(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
+
=
∫