exam 2 winter 18 solutions.pdf-Math 115 ...
exam_2_winter_18_solutions.pdf-Math 115 — Second Midterm —
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exam 2 winter 18 solutions.pdf-Math...
exam_2_winter_18_solutions.pdf-Math 115 — Second Midterm —
##### Page 3
Math 115 / Exam 2 (March 20, 2018.)
page 3
2
. [12 points]
a
. [6 points] Each the following is the graph of an implicit function.
-
2
-
1
1
2
3
4
-
3
-
2
-
1
1
2
3
x
y
Graph I
-
4
-
3
-
2
-
1
1
2
3
4
-
3
-
2
-
1
1
2
3
x
y
Graph II
-
2
-
1
1
2
3
4
-
4
-
3
-
2
-
1
1
2
3
4
x
y
Graph III
Match each of the graphs above to the formula below that gives the slope at each point
on the graph.
(A)
dy
dx
=
3
x
2
2
y
,
(B)
dy
dx
=
(
x
-
1)(
x
+ 2)
2
y
,
(C)
dy
dx
=
x
2
-
1
2
y
,
(D)
dy
dx
=
(
y
-
1)(
y
+ 2)
2
x
.
You do not need to show work in this part.
Solution:
Graph I:
C
,
Graph II:
B
, Graph III:
A

##### Page 4
Math 115 / Exam 2 (March 20, 2018.)
page 4
b
. [6 points] Find
dy
dx
for the implicit function given by
2
x
+
y
+ sin(
x
) cos(
y
)=5
-
x.
Solution:
d
dx
(
2
x
+
y
+ sin(
x
) cos(
y
)
)
=
d
dx
(5
-
x
)
ln(2)2
x
+
y
1+
dy
dx
+
-
sin(
x
) sin(
y
)
dy
dx
+ cos(
x
) cos(
y
)
=
-
1
(
ln(2)2
x
+
y
-
sin(
x
) sin(
y
)
)
dy
dx
+ ln(2)2
x
+
y
+ cos(
x
) cos(
y
)=
-
1
(
ln(2)2
x
+
y
-
sin(
x
) sin(
y
)
)
dy
dx
=
-
(1 + ln(2)2
x
+
y
+ cos(
x
) cos(
y
))
dy
dx
=
-
1 + ln(2)2
x
+
y
+ cos(
x
) cos(
y
)
ln(2)2
x
+
y
-
sin(
x
) sin(
y
)

##### Page 5
Math 115 / Exam 2 (March 20, 2018.)
page 5
3
. [12 points]
The Public Transit Authorities (PTA) are designing rain shelters for their bus
stops. They decide to place a roof in the shape of half a cylinder on four vertical legs of height
y
feet. The four legs are placed in a
rectangle
on the ground with width
x
feet and length
y
feet.
The costs of production are:
\$25 for each foot of the total length of the legs,
\$40 for each square foot of the area of the roof.
The following formulas may be useful in this problem:
the surface area of a cylinder of radius
r
and length
is 2
πr‘
,
the volume of a cylinder of radius
r
and length
is
πr
2
.
x
y
x
y
y
The PTA would like to spend exactly \$5000 on one rain shelter.
a
. [5 points] Find a formula for
y
in terms of
x
.
Solution:
We have that
25
·
(4
y
) + 40
·
1
2
2
π
x
2

y
= 5000
so
y
(100 + 20
πx
) = 5000
,
y
=
5000
100 + 20
πx
.
y
=
250
5+
πx
.
b
. [4 points] Find a formula for the total volume in cubic feet covered by the shelter,
V
(
x
),
if the width of the dashed rectangle has length
x
feet.
Solution:
The volume is
V
=
xy
2
+
1
2
π
x
2
2
y,
and hence
V
(
x
)=
x
·
250
5+
πx
2
+
1
2
π
(
x
2
)
2
·
250
5+
πx
.
c
. [3 points] The PTA wants to make sure that
each
of the sides of the rectangle has length
at least 5 feet, and the height (that is,
y
) of the shelter is at least 8 feet. In the context
of the problem, what is the domain of the function
V
(
x
)?
Solution:
We know that
x
5 and also
y
8. Therefore,
250
5+
πx
=
y
8
250
40 + 8
πx
x
210
8
π
=
105
4
π
.
5
x
105
4
π

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