A particle moves in the plane with given x'(t) and y'(t), at a known position at a specific time. Students find the speed at an instant, the total distance traveled over an interval, an initial coordinate recovered by integration, and the times the particle is moving toward a coordinate axis.
A particle moves in the plane with given x'(t) and y'(t), at a known position at a specific time. Students find the speed at an instant, the total distance traveled over an interval, an initial coordinate recovered by integration, and the times the particle is moving toward a coordinate axis.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
The speed at time $t$ is the magnitude of the velocity vector, $\sqrt{(x'(t))^2+(y'(t))^2}$. With $x'(t)=8t-t^2$ and $y'(t)=-t+\sqrt{t^{1.2}+20}$, evaluate at $t=2$: $$x'(2)=8(2)-2^2=12,\qquad y'(2)=-2+\sqrt{2^{1.2}+20}=2.72201.$$ Then $$\text{speed}(2)=\sqrt{(x'(2))^2+(y'(2))^2}=\sqrt{12^2+(2.72201)^2}=\sqrt{151.4093}=12.305.$$ The speed of the particle at $t=2$ seconds is $\mathbf{12.305}$ (or $12.304$) centimeters per second.
2 points. 1 point for the correct setup of speed as sqrt((x'(2))^2 + (y'(2))^2) with the values substituted; 1 point for the answer 12.305 (or 12.304). A decimal answer must be correct to three places; an answer presented only as the unsimplified radical sqrt(12^2 + (-2+sqrt(2^1.2+20))^2) earns the setup point and, if it is the final answer, must be evaluated for the answer point.
The total distance traveled equals the arc length of the path, the integral of speed: $$\text{distance}=\int_0^2 \sqrt{(x'(t))^2+(y'(t))^2}\,dt=\int_0^2\sqrt{(8t-t^2)^2+\left(-t+\sqrt{t^{1.2}+20}\right)^2}\,dt.$$ Evaluating numerically (calculator): $$\text{distance}=15.902.$$ The total distance traveled over $0\le t\le 2$ is $\mathbf{15.902}$ (or $15.901$) centimeters.
2 points. 1 point for the correct distance integral (integral from 0 to 2 of sqrt((x')^2+(y')^2) dt); 1 point for the value 15.902 (or 15.901). Setting up arc length of x'(t) alone, or omitting the square root, loses the integral point.
Recover $y(0)$ from the known position $y(2)=6$ using the Fundamental Theorem of Calculus. Since $y(2)=y(0)+\int_0^2 y'(t)\,dt$, $$y(0)=y(2)-\int_0^2 y'(t)\,dt=6-\int_0^2\left(-t+\sqrt{t^{1.2}+20}\right)dt.$$ Equivalently, $y(0)=6+\int_2^0 y'(t)\,dt$. Numerically $\int_0^2 y'(t)\,dt=7.173613$, so $$y(0)=6-7.173613=-1.174.$$ The $y$-coordinate of the particle at $t=0$ is $\mathbf{-1.174}$ (or $-1.173$).
3 points. 1 point for the definite-integral expression int_0^2 y'(t) dt (equivalently int_2^0); 1 point for using the initial/known condition y(2)=6 to relate y(0) to the integral; 1 point for the answer -1.174 (or -1.173). The minus sign matters: 6 minus 7.173613 is negative. Writing 6 + int_2^0 y'(t) dt = 6 - 7.173613 is the same computation. Reporting the magnitude 1.174 without the sign does not earn the answer point.
For $2\le t\le 8$ the particle stays in the first quadrant, so $y(t)>0$; the particle moves **toward the $x$-axis** exactly when its $y$-coordinate is **decreasing**, i.e. when $y'(t)<0$. Solve $y'(t)=-t+\sqrt{t^{1.2}+20}=0$ on $[2,8]$. Numerically the only solution is $t=5.222$. Checking signs: $y'(2)=2.722>0$ and $y'(8)=-2.332<0$, so $y'(t)$ changes from positive to negative at $t=5.222$ and stays negative afterward. Therefore the particle is moving toward the $x$-axis for $\mathbf{5.222\le t\le 8}$ (equivalently $5.221\le t\le 8$); the interval may be written open or closed.
2 points. 1 point for considering the sign of y'(t) (recognizing that 'toward the x-axis' means the y-coordinate decreases, y'(t)<0); 1 point for the answer 5.222 <= t <= 8 (or 5.221) with reason. The endpoint t=5.222 is where y'(t)=0; the interval may be open, half-open, or closed. A response based on x'(t) or on speed does not earn the sign point.
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