A polar curve and a circle are given. Students find dr/dtheta at a point, the area of the region inside the curve but outside the circle, the angle corresponding to the point farthest from the y-axis, and the rate at which a particle's distance from the origin changes as it travels along the curve.
A polar curve and a circle are given. Students find dr/dtheta at a point, the area of the region inside the curve but outside the circle, the angle corresponding to the point farthest from the y-axis, and the rate at which a particle's distance from the origin changes as it travels along the curve.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
Curve $C$ is $r(\theta)=2\sin^2\theta$. Differentiate with respect to $\theta$ (chain rule or the identity $2\sin^2\theta=1-\cos 2\theta$): $$\frac{dr}{d\theta}=2\cdot 2\sin\theta\cos\theta=4\sin\theta\cos\theta=2\sin 2\theta.$$ Evaluate at $\theta=1.3$: $$\frac{dr}{d\theta}\Big|_{\theta=1.3}=4\sin(1.3)\cos(1.3)=1.031003\approx 1.031.$$ So at the point on $C$ where $\theta=1.3$, $r$ is changing at about $1.031$ (units of $r$ per radian).
1 point (P1): the response must show differentiation of r and report the correct value 1.031 (accurate to three decimals, rounded or truncated). The exact setup 4 sin(1.3) cos(1.3) also earns the point. Notation that conflates dr/dtheta with r'(theta) (e.g. r'=1.031) is still accepted; an inappropriately rounded answer is not.
Find where $C$ meets the semicircle $r=\tfrac12$ to set the limits: $$2\sin^2\theta=\frac12 \;\Rightarrow\; \sin^2\theta=\frac14 \;\Rightarrow\; \sin\theta=\frac12 \;\Rightarrow\; \theta_1=\frac{\pi}{6}=0.523599,\quad \theta_2=\frac{5\pi}{6}=2.617994.$$ Between these angles $C$ lies outside the circle, so the area inside $C$ and outside $r=\tfrac12$ is $$A=\frac12\int_{\pi/6}^{5\pi/6}\!\left[\big(2\sin^2\theta\big)^2-\left(\tfrac12\right)^2\right]d\theta = 2.066769\approx 2.067.$$ Exact value: $A=\dfrac{7\sqrt3}{16}+\dfrac{5\pi}{12}=2.066769$.
3 points. P2: a definite integral including (r(theta))^2, i.e. integral of (2 sin^2 theta)^2 (with or without d theta). P3: a correct integrand, the difference (2 sin^2 theta)^2 - (1/2)^2 (or two integrals of the two squared radii). P4 (the answer point): the correct limits pi/6 and 5pi/6, the factor 1/2, and the value 2.067 (or 2.066). A symmetry form (1/2) integral_{pi/6}^{pi/2} of the integrand times 2 also earns all three. The 1/2 factor and limits are assessed in P4, not P2/P3.
The $x$-coordinate of a point on $C$ is $x(\theta)=r\cos\theta=2\sin^2\theta\cos\theta$. On $0\le\theta\le\tfrac{\pi}{2}$ the curve lies in the first quadrant ($x\ge 0$), so the point farthest from the $y$-axis is where $x(\theta)$ is greatest. Maximize $x$ by setting its derivative to zero (the given $\tfrac{dx}{d\theta}=4\sin\theta\cos^2\theta-2\sin^3\theta$): $$\frac{dx}{d\theta}=2\sin\theta\big(2\cos^2\theta-\sin^2\theta\big)=0.$$ On $(0,\tfrac{\pi}{2})$, $\sin\theta\neq 0$, so $2\cos^2\theta=\sin^2\theta\Rightarrow\tan^2\theta=2\Rightarrow\theta=\arctan\sqrt2=0.955317\approx 0.955.$ Compare $x$ at the critical value and the endpoints (candidates test): $$x(0)=0,\qquad x(0.955317)=0.769800,\qquad x\!\left(\tfrac{\pi}{2}\right)=0.$$ The largest $x$ occurs at $\theta=0.955$, so that is the angle for the point farthest from the $y$-axis.
3 points. P5: considers dx/d theta = 0 (discussing sign change of dx/d theta or 'critical points of x(theta)' also earns it; merely stating theta=0.955317 does not). P6 (justification): a global argument, e.g. a candidates test correctly evaluating x at theta=0, 0.955317, and pi/2 (correct to the first decimal), or a sign analysis of dx/d theta plus the fact that 0.955 is the only interior critical point. A purely local First/Second Derivative Test does not earn P6 but is still eligible for P7. P7 (answer): theta=0.955 (=arctan sqrt2 = arccos(1/sqrt3) = arcsin(sqrt(2/3))), to three decimals.
The particle's distance from the origin is $r$, and $\tfrac{d\theta}{dt}=15$. By the chain rule, $$\frac{dr}{dt}=\frac{dr}{d\theta}\cdot\frac{d\theta}{dt}.$$ Using $\tfrac{dr}{d\theta}\big|_{\theta=1.3}=1.031003$ from part (A), $$\frac{dr}{dt}\Big|_{\theta=1.3}=1.031003\cdot 15=15.465041\approx 15.465.$$ When the particle is at $\theta=1.3$, its distance from the origin is increasing at about $15.465$ (units per unit time).
2 points. P8: the chain-rule setup dr/dt = (dr/d theta)(d theta/dt), shown symbolically or numerically (e.g. 1.031*15 or [part A answer]*15), in one or more steps. P9 (answer): the correct value 15.465 to three decimals. A response of 1.031*15 earns both P8 and P9.
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