Download Math 1131Q Grading Rubric Fall 2010 Here is the grading rubric for

Math 1131Q
Grading Rubric
Fall 2010
Here is the grading rubric for the
ECE Final Exam Core Questions for
Math 1131Q, Calculus I
Fall 2010
I have included grading directions for partial credit in blue
and the mathematical answers in red. So it would be better
to print this out on a color printer if possible.
Let me know if you have any problems with this or understanding what was done on the Storrs campus or if I have
made some mistakes along the way or if I haven’t taken something into account.
Core Final Exam Rubric
2
Math 1131Q
1. (3 pts each) For each part, if the statement is always true, circle the printed capital T. If the statement
is sometimes false, circle the printed capital F. Justify your answer.
The
idea is that there is one point for the correct answer and
upExam
to 2 points for a correct
Mathgeneral
1131
Final
justification, but if the justification is completely wrong where the student’s choice of true or false is
clearly
chosen
then
the one
a correct
orstatement
false is is
NOT given.
1. For each
part,at
if random,
the statement
is always
true,point
circle for
the the
printed
capital T.true
If the
sometimes false, circle the printed capital F. Justify your answer.
(a) If f 0 is the graph pictured, then f is increasing at the point B.
(a) If f 0 is the graph pictured, then f is increasing at the point A.
T F
[3]
T
F
y
graph of f ’
B
Math 1131
x
A
Final Exam
1. For each0 0 part, if the statement is always true, circle the printed capital T. If the statement is
AsIfff(B)
< graph
0,
thepictured,
function
f fisis decreasing
Bpoint A.
(b)
isfalse,
the
then
concave
upyour
atat
the
T F
sometimes
circle
the
printed
capital
F. Justify
answer.
(b) If
fy 0
is the graph pictured, then f is concave up at the point A.
of fgraph
’
(a) If f 0 graph
is the
pictured, then f is increasing at the point A.
y
B
x
B
x
[3]
T F
[3]
T F
[3]
T
F
A graph of f ’
A
(c) An antiderivative of f (x) =
2x
x2
is F (x) =
.
cos x
sin x
Since
is increasing
at then
A, ff is(A)
> 0, up
and
so the
(b)
If f 0 fis the
graph pictured,
concave
at the
pointfunction
A.
00
00
y
(c) An antiderivative
of f (x) =
graph of f ’
(d) If A(x) =
Z b
F is concave
up at A.
T F
[3]
x2
2x
is F (x) =
.
cos x
sin x
f (t) dt,
x then A0 (x) = f (x).
T
T F
F
[3]
B
a
The Aantiderivative
of a quotient is not the quotient of the antiderivatives. This can be verified
2x sin x − x2 cos x
2x
0
here by noting F (x) =
6=
= f (x).
(sin x)2
cos x
Z b
2x
x2
(c) An antiderivative
of f (x) =
is F (x) =
.
T F
[3]
0
cos
x
sin
(d) If A(x) =
f (t) dt, then A (x) = f (x). x
T F
(e) If v(t) > 0 a
is the velocity of a car that is speeding up, then a right-endpoint
Riemann sum overestimates the car’s displacement.
[3]
T F
Since a and b are constants, we have A(x) = F (b) − F (a) (where fZ = F 0 ) is constant. Thus
x
d
A0 (x) = 0. ZThe Fundamental Theorem of Calculus states f (x) =
f (t) dt.
b
dx a
f (t) dt, then A0 (x) = f (x).
T F
(d) If A(x) =
[3]
(e) If v(t) > 0 isa the velocity of a car that is speeding up, then a right-endpoint
14x + 7
14
Riemann
sum
overestimates
T F
(f)
L’Hôpital’s
Rule
tell us that lim the car’s
[3]
= displacement.
lim
=7
T F
x→1
x2 − 1
x→1
2x
R
Since the displacement is given by v dx, the car is speeding up, v is increasing. This with v > 0
gives
us >that
forvelocity
a fixedof rectangle
the velocity at the right-endpoint
is
(e)
If v(t)
0 is the
a car that isinterior
speeding of
up,the
theninterval,
a right-endpoint
[3]
Riemann
sumthe
overestimates
T F out and above the
greater
than
velocity the
at car’s
the displacement.
left endpoint. Thus the rectangle reaches
curve v and hence, an overestimate is obtained.
Page14x
1 of+7 7 = lim 14 = 7
T F
(f) L’Hôpital’s Rule tell us that lim
x→1 x2 − 1
x→1 2x
14x we
+ 7 need the
14 limit of the numerator and limit of the denominator
In L’Hôpital’s
order to use
Rule,
(f)
RuleL’Hôpital’s
tell us that lim
[3]
= lim
=7
T F
x→1 x2 − 1
x→1 2x
to both be zero (or infinite). Here the limit of the numerator is 21. Thus, L’Hôpital’s Rule does
not apply.
Indeed, the limit does not exist as lim
x→1−
Page 1 of 7
14x + 7
14x + 7
= −∞ and lim
= +∞.
x2 − 1
x→1+ x2 − 1
Core Final Exam Rubric
3
Math 1131Q
2. (2 pts each) For each part, circle the correct answer. There is exactly one correct answer for each
question. There is no need to justify your answer.
give 0 or 2 points since justification is NOT required.
(a) If f is one-to-one and f (5) = 2, f (2) = 6, f 0 (5) = 9, and f 0 (2) = 7, then (f −1 )0 (2) is:
(A) 1/9
(B) 1/7
(C) 2/9
(D) 1/6
(E) none of the above
We have (f −1 )0 (2) = 1/f 0 (5) = 1/9 as f (5) = 2
(b) The linear approximation of f (t) = −2t2 + 3t + 2 at a = 3 is:
(A) y = −7 − 3(t − 9)
(B) y = −7 − 9(t − 3)
(D) y = −9 − 3(t − 7)
(E) y = −7 − 9(t − 7)
(C) y = −9 − 7(t − 3)
We find f 0 (t) = −4t + 3 so f 0 (3) = −9. Also f (3) = −7.
(c) If
f 0 (t)
< 0 for all t, then the graph of y = A(x) =
Z
x
f (t) dt is:
0
(A) increasing
(B) decreasing
(C) concave up
(D) concave down
(E) none of the these
By the Fundamental Theorem of Calculus, we have A0 (x) = f (x). Hence A00 (x) = f 0 (x) < 0
and so the graph of y = A(x) is concave down.
(d) A tank filled with 20 cubic feet of water develops a leak. If water is leaking at the rate of 2e−2t
cubic feet per minute, how much water remains after 3 minutes?
(A) 1 −
1
e6
The tank loses
(B) 18 −
Z
0
3
2
e6
(C) 19 +
1
e6
(D) 19 −
1
e6
(E) none of these
2e−2t dt cubic feet of water from time t = 0 to time t = 3 minutes. This
definite integral evaluates to 1 − e−6 , so 19 + e−6 cubic feet remain.
Core Final Exam Rubric
4
Math 1131Q
3. (5 pts) Find A0 (x) if A(x) =
Z
√
x
tet dt.
18
We have, by the Chain Rule and the Fundamental Theorem of Calculus:
#
"Z √
x
d
t
0
te dt
A (x) =
dx 18
√ √ d √ =
xe x ·
x
dx √ √ 1
√
xe x
=
2 x
=
1
2
1
1
1 √x
e
2
point if they realize they need the FundamentalTheorem
of Calculus.
√ √x more points if they use it correctly and get the
xe
part.
more point if they realize they need the chain rule.
more point if they correctly use the Chain Rule to multiply by 2√1 x .
They do not need to simplify their answer to 12 e
√
x,
but one hopes they do. ^
¨
x
.
ln(sin−1 (3x))
By the Quotient Rule and the Chain Rule, we have
4. (8 pts) Find f 0 (x) if f (x) =
(ln(sin
−1
f 0 (x) =
(3x))) · (1) − (x) ·
2 more points for
1
sin−1 (3x)
√ 1 2
1−(3x)
2 more points for finally multiplying by
sin−1 (3x)
2
ln(sin−1 (3x))
2 points for the correct use form of the Quotient Rule:
2 more points for
1
d
dx (3x)
= 3.
·√
denom. ·
1
1−(3x)2
·3
d
dx (num.)
.
− num. ·
(denom.)2
d
dx (demon.)
Core Final Exam Rubric
5
Math 1131Q
1
for x in an interval not containing 1.
x ln x
We make the u-substitution u = ln x. Then du = x1 dx and so dx = x du. Hence
Z
Z
1
1
1
dx =
· dx
x ln x
ln x x
Z
1
du
=
u
= ln |u|
5. (6 pts) Find an antiderivative of g(x) =
=
ln |ln(x)|
Since we are asking for “an” antiderivative, the presence or absence of a constant (either fixed or
arbitrary) should not effect the score.
2
1
2
1
points for using the correct u-substitution. R
more point for substituting correctly, getting u1 du.
more points for integrating correctly, getting ln |u|, only 1 point if they forget the absolute values.
final point for substituting back and giving their answer in terms of x.
Z √
5/2
6. (6 pts) Evaluate the definite integral
2x cos(πx2 )dx.
0
p
Then du = 2πx dx , u(0) = 0 and u( 5/2) = 5π/2. Hence
Z √5/2
5/2
1
cos(πx2 )2πx dx
2x cos(πx2 )dx =
π
0
Z 5π/2
1
=
cos(u) du
π
0
5π/2
1
=
sin(u)
π
We make the u-substitution u =
Z √
0
πx2 .
0
=
=
=
1
) − sin(0)
sin( 5π
2
π
1
(1 − 0)
π
1
≈ 0.3183.
π
up to 3 points if they make the correct u-substitution, changing the limits of integration and arriving
Z 5π/2
1
correctly at
cos(u) du.
π
0
up to 3 more points for correctly finding the antiderivative and evaluating it at the appropriate limits
1
of integration correctly, arriving at a final answer as .
π
5π/2
√5/2
They have to show that they get π1 by correctly evaluating π1 sin(u)
or π1 sin(2πx2 )
if they
0
0
5π/2
1
substituted back before evaluating. They get 0 points if they go directly from the π sin(u)
0
directly to π1 . This would indicate numerically evaluating the integral and recognizing the answer as
the reciprocal of π.
Core Final Exam Rubric
6
Math 1131Q
7. (6 pts) Find the derivative of f (x) =
√
3x + 1 using the limit definition of the derivative.
Note: No credit will be awarded for finding the derivative using any other method. Be careful and
write all the essential mathematics. Do not abbreviate your work.
We compute
f (x + h) − f (x)
h
p
√
3(x + h) + 1 − 3x + 1
= lim
h→0
h
p
p
√
√
3(x + h) + 1 − 3x + 1
3(x + h) + 1 + 3x + 1
= lim
·p
√
h→0
h
3(x + h) + 1 + 3x + 1
(3(x + h) + 1) − (3x + 1)
= lim p
√
h→0 h( 3(x + h) + 1 + 3x + 1)
3
= lim p
√
h→0
3(x + h) + 1 + 3x + 1
f 0 (x) =
=
lim
h→0
3
√
.
2 3x + 1
They need to show they are manipulating a limit, so if lim is not consistently there, they lose 1
h→0
point.
2 points for correctly getting to lim
h→0
√
√
3(x+h)+1 −
h
3x+1
√
√
3(x+h)+1 + 3x+1
√
2 more points for using the appropriate algebraic “trick” in multiplying by √
.
3(x+h)+1 +
1 more point for multiplying it out and simplifying to √
1 last point for taking the limit as h → 0 yielding the
3 √
3(x+h)+1 + 3x+1
3
.
answer 2√3x+1
3x+1
.
8. (8 pts) An inverted conical tank (i.e., point-down, like an ice cream cone) is 10 feet high and has a
diameter of 12 feet at its top. The tank is filled with water at a rate of 4 cubic feet per second. At
what rate is the height of the water changing when the water is 7 feet deep?
1
Note: The volume V of a cone of height h and radius r is given by V = πr2 h.
3
2
3
1 2
1
3
3
By similar triangles, we have r = h. Thus, we have V = πr h = π
h h = πh3 .
5
3
3
5
25
dV
9
dh
= πh2 .
Differentiating this with respect to time, we find
dt
dt
25
dh
9
dh
100
2
Putting in h = 7 and dV
and so
=
≈ 0.07218.
dt = +4 yields 4 = 25 π(7)
dt
dt
441π
3 points for realizing the similar triangle relationship between h, r, 10 and 6.
9
2 dh
3 points for simplifying V using this relationship and differentiating correctly to get dV
dt = 25 πh dt .
100
0
2 more points for plugging in the values, V = 4, h = 7 and correctly solving for h = 441π .
If instead they implicitly differentiated the original volume formula, then use the following:
3 points for realizing the similar triangle relationship between h and r, 10 and 6 and the resulting
relationship between r0 and h0 (i.e., r0 = 35 h0 ).
2
dr
1
2 dh
3 more points for correctly differentiating to get dV
dt = 3 πr dt h + 3 πr dt .
21
3 dh
2 more points for plugging in the values,. V = 4, h = 7, r = 5 and dr
dt = 5 dt and solving correctly
100
for dh
dt = 441π .