], for [3, + ∞].

L = - (1 / 9) · [√(2³) - √(1³)]

L = - (1 / 9) · (2√2 - 1)

L = (1 - 2√2) / 9

The length of the spiral when r = 3 is (1 - 2√2) / 9 units.

To learn more on arc lengths: https://brainly.com/question/16403495

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