Download Exam 1 - Solutions (10 points) 1. Find the equation of a line through

MATH 131
Oehrtman
Fall 2011
Exam 1 - Solutions
(10 points)
1. Find the equation of a line through the point
(
)
3 2, π 6 with slope −2 .
Using the point-slope form of a line, we can write down y =
π
6

− 2  x −

3
.
2 
(15 points)
2. A calculus textbook dropped from the top of a 240 ft. tall building speeds up as it falls. It strikes
the ground 3.87 seconds later.
a. Draw a graph of the height of the book above the ground (measured in feet) as a function of the
time since it was dropped (measured in seconds). Include labels and units on your axes. Mark
the points on your graph corresponding to each whole second after the book is dropped. Choose
reasonable numerical values for the height these points and clearly add them to your graph.
b. Using the numerical values from your graph, compute the average speed of the textbook
starting one second after it was dropped until three seconds after it was dropped.
Average speed =
∆h 96 ft − 224 ft
=
= −64 ft s , or 64 ft/s downward.
3 s −1 s
∆t
c. Explain the meaning of the average speed you found in Part b.
Although the book is speeding up, an object would have to travel at the constant speed of 64
ft/s in order to fall the same distance, 128 ft, in the same amount of time, 2 s, as the book.
MATH 131
Oehrtman
Fall 2011
(20 points)
3. The point P is rotating around a circle of radius 2 as shown in the figure below. The angle ߠ, in
radians, is given as a function of time t, in seconds, by the graph.
y
P
2
ߠ
x
)
d
a
r
(
ߠ
A
t (sec)
a. Estimate the x and y coordinates of P when t = 1 .
Looking at the graph of θ vs. t, we see that θ (1) ≈ 4 radians, so point P will have traveled 4 out
of the 2π radians around the circle (63.662% of the way). So P will be in the 4th quadrant and a
reasonable guess might be x ≈ 1.3 and y ≈ −1.5 .
b. Estimate the time(s) at which P is at the bottom of the circle (point A).
At point A, θ =
3π
≈ 4.7 . From the graph we see this happens at times t ≈ 1.3 and t ≈ 3.7 .
2
c. Describe in words the motion of the point P on the circle.
P starts at the left of the circle at point (−2,0) and starts to move quickly around the circle in a
clockwise direction. It slows down as it moves until it comes to a stop just as it completes one
full revolution. The point then turns around and starts to move counterclockwise, speeding up
until it reaches point (−2,0) again.
d. Sketch a graph of the y-coordinate of P as a function of time. Number and label the axes.
y
t (sec)
MATH 131
Oehrtman
Fall 2011
(40 points)
4. The amount A in mg of the cough suppressant hydrocodone bitartrate that is left in the body t hours
after taking the recommended dose is given by A(t ) = 7(0.85)t . Include units on all answers.
a. What is the amount of hydrocodone bitartrate in the recommended dose?
7 mg
b. What percent of the drug leaves the body each hour?
15%
c. How much of the drug is left in the body 2 hours after the dose is administered?
A(2) = 7(0.85)2 ≈ 5.0575 mg
d. What is the average rate at which the drug leaves the body during the first 2 hours?
∆A 7 − 5.0575
=
≈ 0.97125 mg/hr
2
∆t
e. Explain the meaning of the average rate of change that you found in Part d.
The hydrocodone bitartrate is leaving the body at a decreasing (nonconstant) rate. If it did leave
the body at a constant rate of 0.97125 mg.hr, there would be the same decrease in the amount
of the drug, 1.9425 mg, in the same amount of time, 2 hr.
f. How long does it take for only 10% of the drug to remain in the body?
(0.85)t = 0.1 when t =
ln(0.1)
= 14.1681 hours
ln(0.85)
MATH 131
Oehrtman
Fall 2011
g. What is the continuous rate at which the drug leaves the body (expressed as a percent per
hour)?
To rewrite A(t ) = 7(0.85)t in the form A(t ) = 7e kt , we need ek = 0.85 so
k = ln(0.85) = −0.16252 . So the drug is leaving the body at the continuous rate
of 16.25% per hour.
h. What is the instantaneous rate at which the drug leaves the body when there is 5 mg remaining?
Using the result from part g, 16.25% of 5 mg is 0.8126 mg. So the drug will be leaving the
body at the instantaneous rate of 0.8126 mg/hr.
(15 points)
e x − e3
, which has a removable discontinuity (a hole) when x = 3 . Let h = lim f ( x ) .
x →3
x −3
Without any fancy limit calculations, we can only approximate the value of h. For what values of x
can you be sure that f ( x) will approximate h to within an accuracy of 0.1? Explain your
reasoning.
5. Let f ( x) =
Using x = 2.9 and x = 3.1 , we get approximations for h of f (2.9) = 19.114 and f (3.1) = 21.124 .
Using these an overestimate and underestimate, respectively, we can only be sure that the error is
less than 21.124 − 19.114 = 2.01 . So we cannot say these are within 0.1 of the actual value of h.
Using x = 2.99 and x = 3.01 , we get approximations for h of f (2.99) = 19.985 and
f (3.01) = 20.186 . Using these an overestimate and underestimate, respectively, we can only be sure
that the error is less than 20.186 − 19.985 = 0.201 . So we cannot say these are within 0.1 of the
actual value of h.
Using x = 2.999 and x = 3.001 , we get approximations for h of f (2.999) = 20.076 and
f (3.01) = 20.096 . Using these an overestimate and underestimate, respectively, we can be sure that
the error is less than 20.096 − 20.076 = 0.02 which is less than the required 0.1 error bound.