A particle moves on a parametric path with a given dx/dt and an explicit y(t), starting from a known position. Students find the acceleration vector at an instant, the first time the speed reaches a target value, the slope of the tangent to the path and an x-coordinate, and the total distance traveled over the interval.
A particle moves on a parametric path with a given dx/dt and an explicit y(t), starting from a known position. Students find the acceleration vector at an instant, the first time the speed reaches a target value, the slope of the tangent to the path and an x-coordinate, and the total distance traveled over the interval.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
The acceleration vector is $\langle x''(t),\,y''(t)\rangle$. Differentiate the given velocity components once more. For $x$: we are given $\dfrac{dx}{dt}=e^{\cos t}$, so $$x''(t)=\frac{d}{dt}\left(e^{\cos t}\right)=-\sin t\,e^{\cos t},\qquad x''(1)=-\sin(1)\,e^{\cos 1}=-1.444.$$ For $y$: $y(t)=2\sin t\Rightarrow y'(t)=2\cos t$, so $$y''(t)=\frac{d}{dt}(2\cos t)=-2\sin t,\qquad y''(1)=-2\sin(1)=-1.683.$$ The acceleration vector at $t=1$ is $$\mathbf{a}(1)=\langle -1.444,\,-1.683\rangle.$$
2 points: 1 for x''(1) with setup (the derivative d/dt(e^{cos t}) evaluated at t=1, giving -1.444), 1 for y''(1) with setup (d/dt(2cos t) at t=1, giving -1.683). The exact components are <-e^{cos 1} sin 1, -2 sin 1>. Writing the symbolic derivatives without evaluating at t=1 earns 1 of 2; an unsupported correct vector earns 1 of 2. Components must be labeled if listed separately. Calculator must be in radian mode.
The speed is the magnitude of the velocity vector: $$\text{speed}(t)=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=\sqrt{\left(e^{\cos t}\right)^2+(2\cos t)^2}.$$ Set the speed equal to $1.5$ and solve on $0\le t\le\pi$: $$\sqrt{\left(e^{\cos t}\right)^2+(2\cos t)^2}=1.5.$$ Solving numerically gives two solutions on the interval, $t=1.254472$ and $t=2.358077$. The **first** time the speed equals $1.5$ is $$t=1.254.$$
2 points: 1 for setting the speed expression sqrt((e^{cos t})^2 + (2cos t)^2) equal to 1.5 (an implied equation is acceptable), 1 for the answer t = 1.254. A response need not mention the second solution t = 2.358. 'Speed = 1.5' alone does not earn the setup point. t = 1.254 with no equation earns neither point. Radian mode required (degree mode yields no solution).
**Slope of the tangent line at $t=1$.** For a parametric path, $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2\cos t}{e^{\cos t}},\qquad \frac{dy}{dx}\Big|_{t=1}=\frac{2\cos 1}{e^{\cos 1}}=0.630.$$ **$x$-coordinate at $t=1$.** Recover $x$ from $dx/dt$ by integrating from the known initial position $x(0)=1$: $$x(1)=x(0)+\int_0^1\frac{dx}{dt}\,dt=1+\int_0^1 e^{\cos t}\,dt=1+2.341575=3.342.$$
3 points: 1 for the slope with supporting work — the response must communicate dy/dx = (dy/dt)/(dx/dt), giving 0.630 (or 0.629); 1 for the x(1) setup, the definite integral int_0^1 e^{cos t} dt (with or without the initial condition 1+); 1 for the value x(1) = 3.342 (or 3.341). A bare slope value with no dy/dt over dx/dt structure does not earn the slope point. Radian mode required.
Total distance traveled equals the arc length, the integral of speed over the time interval: $$\text{distance}=\int_0^{\pi}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt=\int_0^{\pi}\sqrt{\left(e^{\cos t}\right)^2+(2\cos t)^2}\,dt.$$ Evaluating numerically, $$\text{distance}=6.035.$$
2 points: 1 for the correct definite integral with the speed integrand on [0, pi], 1 for the answer 6.035 (or 6.034). Parenthesis errors were already assessed in part (b) and do not affect part (d). An incorrect speed function imported from (b) earns the integral point but not the answer point. An unsupported 6.035 earns neither point. Radian mode required.
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