AP@ Calculus BC
Free-Response Scoring Guidelines
Ouestion
1
The rate at which raw sewage enters â t¡eatment tank is given
by E(r) =
850 +
f -^2\
$ | gallons
\>
)
Tl5cosl
for 0 < t < 4 hours. Treated sewage is removed from the tank at the constant rate of 645
gallons per hour. The treâtment tank is empty at time I = 0.
per hour
(a) How many gallons of sewage
enter the treatment tank during the time interval 0 < ¡
I
4 ? Round
your answer to the nearest gallon.
(b) For 0 < / < 4, at what time I is the
amount of sewage in the treatment rank greatest? To the nearest
gallon, what is the maximum amount of sewage in the tank? Justiô/ your answers.
1$
For O < t < 4, rhe cost of treating the raw sewâge that enters the tank at rime I is (0.15 - 0.02r)
'dollars per gallon. To the nearest dolla¡, what is the total cost of feating all the sewâge that enters
the tank during the time interval O <
øl
t < 4?
integral
at 3981 sallons
ffzçt¡ =
ans\.ver
(b) Let ,S(t) be the amoùnt of sewage in the treatment tank at
time f. Then S'(t) = B1¡1- 6a5 and S'þ) = I y¡"n
E(t) = 6a5. On the interval O<t<4, E(t) = 645
.when ¡ = 2.309 atJ,d t = 3.559.
r
(hours)
0
2309
3.559
4
amount of sewage in featment tank
0
dr
-
6a5(2.309)
= t63i.t.B
n1,¡ a,
-
64s(3.ss9)
=
lo"otu,,)
!"tn
3981.022
-
645(4)
=
1228.520
1401.022
The amount of sewage in the treatment tank is greatest at
t = 2.309 hoùrs. At that time, the amount of sewage in
the tank, rounded to the ne est gallon, is 1637 gallons.
(c) Total cosr =
o.ozt) E(t)
lr-to.tt -
dr
The total cost of treating the sewage that enters the tank
during the time interval 0 < t < 4, to the nearest dollar,
is $47 4.
E(l) = 645
1:identifies t = 2.309 as
1 : sets
4l
a
candidate
I : amount of sewage
I : conclusion
aÍ.
t = 2.309
Calculus BC
Free-Response Scoring Guidelines
AP@
Ouestion 2
A particle moving along
a
(¡(¡), l(¡))
cuwe in the x)-plane has position
& t6 -)^\r/¡
-
dt-\t
"
at time
/ > 0, where
dv
At time / = 3, the particle is at rhe point (5, 4).
(a) Find the speed of
(b) Write
fhe particle at time
/ = 3.
an equation for the line tangent to the path of the
¿,.
(c)
/ at which the particle is farthest to the dght? Ifyes, explain why and give the value
and the x-coordinate of the position of the particle at that time. If no, explain why not.
Desc¡ibe the behavior of the path of the particle as
f
increases without boùnd.
I
(b)
3.
Is there a time
of /
(d)
paficle at time / =
At¡=',*=#=+=-í
= -0.149
: answer
[1:slope
".
' l I: equation of tangent line
An equation for the line tangent to the path at the
poinr (5,4) is y
(c)
Since
t)
-4 = -4t'e
ff , o fo, 0 < t <2 una ff
the particle is farthest to the right at
¡-coordinate of the position is
.xrlt = 5 +
(d) since
t\#
the particle
f' E - t\"'
\¡ )
Jt
= (-3)'/'
dr =
un¿
<0 lor ¡ >2,
t = 2. The
).let
,,y#
o1
5.js2
= o,
will continue moving farther to the left
while approaching
a
4:
horizontal asymptote.
,. { t: behavior of -r
ll:behaviorofy
Calculus BC
Free-Response Scoring Guidelines
AP@
Ouestion 3
I
(minutes)
/1(r) ('c)
0
4
8
12
t6
65
68
73
80
go
The temperature, in degrees Celsius ("C), ofan oven being heated is modeled by an increasing
differentiable function 11 of time l, where f is measured in minutes. The table above gives the
temperatùre as recorded every 4 minutes over a l6-rninute period.
(a)
Use the datâ in the table to estimale tfe instantaneous rate at which the temperature of the oven is
changing at time t = 10. Show the crfìputations that lead to your answer. Indicate units of measure.
(b) Write an integral expression in terms of' ll for the average temperature of the oven between time
f = 0 and time / = i6. Estimate the average temperature of the oven using a left Riemann sum with
(c)
four subintewals of equal length. Show the computations that lead to your answer.
Is your approximation in part (b) an uderestimate or an overestimate of the average temperature?
Give a reason lor your answer.
(d) Are
the daø in the table consistent with or do they contradict the claim that the temperature of the
oven is inüeasing at an increasing rate? Give a reason for your answer.
(a) H'(ro) = Htll) _lrtst
(b)
Average temperature is
ltu
¡16
16 Jn
= f,"c I
^;,
difference quotient
2:
answer with units
r
H(r) dt
4.286
16
=
71
.ló
*1,
a(r)at
left Riemann sum
n1,¡ a, = 4.(6s + 68 + 73 + 80)
Average temperature
(c)
I
=Y
answer
.5"C
The left Riemann sum approximation is an underestimate of the
integral because the graph of Il is increasing. Dividing by 16
will not change the inequality, so 7l.5oC is an underestimate of
I
: answer with reason
the average temperature.
(d) If
a continuous flnction is inc¡easing at an increasing rate, then
the slopes of the secant lines of the graph of the function are
increasing. The slopes of the secant lines for the four intervals in
therableare
i i i
ana
loQ.
respectively.
Since the slopes are increasing, the data âre consistent with
the claim.
OR
By the Mean Value Theorem, the slopes are also the values
of H'(c¡) fo¡ some times I 1c2 1ca < c4, respectively.
Since these derivâtive values are positive and increasing, the
data are consistent with the claim.
-39-
I
: considers slopes
I
I
: explanation
of
four secant lines
3:
: conclusion consistent
with explanation
Gatculus BC
Free-Response Scoring Guidelines
AP@
Ouestion 4
Graph
Letlbe
0
/
of
/(x) = (lnx)(sinx). The figure above shows the graph of/
g is defrned by gþ)=l: f(t)dtfor0<x3Ztt.
the function given by
<x.3Ztt.
The function
for
(a) Find s(1) and s'(i).
(b) On what intervals, if any, is g increasing? Justify your answer.
(c) For 0 < x 1 2x, find the value of ¡ at which g has an absolute minimum. Justify your ânswer.
(d) For 0 < x <2x, is the¡e a value of ;r at which the graph of g is tangent to the;r-axis? Explain why
or why not.
.t
1at s(r) =
(b). Since
l,Í(,),t, -
e'G) =
0 and
f(r),
s'(l)
-
"r(l)
g is increasing on
-
I : s(l)
I : s'(l)
0
the interval
2:
13x3nbecause/(x) > 0 for 1<x<2.
(c) For 0 < x < 2tc, C'þ)= f(r) = 0 when ;r = l, z
g' = / changes from negâtive to positive only at
¡ = 1. The absolute minimum must occur at r = I
I
s(2o\ =
J,
.n
identifies
I
and
2z
as candidates
indicates that the graph of g
or at the right endpoint. Since g(1) = 0 and
.)Ì
:
decreases, increases, then decreases
.1n
1 :justifies
fQ) ú = ), f(r) ù + J, flr) dt < 0
g(2ft) < g(1)
1 : answer
by comparison of the two areas, the absolute
minimum occurs at r = 2n.
(d)
I
Yes, the graph of is tangent to the x-axis at x
since s(1) = 0 and C'(1) = 0.
:
1
2:
1 : answer of "yes"
I
:explanation
with
¡
=I
Calculus BC
Free-Response Scoring Guidelines
AP@
Ouestion
/ be the function
)yyf a) = z.
Let
(a.¡ Find the value
satisfying
ot JOliç+*
f'(x) = 4x - zxf(x)
-Z*¡1r)) rtr.
(b)
Use Euler's method to approximate
(c)
Find the
particul
condition
@)
/(0) =
solution y =
=
f(
!dx:= +, 4ry
wirh the initiaÌ
I
I
: use
of FTC
: answer from
limiting process
= -3
/(o) /'(o) (-+) = .0
'
(-
j)
=
'
:
Euler's method with two steps
: Euler's approximation
1 : separates variables
1 : antiderivatives
-!nl+-zyl=|," *e
lnla-zyl= -x2
2y-4=Ce-'2
1 : constant of integration
1 : uses initial condition
+B
1 : solves for
y
max2f 5 [l-1-0-0-0] if no
=6
constant of integration
Y
=2
to
=,( +).'(-j) (-+)='., (-j)
{Ò U}rrav = xax
C
rnr*"..
= 0, with two stepsoofequal size.
to the differential eqùation
ff r'rÐ o, = ;y¿!!r'ata" = ¡afrc,li]
/(
;
/(0) = 5 an¿
5.
nmf(a)-/(o)) =2-5
(b)
stafting ât
I-Øx-zxf(x))dx
=
for all real numbers .x, with
Strow rhe work rhar leads to your
/(-1),
¡(x)
5
for all ¡eal numbers
¡
.f
(-l)
Calculus BG
Free-Response Scoring Guidelines
AP@
Ouestion 6
I
The function
is continuous for all real numbers
.r and is defined by
forx*0.
,1*¡=sr*(411
x
(a)
Use L'Hospital's Rule to find the value of
(b)
Let / be the function given by /(-r) = cos(2:r). Write the first four nonzero terms and the general
term of the Taylor series for / about .r = 0.
g(0). Show
the work that leads ro your answer.
(c) Use your answer from part (b) to write the first three nonzero terms and the general term of
Taylor series for I about r = 0.
g
(d) Determine whether
has a relative minimum, a relative maximum, or neither at
¡
your answer.
(a)
the
= 0. Justify
Since g is continuous,
=l'"lgÚ|1
+0
s(0)=liqe{x)
n+u
.r
-.
Itm*-2sin(Zx\
r-iO ¿x
,.
(b)
(c)
-4cos(2x)
(2r)a (2r)6
41 - 61
. 16 q 64 ¡
ï4!x-6lx+
-.
cos(2rc)
qlx) =
(d) s'(¡)
-
4 16¡
21+ 4tx- -
2.16
64
6't
4.64
-fi,-;t
+ ... +
a
t
I
Therefore, g has
Derivative Test.
^2t1
.+(-t)"f;¡.x'z"
+.
^2t1
(-r)'frr.x')"-'z
+
.so8'(0) =0.
,,1
t
:
first t\'r'o terms
:
next two lerms
: general term
: first three terms
: general term
; answer
:justification
2.16_ 3.4 64 t I
e"þ)= ii
^ ,zti
ç\'l#+
,
+.
-!*,,
a
.... so s,,(0) > 0.
relative minimum at
r
= 0 by the Second
o
a