• A surveillance hot air balloon is drifting along when it’s
cameras detect a strange object in the distance. The
worker at the control tower realizes that the object is a
spaceship! However, before he can get a closer look, the
equipment mysteriously fails… the man decides to use
the information transmitted in the instant before the
breakdown to calculate the speed of the UFO, so he can
send someone to investigate. The balloon was 2km below
and 6km to the West of the UFO. The balloon was moving
upward at a rate of 144 km/h, and the distance between
the two objects was increasing at a rate of 2016/√10 km/h.
How fast was the UFO travelling in km/h?
x = 6km
dx/dt = ?
-144km/h z = 2 √10 km
dz/dt = 2016/ √10
1. To solve this problem, 2. You can then use
you must first find an this to find the
equation to relate x, y, and distance z.
z. x² + y² = z² 6² + 2² = z²
z = 2 √10 km
3. The next step is to get an x² + y² = z²
equation that relates the
rates of change of the three
distances. 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)
You can do this by implicit
differentiation, and by using the chain
x(dx/dt) + y(dy/dt) = z(dz/dt)
rule: F’[f(g(x)] = f’[g(x)]*g’(x)
(If you consider x² to be a composite
of functions, then the derivative of
the inner function ‘x’ in relation to
time is dx/dt, and the same is true for (6)(dx/dt)+(2)(-144) = (2 √10 )(2016/
y and z.)
Now you have an 6(dx/dt) = 4320
expression relating all the
velocities, and dx/dt = 720 km/h
all you have to do is plug in
the values given in the Answer: The UFO is travelling East
problem to solve for the at a
• What is the maximum possible area of a
rectangle, with sides parallel to the
coordinate axis, bound by the graphs of
the following functions (to four decimal
a(x) = -(1/2)x² + 2
b(x) = 2x – 4
c(x) = -2x – 4
1. The first step is
to draw a diagram
of the problem.
whose area you
want to maximize
2. To maximize
the area, we must
first find an
area. case, the rectangle is
symmetrical on either side of the
y-axis. If we call the distance from
x = 0 to one of the vertices ‘x’, the
3. The area of the rectangle can be written
entire base can be written as ‘2x’.
as: A(x) = base * height
The height ‘y’ of the rectangle is
the distance from the top
function to the bottom function.
Written in terms of x, it is a(x)- A(x) = -x³-4x²+12x
We are looking for the A(x) = -x³-
maximum area. One way to 4x²+12x
do this is using the derivative
of A(x). A’(x) = -3x²-
x = -b ± √(b²-4ac)
Use the quadratic 2a
formula to find the values
of x where A’(x)=0. The x = -(-8) ± √((-8)²-4(-3)(12))
maximum value A(x) will 2(-3)
be found at one of these
x = 8 ± √(64+144)
x = 8 ± 4 √13
x = 4 + 2 √13 x= 4-2
x = 4 + 2 √13 x= 4-2
Now it is important to remember
the context of the problem. We
are looking for an area, so the
value of x (which describes the
length of the sides) can’t be ≈ -3.7370 ≈
Just to confirm that the
remaining zero is a maximum,
wemaximum on the parent function
do a line test.
is denoted by a zero at some value
‘x=a’ on the derivative, where There is a maximum on the
when x<a, f’(x) >0, and when x>a, parent function at x = 4 - 2
We are answering the question
x= 4 - 2 √13
‘what is the greatest possible
area of the rectangle?’ What we
is the value of x at which this
maximum is found.
A(x) = -x³-4x²+12x
To find the final answer,
substitute that x-value back into = -( 4 - 2 √13 ) ³ -4( 4 - 2 √13 ) ²+12 ( 4 -
the equation that describes the 2 √13 )
area of the rectangle.
The maximum possible area of the rectangle is 7.0354u².
Since we are not looking for the exact value
of the maximum area, there is a much
quicker way of finding the answer.
3.The initial steps are the same: make a
graph and use it to find an equation A(x)
that describes the area of the rectangle.
A(x) = -x³-4x²+12x
6.Now you can simply use your calculator to
find the maximum value.
Graph the equation A(x), and
press 2nd Calc, 4:maximum. The y-value
you get is the maximum value. This is also
a good way to check that you did the
A(x) = 7.0354u²
• What is the volume of a solid created
when the area bound by the graphs of
f(x)=x²+2 and g(x)=x+4 is revolved around
the x-axis, from x=0 to x=3.
When the graphs are rotated around the x-axis, the cross-sections
of the solid will form cylinders with radius equal to the value of the
upper graph. The cylinders will contain holes with radius equal to the
value of the lower graph. For this reason, the area has to be found
on two different intervals.
The length of the cross sections in
this case (the change in x) are
decreased until they approach
zero. So they essentially form an
infinite number of washer shapes
like the ones at right. If you add up
the areas of all the washers, you’ll
know the volume of the solid.
1. Find out where the two
functions intersect, to determine
the intervals on which you must
x²+2 = x+4
integrate. In this case, as you
are only looking for the area
x = -1 , 2
from [0,2], the intersection at
x=2 is the only one that’s
V₁ = ² πr² - πr²
Now you need an equation ₀
describing the area of the washers.
=π₀ ² (x+4)² - (x²+2)²
The area of a circle is πr². The area
of the washers is the total area of
the circle (with radius equal to the
upper function) minus the area of V ₂ = π ³ (x²+2)² - (x+4)²
the hole (radius equal to the lower
function). By substituting the
appropriate functions in for r, you
get two equations, on the intervals
[0,2] and [2,3].
Now solve the integrals: V= π (x+4)² - (x²+2)² V= π ³
₀² (x²+2)² - (x+4)²
= π₀ ² -x⁴-3x²+8x+12 = π₂ ³ x⁴+3x²-8x-12
₀ = π[x⁵/5+x³-4x²-12x]³
= π(128/5 - 0) = π[18/5 – (-128/5)]
The total volume of the solid = 128π/5u³ + 146π/5u³
is the sum of the two
volumes. = 274π/5u³
The volume of the solid is 274π/5u³.
1. Your calculator can be used to find the 1. V= π ₀ ² (x+4)² - (x²+2)²
volume of the solid as well. Recall that
the volume of the first segment of the
solid is expressed by the equation
2. (x+4)² - (x²+2)²
2. Plug the equation describing the area
of the washers into Y1.
3. Press 2nd Calc, 7 to find the integral.
Enter the appropriate interval [0,2]. The
calculator will shade in that area on the
graph, and give you the value 25.6.
4. 25.6 π u³
4. Multiply that value by π, as it was not
included in the integral. This is the
volume of the yellow segment of the
solid. 5. 29.2 π u³
5. Repeat steps 1-4 using the equation
for the volume of the green segment. 6. 54.8 π
• A magical imp from an alternate
dimension is moving through various
planes of existence (or sometimes non-
existence, alternate universes can get
pretty strange). His acceleration is
described by the equation
dy/dx = cos²x · sinx · y
Find an equation that describes the imp’s
velocity if at time zero he is moving at 2e⅓
1. dy/dx = cos²x · sinx · y 1. To antidifferentiate this equation, the
first step is to separate the variables.
1/y dy = cos²x · sinx dx
2. Now, taking the integral of the left side
is simple. However the right side requires
2. 1/y dy = cos²x · sinx dx
the use of substitution.
Let u = cos(x)
3. ln|y|+ c = u² du
The derivative of cos(x) is sin(x), so
Let du = sin(x) dx
4. ln|y|+ c = u³/3 + c 3. Substituting u and du into the equation,
we get a simpler expression to
5. ln|y| = (cosx)³/3 + c antidifferentiate.
4. Remember to include +c !
cos³x + c
6. y = e^ 9.Substitute cos(x) back into the equation,
and combine the c values onto one side.
7. y = e · 10.Isolate y by making the entire left side
an exponent of e.
11.Multiplying two powers of the same
y=C e^ 3 base means the exponents are added, so
working that fact backwards, you can
remove e^c, resulting in a constant C.
cos³x 1. To find the value of C,
1. v(t) = C e^ simply plug in the
cos³(0) coordinates given in the
(2)e ⅓ = C e^ question (0, 2e⅓).
2e ⅓ = Ce⅓
2. This works out to C =
2 C=2 2
v(t) = 2e^ 3. Now just plug in the
value for C back into the
The equation describingcos³x
imp’s velocity is v(t) = 2e^ 3
Well, it’s done. I must admit, I really procrastinated on this… most of it was
done in the past 3 days. That leads me to the first thing I’ve learned.
Select deadlines wisely when given the opportunity.. And stop
procrastinating. now, on with the show.
The concepts I selected were based on the types of problems we would
come across near the ends of units. I was trying to use word problems
as they tend to cover several different aspects of a unit and generally
also span over other units in some way. I was also looking for things
that I needed to brush up on, as the exam is very VERY close now.
I think my project covers a fairly good span of the course. It’s pretty easy to
do so in this course, as each chapter really builds on the concepts of
earlier chapters. I don’t think that’s the case as much in 30S and 40S.
As for learning from this assignment, I wouldn’t say it really increased my
knowledge a lot, but I think I picked up some details I hadn’t the first
time around. I also think that it was a good way to solidify my
understanding of specific topics and study for the exam. I’m a bit
worried that I’ve accidentally made some glaring errors, and if that’s the
case then I guess it wasn’t very valuable. However that would be
entirely my fault for not leaving enough time to get feedback from