AP Calculus BC Fun 8 B Determine Values for Positions and Rates of Change in Problems Involving Planar Motion Questions

$\sqrt{\left(x^{\prime}(2)\right)^{2}+\left(y^{\prime}(2)\right)^{2}}$

Setup for speed 1 point

=12.3048506

The speed of the particle at time t=2 seconds is 12.305 (or 12.304 ) centimeters per second.

Answer 1 point

Scoring notes:

- The first point is earned for the expression $\sqrt{\left(x^{\prime}(2)\right)^{2}+\left(y^{\prime}(2)\right)^{2}}, \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}}$, or

equivalent.

- A response that presents just the exact answer, $\sqrt{144+\left(-2+\sqrt{2^{1.2}+20}\right)^{2}}$, earns both points.

- The second point is earned only for the answer 12.305 (or 12.304 ) regardless of whether the first

point is earned.

- A response that includes a linkage error, such as

$\sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}}=\sqrt{144+\left(-2+\sqrt{2^{1.2}+20}\right)^{2}}$ or $\sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}}=12.305$ (or 12.304 ),

earns at most 1 of the 2 points.

- Missing or incorrect units will not affect scoring in this part.

Total for part (a) 2 points

$\int_{0}^{2} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t$

Integral

1 point

=15.901715

The total distance traveled by the particle over the time interval $0 \leq t \leq 2$ is 15.902 (or 15.901) centimeters.

Answer

1 point

Scoring notes:

- The first point is earned only for an integral of $\int_{0}^{2} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t$ (or the mathematical

equivalent), with or without the differential.

○ Note: $\int_{0}^{2} \sqrt{x^{\prime}(t)^{2}+y^{\prime}(t)^{2}} d t$ is not read as a parenthesis error.

- The second point is earned only for an answer of 15.902 (or 15.901), regardless of whether the

first point is earned.

- Missing or incorrect units will not affect scoring in this part.

Total for part (b) 2 points

$y(0)=6+\int_{2}^{0} y^{\prime}(t) d t=6-7.173613=-1.173613$}

Definite integral

1 point

Uses initial condition

1 point

The y-coordinate of the position of the particle at time t=0 is
-1.174 (or -1.173).

Answer

1 point

Scoring notes:

- An answer of -1.174 (or -1.173) with no supporting work does not earn any points.

- The first point is earned for either of the definite integrals $\int_{2}^{0} y^{\prime}(t) d t$ or $\int_{0}^{2} y^{\prime}(t) d t$.

- The second point is earned for any of:

○ $y(2) \pm \int_{2}^{0} y^{\prime}(t) d t, 6 \pm \int_{2}^{0} y^{\prime}(t) d t$,

○ $y(2) \pm \int_{0}^{2} y^{\prime}(t) d t, 6 \pm \int_{0}^{2} y^{\prime}(t) d t$,

○ $y(2) \pm 7.173613$, or $6 \pm 7.173613$.

- A response that attempts to evaluate $\int y^{\prime}(t) d t$ does not earn the first or the third point.

○ Such a response can earn the second point by attempting to solve for the constant of integration

by presenting an expression as an antiderivative for $y^{\prime}(t)$, evaluating this expression at t=2,

and setting this expression equal to 6.

- A response that reverses the limits of integration, e.g., $y(0)=6+\int_{0}^{2} y^{\prime}(t) d t$ or 6+7.173613,

earns the second point but does not earn the third point.

- In order to earn the third point, a response must have earned at least 1 of the first 2 points.

- A response containing any linkage error can earn at most 2 of the 3 points. For example:

○ Equating two unequal quantities: $\int_{2}^{0} y^{\prime}(t) d t=-1.174, \int_{2}^{0} y^{\prime}(t) d t=6-7.173613$,

$6+\int_{0}^{2} y^{\prime}(t) d t=6-7.173613$, or $6+\int_{0}^{2} y^{\prime}(t) d t=6+7.173613=-1.173613$

○ Equating an expression to a numerical value: $y(t)=6+\int_{2}^{0} y^{\prime}(t) d t=-1.174$

- Missing differentials (d t):

○ Unambiguous responses of $y(2)+\int_{2}^{0} y^{\prime}(t), y(2)-\int_{0}^{2} y^{\prime}(t), 6+\int_{2}^{0} y^{\prime}(t)$, or $6-\int_{0}^{2} y^{\prime}(t)$ earn

the first 2 points and would earn the third point for the correct numerical answer.

○ Unambiguous responses of $y(2)+\int_{0}^{2} y^{\prime}(t)$ or $6+\int_{0}^{2} y^{\prime}(t)$ with reversed limits of integration

and missing differential earn the first 2 points but cannot earn the third point.

○ Ambiguous responses of $\int_{2}^{0} y^{\prime}(t)+y(2),-\int_{0}^{2} y^{\prime}(t)+y(2), \int_{2}^{0} y^{\prime}(t)+6$, or $-\int_{0}^{2} y^{\prime}(t)+6$

earn the first point, do not earn the second point, but do earn the third point if a correct numeric

answer is provided. If no numeric answer is given, none of these responses earn the third point.

○ Ambiguous responses of $\int_{0}^{2} y^{\prime}(t)+y(2)$ or $\int_{0}^{2} y^{\prime}(t)+6$ with reversed limits of integration and

no differential earn 1 out of 3 points.

- If a response provides work for both the x - and y-coordinates, the work for the x-coordinate will

not affect scoring.

- However, a response that reports only a completely correct x-coordinate of the particle's position

at time t=0 with all supporting work, e.g., $x(0)=3+\int_{2}^{0} x^{\prime}(t) d t=-\frac{31}{3}=-10.333$, earns 2 out

of 3 points.

Total for part (c) 3 points

Because y(t)>0 when $2 \leq t \leq 8$, the particle will be moving
toward the x-axis when $y^{\prime}(t)<0$. This occurs when
5.222 (or 5.221) <t<8.}

Considers sign of
$y^{\prime}(t)$

1 point

Answer with reason

1 point

Scoring notes:

- The first point can be earned by stating $y^{\prime}(t)=0, y^{\prime}(t)<0, y^{\prime}(t)>0$, or t=5.222 (or 5.221).

Note: $y^{\prime}(t)$ may be written as $\frac{d y}{d t}$.

- The second point cannot be earned without the first.

- To earn the second point, a response must identify the correct interval (and no additional intervals

in [2, 8]) and explicitly state the need for $y^{\prime}(t)<0$. The interval can be open, closed, or half-open.

Total for part (d)

for question 2

9 points