Mathematical Immaturity

14.21 Miscellaneous Review Exercises

• A curve is given by a polar equation $r = f(\theta).$ Find $f$ if an arbitrary arc joining two distinct points of the curve has arc length proportional to (a) the angle subtended at the origin; (b) the difference of the radial distances from the origin to its endpoints; (c) the area of the sector formed by the arc and the radii to its endpoints.
• Recall from Section 14.19, Exercise 4 where we proved that for a polar function $r = f(\theta),$ its arc length from $\theta = a$ to $\theta = b,$ where $a \leq \theta \leq b \leq a + 2\pi,$ is given by: \begin{align*} \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta \end{align*}
• (a) For a polar function $r = f(\theta),$ its arc length from $\theta = a$ to $\theta = b,$ where $a \leq \theta \leq b \leq a + 2\pi,$ is given by: \begin{align*} \int_a^b \sqrt{f^2(\theta) + \left[f'(\theta)\right]^2}\,d\theta \end{align*} Then, if an arbitrary arc joining points $(a, f(a))$ and $(x, f(x))$ is proportional to the angle $x$ distended from the origin, we have \begin{align*} \int_a^x \sqrt{f^2(\theta) + \left[f'(\theta)\right]^2}\,d\theta &= kx \end{align*} for some $k > 0.$ Using the First Fundamental Theorem of Calculus and differentiating with respect to $x,$ we have $k = \sqrt{f^2(x) + \left[f'(x)\right]^2},$ or equivalently, \begin{align*} k^2 &= f^2(x) + \left[f'(x)\right]^2 \end{align*} Differentiating once more, we see that $f'(x)\left[f(x) + f''(x)\right] = 0.$ This relation is satisfied in two possible ways: 1. $f'(x) = 0.$ In this case, $f(x)$ is constant and the integral for arc length becomes \begin{align*} \\ \int_a^{\theta} \sqrt{f^2(x) + \left[f'(x)\right]^2}\,dx &= \int_a^{\theta} f(x)\,dx \\ \\ &= F(\theta) - F(a) \\ \\ &= k\theta \end{align*} where $F'(\theta) = f(\theta).$ Then, differentiating $F$ with respect to $\theta$ gives us $f(\theta) = k.$ 2. $f'(x) \neq 0.$ In this case, $f''(x) + f(x) = 0.$ But recall from Theorem 8.6 (c) that for $b > 0 $ the the second order equation $y'' + by = 0$ is satisfied uniquely by \begin{align*} y &= c_1\cos cx + c_2\sin cx \end{align*} where $c_1$ and $c_2$ are arbitrary constants, and $b = c^2.$ But in this case, $b = 1,$ which gives us \begin{align*} \\ f(x) &= c_1\cos x + c_2\sin x \end{align*} Where $c_1^2 + c_2^2 = k^2.$ Then, setting $c_1 = 0,$ $c_2 = k,$ and $C = 0,$ we can then use the identity $\sin(x + C) = \sin x \cos C + \cos x \sin C$ to rewrite $f(x) = k\sin(x + C).\, \blacksquare$ (b) For a polar function $r = f(\theta),$ its arc length from $\theta = a$ to $\theta = b,$ where $a \leq \theta \leq b \leq a + 2\pi,$ is given by: \begin{align*} \int_a^b \sqrt{f^2(\theta) + \left[f'(\theta)\right]^2}\,d\theta \end{align*} Then, if an arbitrary arc joining points $(a, f(a))$ and $(x, f(x))$ is proportional to the difference of the radial distances from the origin, we have: \begin{align*} \int_a^x \sqrt{f^2(\theta) + \left[f'(\theta)\right]^2}\,d\theta &= k\left[f(x) - f(a)\right] \end{align*} for some $k > 0.$ Using the First Fundamental Theorem of Calculus and differentiating with respect to $x,$ we have $\sqrt{f^2(x) + \left[f'(x)\right]^2} = kf'(x),$ or equivalently, \begin{align*} \\ f^2(x) + \left[f'(x)\right]^2 &= k^2\left[f'(x)\right]^2 \end{align*} Differentiating once more and combining terms, we get \begin{align*} \\ f'(x)\left[f(x) + f''(x)\right] &= f'(x)k^2f''(x) \end{align*} If $f'(x) = 0$ then $f(x)$ is constant for all $x$ and the difference $f(\theta) - f(a)$ is trivially zero. Suppose then that $f'(x) \neq 0.$ If $k = 1,$ we have $f(x) = 0.$ Otherwise, we can see that $f(x)$ is a scalar multiple of $f''(x),$ where \begin{align*} \\ f''(x) + \frac{1}{1 - k^2}f(x) &= 0 \end{align*} Following the result of Theorem 8.6 for the second order equation $y'' + by = 0,$ we have two cases: 1. $0 < k < 1.$ In this case, $b > 0$ and $y = f(x)$ can be solved for arbitrary constants $c_1$ and $c_2$ as $$y = c_1\cos cx + c_2\sin cx$$ where $c = \sqrt{1/(1 - k^2)}.$ 2. $k > 1.$ In this case, $b < 0$ and $y = f(x)$ can be solved for arbitrary constants $c_1$ and $c_2$ as \begin{align*} f(x) &= c_1e^{cx} + c_2e^{-cx} \end{align*} where $c = \sqrt{1/(k^2 - 1)}.$ Setting $c_1 = C$ and $c_2 = 0,$ we get \begin{align*} \\ f(x) &= Ce^{x/\sqrt{k^2 - 1}} \quad \blacksquare \end{align*} (c) For a polar function $r = f(\theta),$ its arc length from $\theta = a$ to $\theta = b,$ where $a \leq \theta \leq b \leq a + 2\pi,$ is given by: \begin{align*} \int_a^b \sqrt{f^2(\theta) + \left[f'(\theta)\right]^2}\,d\theta \end{align*} Then, if an arbitrary arc joining points $(a, f(a))$ and $(x, f(x))$ is proportional to the area of the sector formed by the arc and the radii to its endpoints, we have: \begin{align*} \int_a^x \sqrt{f^2(\theta) + \left[f'(\theta)\right]^2}\,d\theta &= \frac{k}{2}\int_a^x f^2(\theta)\,d\theta \end{align*} for some $k > 0.$ Using the First Fundamental Theorem of Calculus and differentiating with respect to $x,$ we have $\sqrt{f^2(x) + \left[f'(x)\right]^2} = \frac{k}{2}f^{2}(x),$ or equivalently, \begin{align*} \\ f^2(x) + \left[f'(x)\right]^2 &= \frac{k^2}{4}f^{4}(x) \end{align*} Differentiating once more, setting $y = f(x),$ we get \begin{align*} \\ 2yy' + 2y'y'' &= k^2y^3y' \end{align*} This gives us two possible cases 1. $y' = 0.$ In this case, $y$ is a constant $C$ for all $\theta$ and the relation between arc length and sector area becomes \begin{align*} \\ C\int_a^x \,d\theta &= C^2\frac{k}{2}\int_a^x\,d\theta \end{align*} And, assuming $C \neq 0,$ we find that $f(\theta) = C = 2/k.$ 2. $y' \neq 0.$ In this case, we can return to the first-order equation, setting $r = f(\theta)$ to give us \begin{align*} \\ \left(\frac{dr}{d\theta}\right)^2 &= \frac{k^2}{4}r^4 - r^2 \\ \\ &= r^2\left(\frac{k^2}{4}r^2 - 1\right) \end{align*} Taking the square root of both sides gives us the separable differential equation \begin{align*} \\ \frac{dr}{d\theta} &= r\left(\frac{k^2}{4}r^2 - 1\right)^{1/2} \end{align*} Separating the variables and setting up the integrals, we get \begin{align*} \\ \int \frac{dr}{r\sqrt{(k^2r^2/4) - 1}} &= \int d\theta \end{align*} Now, recall from Section 6.22 Exercise 4 that \begin{align*} D\, \text{arcsec}\,x &= \frac{1}{|x|\sqrt{x^2 - 1}} \end{align*} This means that if $x = kr/2$ and $dx = kdr/2,$ the left-hand side becomes \begin{align*} \\ \int \frac{dr}{r\sqrt{(k^2r^2/4) - 1}} &= \frac{2}{k}\int \frac{dx}{\frac{2x}{k}\sqrt{x^2 - 1}} \\ \\ &= \int \frac{dx}{x\sqrt{x^2 - 1}} \\ \\ &= \text{arcsec}\,x + C \\ \\ &= \text{arcsec}\left(\frac{kr}{2}\right) + C \end{align*} Then, evaluating the right-hand side and rearranging terms, we see that \begin{align*} \\ \text{arcsec}\,\frac{kr}{2} &= \theta + C \\ \\ \frac{kr}{2} &= \sec\left(\theta + C\right) \\ \\ r &= \left(\frac{2}{k}\right)\sec\left(\theta + C \right) \quad \blacksquare \end{align*}