1
151 WebCalc Fall 2002-copyright Joe Kahlig
In Class Questions
MATH 151-Fall 02
November 5
1. A picture supposedly painted by Vermeer (1632-1675) contains 99.5% of its carbon-14 (half life
of 5730 years). From this information, can you decide whether or not the picture is a fake?
Explain your reasoning.
First find an equation for the amount of carbon-14 remaining after x yrs. Use the points (0, 100)
and (5730, 50)
The formula is y = Ao ekx Since Ao is the initial amount and that is 100. Now plug in a 5730 for
x and 50 for y and solve for k.
50 = 100e5730k
.5 = e5730k
ln(.5) = 5730k
k = ln(.5)/5730 ≈ −0.000120968094
equation: y = 100e−0.000120968094t
At this point you have two different methods to approach the problem.
Method I: Vermeer died in 1675. That is 2002 − 1675 = 327 years ago. Now plug in 327 for x
to figure out what percentage of C-14 the painting should have. Doing this gives 96.121557%.
Since the painting contains 99.5% of its C-14, it is much younger.
Method II: plug in 99.5 for y and solve for x. This gives x = 41.436891. This means the painting
is around 41.4 years old. Vermeer died over 300 years ago.
In either case, the painting is a fake.
2. A tank contains 100L of brine with a concentration of 0.2 kg of salt per liter. Pure water enters
the tank at a rate of 10 L/min and the resulting solution, which is stirred continuously, runs out
at the same rate.
To answer both of these questions we need a formula. Let y = the amount of salt, in kg, at time
t.
dy
dt
= the rate of change of salt over time.
Noe dy
dt = rate of salt in - rate of salt out. Since only pure water is being pumped into the tank,
the rate of salt in is 0 kg/min.
The rate of salt out =
ykg
100L
∗
10L
1min
−1
0
Thus dy
dt = 10 y and any time you have an equation of the form y = ky, then the solution to
this differential equation is of the form y = Ao ekt , with Ao being the initial value. The initial
amount of salt in the water is 100L ∗ 0.2kg/L = 20kg.
The equation is y = 20e−0.1t
(a) How many kilograms of salt remain after half an hour?
y(30) = .9957kg.
(b) When will the concentration reach 0.15 kg/L?
Solve 100 ∗ 0.15 = 20e−0.1t
15 = 20e−0.1t
.75 = e−0.1t
ln(.75) = −0.1t
t=
ln(.75
−.01
≈ 2.8768min
2
151 WebCalc Fall 2002-copyright Joe Kahlig
3. Find the exact expression.
5π
π
(a) arcsin sin
=
4
4
First, 5π
is
in
the
third
quadrant. The Arcsine function must return an answer that is
4
−π
π
between 2 and 2 (quadrant 1 or quadrant 4). We now get the reference angle, which is π4 .
Now using the fact that sine is positive in both the first and the third quadrants we know
that our answer should be in the first quadrant. See graph of 13 part A that is below.
13 part B
13 part A
13 part C
3pi/4
5pi/4
pi/4
reference
angle
reference
angle
7pi/4
reference
angle
reference
angle
−pi/4
5π
3π
(b) arccos cos
=
4
4
First, 5π
is
in
the
third
quadrant. The Arccosine function must return an answer that is
4
between 0 and π( quadrant 1 and quadrant 2). We now get the reference angle, which is
π
4 . Now using the fact that cosine is negative in both the second and the third quadrants
we know that our answer should be in the second quadrant. See graph of 13 part B that is
above. 7π
−π
(c) arctan tan
=
4
4
First, 7π
is
in
the
fourth
quadrant. The Arctanget function must return an answer that is
4
−π
π
between 2 and 2 (quadrant 1 or quadrant 4). We now get the reference angle, which is π4 .
Since we are in the fourth quadrant, then our answer is just the negative of the reference
angle.
√
13
2
(d) sec arctan
=
3
3
There is some angle θ such that tan θ = 23 as shown by the figure below(13 part D). From
√
this picture it is easy to see that the hypotenuse has length of 13
13 part D
13 part E
√
13
θ
1
2
θ
√
1 − x2
x
3
p
(e) sin(2 cos −1 x) = 2x 1 − x2
There is some angle θ such that cos θ = x1 . Now we need the trig identity
sin(2θ) = 2 sin(θ) cos(θ). Using this identity and the figure above (13 part E) we can get
our answer.
4. Take the derivatives of the following.
(a) y = 4 arcsin(7 − x)
1
−4
p
y0 = 4 ∗ p
∗
(−1)
=
1 − (7 − x)2
1 − (7 − x)2
3
151 WebCalc Fall 2002-copyright Joe Kahlig
(b) y = arccos(4x2 )
−1
−8x
p
y0 = p
∗
8x
=
1 − (4x2 )2
1 − (4x2 )2
(c) y = x3 tan−1 (3x)
y 0 = 3x2 tan−1 (3x) + x3 ∗
1
3x3
2
−1
∗
3
=
3x
tan
(3x)
+
1 + (3x)2
1 + (3x)2
(d) y = cot−1 (e2x )
−1
−2e2x
2x
y0 =
∗
2e
=
1 + (e2x )2
1 + (e2x )2