Note Solutions to both exams are given (since the two versions are identical except for the order). The problem numbers are ordered for the white (version A) exam with a note at the beginning of the solution for the corresponding problem for the blue (version B) exam.
White 1c Blue 4c
Since we have an indeterminate quotient, 0 / 0, we may apply
l'Hospital's rule. Then use a limit fact from Section 3.4 (or use
l'Hospital's rule again if you prefer).
cos(x) - 1 sin(x)
lim ---------- = lim - 1/2 ------
x -> 0 2 x -> 0 x
x
/ sin(x)\
- 1/2 | lim ------| = -1/2
\x -> 0 x /
White 2a Blue 2a
Plug and chug; don't forget the Chain Rule!
2
f := arctan(x )
x
Df := 2 ------
4
1 + x
White 3d Blue 1d
To find the absolute maximum and minimum values of f(x)=x^3 - 3*x + 1
on the interval [-1, 3], first find the critical values of f which lie
in the stated interval. (These are where the derivative of f is zero
or undefined. In our case, since f is a polynomial, this amounts to
where f ' is zero.) To obtain the largest and smallest function
values, evaluate f at its critical numbers and at the endpoints of the
interval. For efficiency, we'll evaluate all the function values at
once by mapping f onto a list of x-values! (This parallel list
processing is also available on an HP48G/GX via DOLIST.)
3
f := x -> x - 3 x + 1
2
Df := x -> 3 x - 3
eq := 3 (x - 1) (x + 1) = 0
critical_numbers := {x = 1}, {x = -1}
x_values := [-1, 1, 3]
function_values := [3, -1, 19]
White 4b Blue 3b
STRATEGY: Use the IVT to show there's at least one solution to the
equation f(x)=0, then draw a rough sketch (using some of your Chapter
5 curve analysis skills) to conclude that there is exactly one root.
3
f := x -> x - 3 x + 3
x_vals := [-3, -1]
func_vals := [-15, 5]
Since f(-3) = -15 and f(-1) = 5, we have by the IVT that there is a
root of f in the interval (-3, -1).
2
Df := x -> 3 x - 3
eq := 3 (x - 1) (x + 1) = 0
critical_numbers := {x = 1}, {x = -1}
interior_x_values := [-3, 0, 3]
Df_sign_analysis := [1, -1, 1]
crit_nums := [-1, 1]
local_max_min := [5, 1]
We now know that:
1. the graph of f crosses the x-axis between x=-3 and x=-1;
2. f(-1)=5 is a local max of f;
3. f(1)=1 is a local min of f;
4. f is increasing on (-infinity, -1] and on [1, infinity);
5. f is decreasing on [-1, 1].
This enables us to draw a rough sketch like the one below.
> plot(f(x), x=-3..2, xtickmarks=6, ytickmarks=4);
We conclude that f(x)=0 has exaclty one real solution.
White 5d Blue 7d
By the MVT, there is a number c in (0, 2) such that `f '`(c) = (f(2) -
f(0)) / (`2` - `0`). Solving this equation for f(2) and applying the
other facts of the problem yields f(2) = 2*`f '`(c) + f(0)``<=2*`(3)`
+ 2``=8. That is, f(2)<=8.
White 6c Blue 5c
From the graph of f ' on the exam, we conclude via the First
Derivative Test that there is a local minimum at x=2, since at this
value f ' changes sign from -`` to +.
White 7b Blue 6b
From the graph of f ' on the exam, x=1 is a point of inflection, as
follows. At a point of inflection, f '' changes sign. Equivalently, (f
') ' changes sign. In other words, "the slope of the depicted curve, f
', changes sign." Clearly this occurs at x=1 (and NOT at x=3).
White 8d Blue 8d
The choices all speak of the increase and concavity of the function
f(x)=x*exp(x). Accordingly, we perform sign analyses on f ' and f ''.
The factored form of f ' shows that it changes sign from -`` to + at
x=-1. Thus f is increasing on (-1, infinity). The factored form of f
'' indicates it changes sign from -`` to + at x=-2. Hence f is concave
down on (-infinity, -2).
f := x exp(x)
Df := exp(x) + x exp(x)
Df_factored := exp(x) (x + 1)
D2f := 2 exp(x) + x exp(x)
D2f_factored := exp(x) (x + 2)
White 9e Blue 9e
Determine the list of critical values, then compute f, f ' and f ''
thereat. Here are conclusions via the Second Derivative Test for Local
Extrema (you could also use the First Derivative Test):
4 3 2
f := x -> 1/4 x - 4/3 x + 3/2 x
3 2
Df := x -> x - 4 x + 3 x
2
D2f := x -> 3 x - 8 x + 3
Df_factored_set_to_zero := x (x - 1) (x - 3) = 0
critical_numbers := [0, 1, 3]
function_values := [0, 5/12, -9/4]
derivative_values := [0, 0, 0]
second_derivative_values := [3, -2, 6]
> plot(f(x), x=-1..4, xtickmarks=5, ytickmarks=5);
White 10a Blue 11a
Antidifferentiate (i.e., indefinitely integrate); don't forget the +C!
Then use the numerical data to resolve the constant.
Df := 3 cos(x) + 5 sin(x)
/
|
antidifferentiate := | 3 cos(x) + 5 sin(x) dx
|
/
f := x -> 3 sin(x) - 5 cos(x) + C
eq := -5 + C = 4
sol := {C = 9}
answer := f(x) = 3 sin(x) - 5 cos(x) + 9
White 11e Blue 10e
since the integral from 1 to 5 equals 7 and the integral from
3 to 5 equals 2, we subtract and find that the integral
from 1 to 3 is 7-2 = 5. Adding this value to the integral
from 3 to 8, which is 6, we obtain that the entire
integral from 1 to 8 is 5+6=11.
White 12 Blue 12
Here's the equation for exponential growth or decay, XGD. Substitute
the numerical data into this equation and solve for the decay constant
k. Then use this to define the amount of radon as a function of time.
(Here the initial amount is a symbolic constant.)
XGD := y = y0 exp(k t)
3/5 y0 = y0 exp(5 k)
ksol := 1/5 ln(3/5)
f := t -> y0 exp(1/5 ln(3/5) t)
Now find out how much time must elapse until half the radon remains;
this is called the half-life. The exact answer is 5*ln(`1/2`) /
ln(`3/5`), which we subsequently float to give a feel for how many
days this really is.
1/2 y0 = y0 exp(1/5 ln(3/5) t)
ln(2)
-5 -------
ln(3/5)
half_life := 6.784577245 days
NOTE: You can do this problem automatically on an HP48G/GX. Simply
choose any numerical value for y[0] (say 100 g) and use the SOLVR.
White 13 Blue 14
For the limit Limit((1+2*sin(x))^`1/x`, x=0), we have an indeterminate
power, 1^infinity. Accordingly, first compute the limit L of the
natural logarithm of y=(1+sin(2*x))^`1/x`. The final answer is then
exp(L).
ln(1 + 2 sin(x))
lim ln(y) = lim ----------------
x -> 0 x -> 0 x
Apply l'Hospital's rule.
cos(x)
L = lim ln(y) := lim 2 ------------ = 2
x -> 0 1 + 2 sin(x)
Thus y equals := exp(L) = exp(2)
Maple agrees with our findings.
(1/x)
lim (1 + 2 sin(x)) = exp(2)
x -> 0
White 14 Blue 13
STRATEGY: By implicitly computing the derivative dy / dx, we will
actually have computed Diff(arccos(x), x).
eq := cos(y) = x
y = arccos(x)
d d
-- y = -- arccos(x)
dx dx
Here are the steps involved in the implicit differentiation.
cos(y(x)) = x, Regard y as an implicit funciton of x.
/d \
-sin(y(x)) |-- y(x)| = 1, Differentiate with respect to x.
\dx /
d 1
{-- y(x) = - ---------}, Solve for dy/dx.
dx sin(y(x))
1
dy_dx := - ------
sin(y)
Now express sin(y) in terms of x. While Maple can do this via brute
force, we would draw a right triangle and use trigonometry.
2 1/2
sine_of_y := (1 - x )
d 1
-- y = - -----------
dx 2 1/2
(1 - x )
d 1
finished := -- arccos(x) = - -----------
dx 2 1/2
(1 - x )
White 15 Blue 15
(a) Diagram: Draw a simple sketch of a rectangular box, labeling it
as follows.
LEGEND: x=width, 2*x=length, h=height.
(b) Surface area function
Recall the top is open.
The surface area is (area of base) + (area of left side + area of
right side + area of front side + area of back side).
2
area := 2 x + 6 h x
(c) Volume constraint
2
volume = 2 x h
2
36 = 2 x h
18
hsol := {h = ----}
2
x
(d) Area as a function of width
2 108
S := x -> 2 x + ---
x
(e) Find the dimensions of the box which give the (absolute) minimum
area.
Note that for 03, `S'`(x)>0.
Therefore, S has an absolute minimum at x=3.
108
DS := x -> 4 x - ---
2
x
108
eq := 4 x - --- = 0
2
x
1/2 1/2
xsol := {x = 3}, {x = - 3/2 + 3/2 I 3 }, {x = - 3/2 - 3/2 I 3 }
optimal_width_in_inches := 3
optimal_length_in_inches := 6
optimal_height_in_inches := 2
2
minimum_surface_area := 54, in.
"MOAPF" Supplement for Problems 16 and 17 (or, "Vectors Revisited")
Yes, friends, that "Mother of All Physics Formulas" for position under
constant acceleration. Here we'll derive it once and for all (why
repeat history?), in an n-dimensional vector setting. Recognize that
Problems 16 and 17 are applications of this same formula; only n
differs! In 16, n=2, whereas in 17, n=1 (regarding scalars as
vectors with one component). Not only that, you can use the same
formula in Calc 2 and Calc 3 for motion in space (n=3) or in higher
dimensional theaters of the mind...
dv / dt = a (The derivative of velocity in this case is constant
acceleration. Here dv / dt is a vector derivative.)
v = a t + C (Antidifferentiate. The a t denotes scalar
multiplication; C is a vector constant.)
v(0) = C (Resolve the vector constant C by plugging in
the initial velocity v``[0] at time t=0.)
dr / dt = v = a t + v(0) (We now know the velocity, the
derivative of position, as a function of time.)
r = 1/2 a t^2 + v(0) t + K (Antidifferentiate again; the
first rule of comedy: if it works once, it will work twice!)
r(0) = K (Resolve the vector constant K by plugging in
the initial position r(0) at time t=0.)
r = 1/2 a t^2 + v(0) t + r(0) (Finally, we have position
as a function of time, the "MOAPF.")
Now you can use this formula for any problem involving constant
acceleration (as you will repeatedly in the first few weeks of your
physics classes, perhaps next term). All you need do is change the
constants from problem to problem (the constant acceleration, initial
velocity, and initial position).
White 16 Blue 18
After having derived the MOAPF in the preceding supplement, we simply
fill in the (vector) constants for this particular problem. Here the
dimension is n=2, signifying motion in the plane. (NOTE: You did have
to derive something for this problem to get full credit. The point of
the supplement was that we don't need to derive the formula every time
we do a problem like this: we derived a general formula.)
General formula for position under constant acceleration.
2
r := t -> 1/2 a t + v0 t + r0
Plug `em in!
a := [0, -9.8]
1/2
v0 := [250 3 , 250]
r0 := [0, 0]
[ 1/2 2 ]
Voila!, position = [250 t 3 , -4.900000000 t + 250 t]
White 17 Blue 16
Same drill as 16, but with a different dimension: n=1, or motion
along a straight line.
2
r := t -> 1/2 a t + v0 t + r0
a := -20
r0 := 0
2
r := t -> -10 t + v0 t
When the car has stopped, r(t)=160 and v(t)=r '(t)=0. This gives
us two equations and two unknowns (final time and velocity), which we
then solve.
2
eq1 := 160 = -10 t + v0 t
eq2 := 0 = -20 t + v0
{v0 = 80, t = 4}, {t = -4, v0 = -80}
We discard the solution in which time is negative. Thus the initial
velocity is +80 ft/s in the direction of motion of the car, and the
speed (the magnitude of the velocity) is 80 ft/s, or 54.55 mph.
White 18 Blue 17
Hand approach: multiply the uniform stepsize
h=Delta*x``=(b-a)/n``=(3-1)/4 ``=1/2 by the sum of the function values
at the midpoints of the subintervals.
2 1/2
f := x -> (x + 1)
h := 1/2
midpoints := [5/4, 7/4, 9/4, 11/4]
1/2 1/2 1/2 1/2
func_vals := [1/4 41 , 1/4 65 , 1/4 97 , 1/4 137 ]
1/2 1/2 1/2 1/2
midpoint_rule := 1/8 41 + 1/8 65 + 1/8 97 + 1/8 137
numerical_equivalent := 4.502367463
Here's an automated midpoint rule approximation using Maple's
middlesum (in the student package) and its numerical equivalent.
/ 3 \
|----- |
| \ 2 1/2|
1/2 | ) ((5/4 + 1/2 i) + 1) |
| / |
|----- |
\i = 0 /
numerical_equivalent := 4.502367463
DONE!