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151 WebCalc Fall 2002-copyright Joe Kahlig
Quiz #17
MATH 151 Section
Name:
November 5, 2002
Show all your work. This is due at the start of class on Thursday.
1. The air in a factory is being filtered so that the quantity of pollutant, P (measured in mg/liter),
is decreasing at a rate that is proportional to the amount left. This means it is decreasing
exponentially. If 10% is removed in the first 5 hours, find a formula that gives the percentage
of the pollution that is remaining as a function of time.
The points that you should be using are (0, 100) and (5, 90). (Note: you could use (0, 1) and
(5, .9))
The formula is y = Ao ekt Since Ao is the initial amount and that is 100. Now plug in a 5 for
t and 90 for y and solve for k.
90 = 100e5k
.9 = e5k
ln(.9) = 5k
k = ln(.9)/5 ≈ −0.0210721
Answer: y = 100e−0.0210721t
If you used the other points then your answer is y = e−0.0210721t
2. Simplify the following. Give exact values. Do not give decimal approximations.
(a) cos(arcsin(x + 2)) =
If θ = arcsin(x + 2), then we have sin θ = x + 2. We can use this information and the
Pythagorean theorem to fill in the triangle pictured below and to the left. Now read off
cos θ
p
Answer= cos(arcsin(x + 2)) = 1 − (x + 2)2
1
θ
x+2
5π
3
p
1 − (x + 2)2
(b) tan−1 (tan 5π
3 )=
From the picture, above and on the right, we see that 5π
3 is in the forth quadrant and
π
the reference angle is π3 . arctan x must return a value between −π
2 and 2
−π
Answer= tan−1 (tan 5π
3 )= 3
3. Find the derivatives of the following functions.
π
3
2
151 WebCalc Fall 2002-copyright Joe Kahlig
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(a) y = sin−1 (3x)
3
y 0 = 4 sin−1 (3x)
3
1
3
−1
∗p
∗
3
=
4
sin
(3x)
∗p
2
1 − (3x)
1 − (3x)2
−1
(b) y = xsec (x)
ln(y) = sec−1 (x) ln(x)
The answer for this problem depends on which definition that you use for sec−1 x
Stewart’s definition:
y0
1
1
= √
∗ ln(x) + sec−1 (x) ∗
2
y
x
x x −1
"
ln(x)
sec−1 (x)
y =y √
+
x
x x2 − 1
#
0
"
0
y =x
sec−1 (x)
ln(x)
sec−1 (x)
√
+
x
x x2 − 1
#
Alternate definition
solution:
"
#
−1
ln(x)
sec−1 (x)
0
sec (x)
√
y =x
+
x
|x| x2 − 1
(c) y = arctan(sin(4x))
1
y0 =
∗ 4 cos(4x)
1 + (sin(4x))2
4 cos(4x)
y0 =
1 + sin2 (4x)