Parametric and polar curves: slopes, arc length and area

Worked solutions

Problem 1

1. $\dfrac{dy}{dx} = \dfrac{3t^2 - 3}{2t}$$\tfrac{dy}{dt}$ over $\tfrac{dx}{dt}$.
2. At $t = 2$: slope $\dfrac{12 - 3}4 = \dfrac94$, at the point $(4, 2)$$x = 4$ and $y = 8 - 6 = 2$.
3. Tangent: $y - 2 = \tfrac94(x - 4)$, or $y = \tfrac94x - 7$Point-slope form.
4. Horizontal where $3t^2 - 3 = 0$, $t = \pm1$: the points $(1, -2)$ and $(1, 2)$. Vertical where $2t = 0$, $t = 0$: the point $(0, 0)$At $t = 0$, $\tfrac{dy}{dt} = -3 \ne 0$.
5. Slope $\tfrac94$, tangent $y = \tfrac94x - 7$; horizontal tangents at $(1, \pm2)$, vertical at $(0, 0)$The curve also crosses itself at $(3, 0)$, where $t = \pm\sqrt3$, with two different slopes, $\pm\sqrt3$.

Problem 2

1. $\dfrac{dy}{dx} = \dfrac{3t^2 - 3}{2t} = \dfrac32t - \dfrac3{2t}$Split the fraction before differentiating.
2. $\dfrac{d}{dt}\Big(\dfrac{dy}{dx}\Big) = \dfrac32 + \dfrac3{2t^2}$The derivative of the slope with respect to $t$, not yet with respect to $x$.
3. $\dfrac{d^2y}{dx^2} = \dfrac{\frac32 + \frac3{2t^2}}{2t} = \dfrac{3(t^2 + 1)}{4t^3}$Divide by $\tfrac{dx}{dt} = 2t$: the chain rule, $\tfrac{d}{dx} = \tfrac{1}{dx/dt}\tfrac{d}{dt}$.
4. $\dfrac{d^2y}{dx^2} = \dfrac{3(t^2 + 1)}{4t^3}$, which is $\dfrac{15}{32}$ at $t = 2$Positive, so the curve is concave up there. It is positive for every $t > 0$ and negative for every $t < 0$.

Problem 3

1. $\Big(\dfrac x2\Big)^2 + \Big(\dfrac y3\Big)^2 = \cos^2t + \sin^2t = 1$Eliminate $t$ with the Pythagorean identity: an ellipse with semi-axes $2$ and $3$.
2. $\dfrac{dy}{dx} = \dfrac{3\cos t}{-2\sin t}$, which is $-\dfrac32$ at $t = \tfrac\pi4$, at the point $\big(\sqrt2, \tfrac{3\sqrt2}2\big)$$\cos\tfrac\pi4 = \sin\tfrac\pi4$.
3. Area $= \Big\lvert\displaystyle\int_0^{2\pi}y\,x'(t)\,dt\Big\rvert = \Big\lvert\int_0^{2\pi}3\sin t\cdot(-2\sin t)\,dt\Big\rvert = 6\int_0^{2\pi}\sin^2t\,dt$The curve runs counterclockwise, from right to left along the top, so the integral comes out negative.
4. An ellipse; slope $-\tfrac32$ at $t = \tfrac\pi4$; area $6\pi$$\int_0^{2\pi}\sin^2t\,dt = \pi$. This is $\pi ab$ with $a = 2$, $b = 3$.

Problem 4

1. $x' = 2t$, $y' = 2t^2$, so $x'^2 + y'^2 = 4t^2 + 4t^4 = 4t^2(1 + t^2)$Factor before taking the root.
2. Speed $= 2t\sqrt{1 + t^2}$$t \ge 0$ here, so $\sqrt{4t^2} = 2t$.
3. $L = \displaystyle\int_0^{\sqrt3}2t\sqrt{1 + t^2}\,dt = \frac23\Big[(1 + t^2)^{3/2}\Big]_0^{\sqrt3}$Substitute $u = 1 + t^2$, $du = 2t\,dt$.
4. $\tfrac23(8 - 1) = \tfrac{14}3 \approx 4.667$The curve runs from $(0, 0)$ to $(3, 2\sqrt3)$, and the chord between them is $\sqrt{21} \approx 4.583$.

Problem 5

1. $x'^2 + y'^2 = (1 - \cos t)^2 + \sin^2t = 2 - 2\cos t = 4\sin^2\tfrac t2$The half-angle identity $1 - \cos t = 2\sin^2\tfrac t2$.
2. $L = \displaystyle\int_0^{2\pi}2\Big\lvert\sin\frac t2\Big\rvert\,dt = \Big[-4\cos\frac t2\Big]_0^{2\pi} = 4 + 4$$\sin\tfrac t2 \ge 0$ on $[0, 2\pi]$, so the absolute value can go.
3. Area $= \displaystyle\int_0^{2\pi}y\,x'(t)\,dt = \int_0^{2\pi}(1 - \cos t)^2\,dt = \int_0^{2\pi}\big(1 - 2\cos t + \cos^2t\big)\,dt$$x$ increases from $0$ to $2\pi$, left to right.
4. $= 2\pi - 0 + \pi$$\int_0^{2\pi}\cos^2t\,dt = \pi$.
5. Length $8$, area $3\pi$The arch is $4$ times as long as the wheel's diameter, and its area is $3$ times the wheel's.

Problem 6

1. $x + y = 1$, with $0 \le x \le 1$The point stays on the segment from $(0, 1)$ to $(1, 0)$.
2. $x' = 2\sin t\cos t = \sin2t$, $y' = -\sin2t$, so the speed is $\sqrt2\,\lvert\sin2t\rvert$$\sqrt{2\sin^22t} = \sqrt2\,\lvert\sin2t\rvert$, not $\sqrt2\sin2t$.
3. Distance $= \sqrt2\displaystyle\int_0^\pi\lvert\sin2t\rvert\,dt = \sqrt2\cdot2$$\sin2t$ is negative on $\big(\tfrac\pi2, \pi\big)$, so split there: each half contributes $1$.
4. It travels $2\sqrt2 \approx 2.828$; the segment it traces has length $\sqrt2$At $t = 0$ it is at $(0, 1)$, at $t = \tfrac\pi2$ at $(1, 0)$, and at $t = \pi$ back at $(0, 1)$: the segment is covered twice. The arc length integral measures distance traveled, which is the length of the curve only if it is traced once.

Problem 7

1. $r^2 = 2r\cos\theta$, so $x^2 + y^2 = 2x$, which is $(x - 1)^2 + y^2 = 1$Multiply by $r$, then convert. A circle of radius $1$ centered at $(1, 0)$.
2. The circle is traced once as $\theta$ runs from $-\tfrac\pi2$ to $\tfrac\pi2$$r = 0$ at both ends, and $r \ge 0$ in between.
3. Area $= \dfrac12\displaystyle\int_{-\pi/2}^{\pi/2}4\cos^2\theta\,d\theta = \int_{-\pi/2}^{\pi/2}(1 + \cos2\theta)\,d\theta$$\cos^2\theta = \tfrac12(1 + \cos2\theta)$.
4. $\pi$The $\cos2\theta$ part integrates to $0$. A circle of radius $1$ has area $\pi$, as it should.

Problem 8

1. Area $= \dfrac12\displaystyle\int_0^{2\pi}(1 + \cos\theta)^2\,d\theta = \frac12\int_0^{2\pi}\big(1 + 2\cos\theta + \cos^2\theta\big)\,d\theta$The cardioid is traced once as $\theta$ goes around once, and $r \ge 0$ throughout.
2. $= \tfrac12(2\pi + 0 + \pi) = \tfrac{3\pi}2$$\int_0^{2\pi}\cos^2\theta\,d\theta = \pi$.
3. $r^2 + r'^2 = (1 + \cos\theta)^2 + \sin^2\theta = 2 + 2\cos\theta = 4\cos^2\tfrac\theta2$The half-angle identity again.
4. $L = \displaystyle\int_0^{2\pi}2\Big\lvert\cos\frac\theta2\Big\rvert\,d\theta = 2\int_0^{\pi}2\cos\frac\theta2\,d\theta = 2\Big[4\sin\frac\theta2\Big]_0^\pi$$\cos\tfrac\theta2 < 0$ for $\pi < \theta < 2\pi$, so use the symmetry about the $x$-axis and double the top half.
5. Area $\tfrac{3\pi}2 \approx 4.712$, length $8$Without the absolute value the length integral would be $\big[4\sin\tfrac\theta2\big]_0^{2\pi} = 0$.

Problem 9

1. $3\sin\theta = 1 + \sin\theta$ gives $\sin\theta = \tfrac12$, so $\theta = \tfrac\pi6$ and $\tfrac{5\pi}6$Where the curves cross. Between them, $3\sin\theta$ is the outer curve.
2. Area $= \dfrac12\displaystyle\int_{\pi/6}^{5\pi/6}\Big(9\sin^2\theta - (1 + \sin\theta)^2\Big)\,d\theta = \frac12\int_{\pi/6}^{5\pi/6}\big(8\sin^2\theta - 2\sin\theta - 1\big)\,d\theta$Outer radius squared minus inner radius squared.
3. $\displaystyle\int_{\pi/6}^{5\pi/6}\sin^2\theta\,d\theta = \frac\pi3 + \frac{\sqrt3}4$, $\displaystyle\int_{\pi/6}^{5\pi/6}\sin\theta\,d\theta = \sqrt3$, $\displaystyle\int_{\pi/6}^{5\pi/6}d\theta = \frac{2\pi}3$$\sin^2\theta = \tfrac12(1 - \cos2\theta)$ for the first.
4. $\tfrac12\Big(\tfrac{8\pi}3 + 2\sqrt3 - 2\sqrt3 - \tfrac{2\pi}3\Big)$The $\sqrt3$ terms cancel.
5. $\pi$The circle $r = 3\sin\theta$ has area $\tfrac{9\pi}4$, so the region is a little under half of it.

Problem 10

1. $r = 1$ and $r' = -\sin\theta = -1$ at $\theta = \tfrac\pi2$The point is $(x, y) = (0, 1)$.
2. $\dfrac{dy}{d\theta} = r'\sin\theta + r\cos\theta = -1 + 0 = -1$The product rule on $y = r\sin\theta$.
3. $\dfrac{dx}{d\theta} = r'\cos\theta - r\sin\theta = 0 - 1 = -1$The product rule on $x = r\cos\theta$.
4. $\dfrac{dy}{dx} = \dfrac{-1}{-1} = 1$As $\theta$ increases past $\tfrac\pi2$ the point moves left and down, along a line of slope $1$.