AP Calculus BC Unit 9 Parametric Equations Polar Coordinates and Vector Valued Functions Questions

Since
$$\frac{dx}{dt}=e^{\cos t},$$$$\frac{d^2x}{dt^2}=-e^{\cos t}\sin t, \qquad x''(1)=-e^{\cos1}\sin1=-1.444407.$$
Also,
$$y(t)=2\sin t \quad\Longrightarrow\quad \frac{d^2y}{dt^2}=-2\sin t, \qquad y''(1)=-2\sin1=-1.682942.$$
Therefore,
$$\mathbf a(1)=\langle-1.444407,-1.682942\rangle.$$

$$\begin{aligned} \text { Speed }=\sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}}=\sqrt{\left(e^{\cos t}\right)^{2}+(2 \cos t)^{2}} 0 \leq t \leq \pi \text { and } \sqrt{\left(e^{\cos t}\right)^{2}+(2 \cos t)^{2}}=1.5 \end{aligned}$$$$\begin{aligned} \sqrt{\left(e^{\cos t}\right)^{2}+(2 \cos t)^{2}} =1.5 \end{aligned}$$

1 point

$\Rightarrow t=1.254472, t=2.358077$

Answer

1 point

The first time at which the speed of the particle is 1.5 is
t=1.254.

Scoring notes:

- A response with an implied equation is eligible for both points. For example, a response of

"Speed $=\sqrt{\left(e^{\cos t}\right)^{2}+(2 \cos t)^{2}}$ and is first equal to 1.5 at t=1.254 " earns both points.

- $\sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}}=1.5$ earns the first point. Speed = 1.5 by itself does not earn the first point.

Both of these responses are eligible to earn the second point.

- A response need not consider the value t=2.358077.

- A response of t=1.254 alone does not earn either point.

- A response with a parenthesis error(s) in either $\left(e^{\cos t}\right)^{2}$ or $(2 \cos t)^{2}$ does not earn the first point

but does earn the second point for the correct answer. Note: $\sqrt{\frac{d x^{2}}{d t}+\frac{d y^{2}}{d t}}$ is not considered a

parenthesis error.

- For the second point, if the differential is missing:

○ $\int_{0}^{1} e^{\cos t}$ earns the second point and is eligible for the third point.

○ $x(1)=\int_{0}^{1} e^{\cos t}$ earns the second point but is not eligible for the third point.

○ $x(1)=1+\int_{0}^{1} e^{\cos t}$ earns the second point and is eligible for the third point.

○ $x(1)=\int_{0}^{1} e^{\cos t}+1$ does not earn the second point but earns the third point for the

correct answer.

- The third point is not earned for a response that presents an incorrect statement, such as

$x(1)=\int_{0}^{1} e^{\cos t} d t=1+2.342$.

- Degree mode: In degree mode, $\frac{d y}{d x}=0.735759$ or 0.012841 and $1+\int_{0}^{1} e^{\cos t} d t=3.718144$.

Total for part

(c)
Find the slope of the line tangent to the path of the particle at time
1.
t =
Find the x -coordinate of the
position of the particle at time
1.
t =
Show the work that leads to your answers.

cos
2cos
t
dy dt
t
dy
dx
dx dt
e
=
=

cos1
1
2cos1
0.629530
t
dy
dx
e
=
=
=

The slope of the line tangent to the curve at
1
t =
is 0.630 (or
0.629).
Slope with supporting
work
1 point

( )
( )
1
1 cos
0
0
1
0
1
3.341575
t
dx
x
x
dt
e
dt
dt
=
+
=
+
=
⌠
⌡
∫

The x -coordinate of the position at
1
t =
is 3.342 (or 3.341).
1 cos
0
t
e
dt
∫

1 point

( )
1
x

1 point

Scoring notes:
•
To earn the first point, the response must communicate;
dy dt
dy
dx
dx dt
=
for example:
o
cos1
2cos1
dy
dx
e
=

o
0.63
dy dt
dx dt =

o
( )
1
1.716526,
x′
=

( )
1
1.080605,
y′
=
slope
0.63
=

o
2cos,
dy
t
dt =
slope
0.63
=

•
A response may import an incorrect expression for
( )
y t
′
or value of
( )
1
y′
from part (a), provided it
was declared in part (a).
•
The second point is earned for a response that presents the definite integral
1 cos
0
t
e
dt
∫
or
1
0
dx dt
dt
⌠
⌡

with or without the initial condition.

AP® Calculus AB/BC 2023 Scoring Guidelines
•
For the second point, if the differential is missing:
o
1 cos
0
t
e
∫
earns the second point and is eligible for the third point.
o
( )
1 cos
0
1
t
x
e
= ∫
earns the second point but is not eligible for the third point.
o
( )
1 cos
0
1
1
t
x
e
=
+ ∫
earns the second point and is eligible for the third point.
o
( )
1 cos
0
1
1
t
x
e
=
+
∫
does not earn the second point but earns the third point for the
correct answer.
•
The third point is not earned for a response that presents an incorrect statement, such as
( )
1 cos
0
1
1
2.342.
t
x
dt
e
=
=
+
∫

•
Degree mode: In degree mode,
0.735759
dy
dx =
or 0.012841 and
1 cos
0
1
3.718144.
t
e
dt
+
=
∫

$\int_{0}^{\pi} \sqrt{\left(e^{\cos t}\right)^{2}+(2 \cos t)^{2}}\,dt$

Integral

1 point

=6.034611

The total distance traveled by the particle over $0 \leq t \leq \pi$ is 6.035 (or 6.034).

Answer

1 point

Scoring notes:

- The first point is earned for presenting the correct integrand in a definite integral.

- Parentheses errors were assessed in part (b) and, therefore, will not affect the scoring in part (d).

- If the integrand is an incorrect speed function imported from part (b), the response earns the first point and does not earn the second point.

- An unsupported answer of 6.035 (or 6.034) does not earn either point.

- Degree mode: In degree mode, the total distance is 10.596835 or 8.536161.

Total for part (d)

for question 2

9 points