Math 1131Q
UConn ECE Core Questions
Fall 2008
Here is the grading rubric for the Math 1131Q
UConn ECE Core Questions, Fall 2008
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.
UConn ECE Core Final Exam Questions
Math 1131Q
2
IMPORTANT: All answers on this and subsequent pages must include either supporting work
or an explanation of your reasoning. These elements are considered part of the answer and
will be graded.
1.
(15 pts) 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. Include justifications (reasons) for your answers – this is a
part of the question and will be graded as an equal part of the question.
All of these are 3 points each. 1 point for a correct answer and up to 2 points for a correct justification
(with 1 point for a reasonably appropriate thought and 2 points for a correct justification). Keep in mind
that there can be more than one correct justification.
(a) f (x) = x3 + 2x + tan x has no local maximum or minimum values.
Justification:
(a)
T
F
the derivative f 0 (x) = 3x2 + 2 + sec2 x ≥ 2 and so there are no places where the derivative is zero
or nonexistent.or since f 0 > 0 (as above), f is always increasing and never turns around. or ...
Example of a 1 point justification: the graph on the calculator is always increasing. This is a
reasonable thought, but how do they KNOW that it continues so off the calculator screen.
(b) If y = π 5 then y 0 = 5π 4
Justification:
(b)
T
F
y 0 = 0 since 5π 4 is just a number. there really aren’t any other justifications that I can come up
with.
Z
(c)
(8y −
y
Justification:
2
) dy = 4y 2 −
1/4
If you differentiate 4y 2 −
8y 3/4
+C
3
8y 3/4
3
+ C, you get 8y −
(c)
3
4
· 8 · y −1/4 = 8y −
T
F
2
y 1/4
Z x2
6
6
d
dt =
(d)
2
dx 1 3 + t
3 + x2
Justification:
(d)
T
F
The Fundamental Theorem says that you have to plug in x2 , not just x and then by the chain rule,
6
you have to multiply by 2x. or The derivative is
· 2x.
3 + (x2 )2
6
6
Examples of 1 point justifications: the answer is
or the derivative is
· 2x
3 + x4
3 + x2
(e) If r(t) represents the rate at which a country’s debt is growing, where t
represents the number of years since 1980, then the
Z 10
total increase in its debt between 1980 and 1990 is given by
r(t) dt.
Justification:
(e)
T
0
Since 1990 is 10 yrs after 1980, the Fundamental Theorem of Calculus gives that the total (net)
increase in debt is the definite integral of the rate of change of debt which is r(t). Here, they should
mention somehow mention that r(t) is the rate of change of debt and that they are looking for the
net change which is a definite integral.
F
UConn ECE Core Final Exam Questions
2.
Math 1131Q
3
(10 pts) A drilling rig 12 miles offshore is to be connected by pipe to a refinery onshore, 20 miles straight
down the coast from the rig. Underwater pipe costs $50,000 per mile and land-based pipe costs $30,000
per mile. Assume that the shore line is along the horizontal axis and the elbow of the pipe is x miles from
the point on land across from the rig (this point is at the origin).
(a) (6 pts) Express the cost of the pipe as a function of x.
√
Cost= 30, 000(20 − x) + 50, 000 x2 + 144
They lose 3 points if they switch the 30,000 and 50,000. If they don’t use a radical in the underwater part
from the rig to land, they lose 3 points. −2 points if they don’t figure out the length of the underwater
piping correctly. A typical mistake was thinking x was the distance from the place where the pipe comes
ashore to the refinery instead of the distance frompthe point perpendicular to the rig to where the pipe
comes ashore. This leads to a cost of C = 50, 000 (20 − x)2 + 144 + 30, 000x which would give them 4
out of the 6 points.
(b) (4 pts) Where should the elbow be located to obtain the least expensive connection?
Here they should be graded consistent with what their answer for part (a) is.
C 0 = −30, 000 + 50, 000 21 (x2 + 144)−1/2 · 2x.
√
50, 000x
So, C 0 = 0 when 30, 000 = √
or when 35 x = x2 + 144
2
x + 144
2 = x2 + 144, so that 16 x2 = 144, which says that x2 = 144·9 = 81, so
That means that 25
x
9
9
16
x = ±9.
if they don’t mention −9, that’s OK. It would be natural for them to omit it in this problem.
We won’t take off for the not testing whether 9 yields a max or min.
so the pipe should come ashore 9 miles from the point perpendicular from the rig or 11
miles from the Refinery.
up to 2 points for finding C 0 correctly, 1 point for setting it equal to 0 and 1 point for
solving.
UConn ECE Core Final Exam Questions
3.
Math 1131Q
4
A skydiver jumps from a helicopter hovering at 6400 feet and descends for 4 seconds before opening her
parachute. Assume that the skydiver falls freely with acceleration -32 ft/sec2 before the parachute opens.
Assume that the initial velocity of the skydiver is 0 ft/sec.
(a) (5 pts) Find the height function for 0 ≤ t ≤ 4 and use it to find at what altitude the parachute opens.
a = −32 so v = −32t + vo and vo = 0. Then s = −16t2 + so , and so = 6400, so s = −16t2 + 6400.
So her height after 4 seconds, when her parachute opens, is h(4) = −16 · 42 + 6400 = 6144 ft.
If they started with a = 32 instead of a = −32, take off 2 point since they doing so, wouldn’t yield a height
function, but s = 16t2 , a distance traveled function and then, the parachute opens after falling 256 ft from
the plane.
So 3 points for getting s correctly, 1 point at each level a, v and s. Then 2 points for the final answer.
(b) (5 pts) After the parachute opens, the skydiver descends with velocity 16 ft/sec. Find the height
function for 4 < t and use it to find the duration of time from when the skydiver opens the parachute until
she lands on the ground.
I’ll use h here to distinguish from the height function s used above:
v = −16, so h = −16t + ho , where ho = s(4) for s as above. So h = −16t + 6144.
Now, h = 0 when 0 = −16t + 6144 or equivalently, 16t = 6144 or t = 384. So she lands on the ground
t = 385 sec.= 6.4 min. after the parachute opens (or 389 sec. ≈ 6.48333 min.) after leaving the helicopter.
up to 3 points for getting h correct and then up to 2 points for solving h = 0 for t and interpreting
it correctly.
4.
(5 pts)
Given the graph of the function f (x) below, sketch the graph of it’s derivative f 0 (x).
1 point for each part, where there are 5 parts: (0, 1), (1, 2), (2, 3), (3, 4), (4, 5) and (5, ∞). Take off a total
of −1 points if they either leave the “dots” out or use “solid dots”.
Don’t worry about the depths or heights of each part, so long as it crosses (or runs along) the x-axis
correctly.
UConn ECE Core Final Exam Questions
5.
Math 1131Q
5
Let f (x) be a continuous function given by the following graph. Suppose F (x) is an antiderivative of f (x),
that is F 0 (t) = f (t) and F (−1) = 2.
f
2_
1_
|
1
|
|
|
|
|
−5 −4 −3 −2 −1
|
|
3
|
4
|
5
|
6
x
(a) (3 pts) At what values of x does F (x) have the local minimum values?
At t = −4 and at t = 0 since F changes from decreasing to increasing exactly as F 0 = f changes from
negative to positive.
1 point for each, but 3 for both values and −1 point for each extra number they put in with a minimum
of 0 overall of course
(b) (3 pts) On what intervals is the graph of F (x) concave up?
over (−5, −3) ∪ (−1, 2) since F is concave up exactly where F 00 > 0. But F 00 = f 0 , so this occurs exactly
where f 0 > 0, which means where f is increasing.
1 point for each interval, 3 points for both - this means both endpoints of the interval. −1 point for any
other intervals listed with a minimum of 0 total points of course. It’s OK if they don’t express their answer
as a union of intervals, but instead simply list the two intervals. Some may write (−5, −3) ∩ (−1, −2)
thinking “and”= ∩. For this, take off only 1 point.
(c) (4 pts) If F (−1) = 2 then use the Fundamental Theorem to determine F (4).
11
2
= 5 12 = 5.5.
Z x
If G(x) =
f (t) dt, then G 0 = F 0 , so G(x) = F (x) + C.
0 Z
−1
Now G(−1) =
f (t) dt = 12 by counting only one “half-square” and realizing that it is under the x-axis
F (4) =
0
and the limits of integration are going from 0 to −1. Since F (−1) = 2, C = − 23 . Finally, F (4) = G(4) +
and G(4) = 4 (again by counting “half-squares”), giving us that F (4) = 4 + 32 = 5 12
3
2
They are clearly not going to explain their reasoning coherently and on this question at this level, it would
be to much to ask. But you should see their reasoning in their calculations. The question is only worth 4
points, so give 1 point if they equate areas with definite integrals, one point for using the grid system to
calculate areas (which you can tell by their answers) and up to 2 points for putting it all together correctly.
6.
(5 pts) The radius r of a cone is increasing at a rate of 6 cm/min, while the height h is decreasing at a
rate of 4 cm/min. At the instant when the radius is 9 cm and the height is 12 cm, how is the volume V
1
changing? Determine the rate of change of V . (Note: V = πr2 h).
3
Since V = 31 πr2 h, V 0 = 23 π r r 0 h + 31 πr2 · h 0
Now, when r = 9, r 0 = 6, h = 12 and h 0 = −4, then V 0 = 432π − 108π = 324π cm3 /min.
up to 2 point for recognizing they need to differentiate using the product and chain rules and that they
are differentiating relative to time. And another point if they do so correctly. Then 1 point if they plug in
the right values for r, r 0 , h and h 0 . If they miss one, don’t take off, but if they miss two, then take off 1
point. Then 1 last point for doing the arithmetic correctly coming to the answer.
UConn ECE Core Final Exam Questions
7.
Let f (x) = ex
2 −4x
Math 1131Q
6
, 0 ≤ x ≤ 5.
(a) (7 pts) Find the local extrema (local maxima and local minima) of f in the interval (0,5) using the
calculus.
2
f 0 = (2x − 4)ex −4x
So f 0 = 0 ⇔ x = 2
Now, either using the second derivative test:
2
f 00 = 2ex −4x + (2x − 4)2 ex2 − 4x
2
= (2 + (2x + 4)2 )ex −4x
Since f 00 (2) = 2 > 0, f (2) = e−4 is a local minimum f -value.
or using the first derivative test:
2
ex −4x > 0 for all x and 2x − 4 < 0 for x < 2 and 2x − 4 > 0 for x > 2,
so f 0 is changing from negative to positive x = 2 and
hence f is changing from decreasing to increasing at x = 2
and so f (2) is a local minimum f -value.
up to 2 points for differentiating correctly,
up to 2 more points for realizing that f 0 = 0 has to be solved and doing so correctly
up to 3 more points for justifying whether f (2) is a local min or max by either using a first or second
derivative test.
(b) (3 pts) Find the largest and smallest values of f (x) for x in the closed interval [0,5]. Give the exact
values; not calculator approximations.
Using the calculations above, they need to find values of f at 0, 2 and 5: f (0) = e0 = 1, f (2) = e−4 and
f (5) = e5 . So f (2) is an absolute minimum and f (5) is an absolute maximum.
1 point for evaluating at each endpoint and the value found in part (a). −1 point if they don’t spell out
what the maximum value and minimum values are.
Z
x
√
8. (a) (6 pts) Evaluate using substitution.
dx
25 − x2
Z
x
√
dx
25 − x2
u = 25 − x2 ,Zdu = −2x dx
−2x
=− 21
dx
sqrt25
− x2
Z
1
=− 12
du
0
Z sqrtu ,
(∗)
=− 12
u−1/2 du
1/2
= −u
√ +C
= − 25 − x2 + C
2 points for the correct u-substitution, 2 points for the re-expressioning in terms of u, 2 points for the
evaluation and subbing back in terms of x. If they left off the +C, do not take off any points.
UConn ECE Core Final Exam Questions
Math 1131Q
Z
(b) (4 pts) Use the Fundamental Theorem and your answer to (a) to find
4
7
3
√
x
dx
25 − x2
They can either take up at (ast) above and alter the limits of integration or use the final answer above
and the limits of integration as given:
Z
3
4
p
x
2
√
dx = − 25 − x 2
25 − x
3
√
√
= −√25 − 9
√+ 25 − 16
= − 16 + 9 = −1
3
x
1
√
dx = −
2
2
25 − x
4
or
Z
4
Z
16
u−1/2 du
9
√ 16
= − u
9
√
√
= − 16 + 9 = −1
This should be graded consistent with what they got for part (a): 2 points for knowing that they need
an antiderivative and which has to be evaluated and 2 more points for doing so correctly including the
arithmetic.
NOTE: If they just give an answer, they get 0 points. They can all use their calculators: