Louisiana State University
Math1550 Test: Oct 31, 2014
Name:
Student ID:
Make sure you show all your work or/and explain your answers
1. Boyle’s Law states that when a sample of gas is compressed at a constant temperature, the pressure
P and volume V satisfy the equation P V = C, where C is a constant. Suppose that at a certain
instant the volume is 1000cm3 , the pressure is 200kP a, and the pressure is increasing at a rate of
20kP a/min. At what rate is the volume decreasing at this instant?
Solution
Differentiating both sides of P V = C with respect to t gives us
P
dP
dV
V dP
dV
+V
=0⇒
=−
.
dt
dt
dt
P dt
dP
dV
1000
= 20kP a/min, we have
=−
(20) = −100.
dt
dt
200
3
Thus, the volume is decreasing at a rate of 100cm /min.
When V = 1000cm3 , P = 200kP a and
2. A rectangular storage container with an open top is to have a volume of 10m3 . The length of this
base is twice the width. Material for the base costs $5 per square meter. Material for the sides
costs $3 per square meter. Find the cost of materials for the cheapest such container
Solution
5
.
w2
The cost is 5(2w2 ) + 3[2(2wh) + 2(hw)] = 10w2 + 18wh so
5
90
2
C(w) = 10w + 18w
= 10w2 + .
2
w
w
90
20
9
3
0
C (w) = 20w − 2 = 2 w −
w
w
2
q
q
q
Then a critical value is w = 3 92 since C 0 (w) < 0 for 0 < w < 3 92 and C 0 (w) > 0 for w > 3 92 ,
q
then at w = 3 92 is an absolute minimum. Thus the minimum cost will be:
Recall V = wlh ⇒ 10 = (2w)(w)h = 2w2 h then h =
C
r !2
r !
90
3 9
3 9
= 10
+ q ≈ $81.77
2
2
3 9
2
3. If f (x) = 3x2 − 2x, 0 ≤ x ≤ 3, evaluate the Riemann sum with n = 6, taking the sample points
to be right endpoints.
Solution
R6 =
6
X
f (xi )
i=1
3−0
6
1
[f (x1 ) + f (x2 ) + f (x3 ) + f (x4 ) + f (x5 ) + f (x6 )]
2 1
1
3
5
=
f
+ f (1) + f
+ f (2) + f
+ f (3)
2
2
2
2
1
15
55
1 189
189
1
− +1+
+8+
+ 21 = ·
=
=
2
4
4
4
2 4
8
=
4. Find the derivative of the following function. Make sure you show/explain your work.
Z π
Z x
√
√
(a) F (x) =
1 + sec tdt = −
1 + sec tdt
x
π
Using FTC we have that
F 0 (x) =
√
Z
(b) G(x) =
1
x
√
1 + sec x
t2
dt
t4 + 1
Z u
du
1
t2
Let u = x, then
= √ thus G(u) =
dt
4
dx
2 x
1 t +1
Using FTC we have that
dG
u2
= 4
du
u +1
thus
√
u(x)2
1
1
x
x
dG du
0
=
· √ = √ · 2
=
G (x) =
4
2
du dx
u(x) + 1 2 x
2(x + 1)
2 x x +1
√
5. Water flows from the bottom of a storage tank at a rate of r(t) = 200 − 4t liters per minute, where
0 ≤ t ≤ 50. Find the amount of water that flows from the tank during the first 30 minutes.
Solution
Since r(t) = 200 − 4t, the amount of water that flows from the tank during the first 30 minutes
is given by,
Z 30
Z 30
30
2
r(t)dt =
200 − 4tdt = (200t − 2t = (6000 − 1800) − 0 = 4200 liters
0
0
0
.
6. Evaluate the integrals
Z
Z
sin t
−du
(a)
dt =
= − arctan u + C = − arctan(cos t) + C
2
1 + cos t
1 + u2
u = cos t =⇒ du = − sin tdt
Z
(b)
ez + 1
dt =
ez + z
Z
du
= ln u + C = ln(ez + z) + C
u
u = ez + z =⇒ du = (ez + 1)dz
3
Z 3
3/2
Z 1
Z 3
Z 3 u−1
1
1
x
du
u
−
1
2u
1
1
√
√2
√ du =
(c)
=
(u 2 − u− 2 )du =
− 2u1/2 =
dx =
2
4
4
3
0 2x + 1
1 u 1 4 u
1
1
3/2
3/2
2·1
1
2·3
2
1
1
1/2
1/2
1/2
1/2
−
=
−2·3
−2·1
2·3 −2·3 − +2
4
3
3
4
3
3
u = 2x + 1 =⇒ du = 2dx,
x=
u−1
,
2
1=2·0+1
and
3=2·1+1