Transcript Karl Castleton - Colorado Mesa University

Karl Castleton
Research Scientist
Pacific Northwest National
Laboratory
What is this good for?
• Intent: To produce a set of practical calculus
problems that can be used at certain points in a
typical series of calculus courses
• Assumptions:
– Not all students in a calculus class are Math
majors
– Many students (even Math majors) benefit from
practical hands on “experiments” in calculus
Inspiration
• Car talk hosts Tom and Ray Magliozzi
say “ooooh this requires calculus”
– A practical question leads to calculus.
– What other practical questions lead to
calculus?
– Wouldn’t it be fun to have a list of such
questions?
Any examples from the
audience?
• I know many of these might be characterized
as engineering questions but a well placed
concrete example helps many understand the
math concept better.
• What problems seem to satisfy students
when they ask you ”What is it good for?”
• I could not seem to think of a concrete
example for series.
• Please bring up any examples that strike you
as I speak.
A diesel truck driver needs to
know how much fuel is in the tank
The area above the
1/4 mark should be
equal to the area
1/2
below. Only 1/4 of
E
the tank need be
The fuel gage is broken.
considered.The Area
He wants to use a dip
of an angular
stick. The mark at full,
segment will be
1/2 and empty are easy.
useful.
But where does the 1/4
and 3/4 go?
Area A+B =C -B
F
Lets name the angle t
t
A
B
C
Diesel Tank continued
C-B=B+A
t
A
B
tr2 - ½ r2cos(t)sin(t)=
½ r2cos(t)sin(t)+(pi/2-t)r2
assume r=1 (for simplicity)
C
t=cos(t)sin(t)+(pi/2-t)
arrange so t is on one side
pi/2=2t-cos(t)sin(t)
Not very satisfying! But
where is the calculus?
Were Tom and Ray wrong
as they so often are?
The answer from above is
roughly 30% of r for ¼
and 70% for ¾ found by
experimentation.
Well the assumption that tr2
is the area of C & A is
essentially a calculus result.
But the hand check clearly
is. Put the 1/4 circle on a
grid. Count the total in the
quarter. Now count until
you reach half that number.
Split the remaining amount.
Want a better answer use a
finer grid. (Clearly the mark
of Calculus)
1/4 approx.
So what other questions?
•
•
•
•
How many sprinkler heads?
Getting the most inside the fence.
Measure totals with sampled rates?
What’s going to happen in the future?
A wacky gardener wants to know how
many heads to put in his garden.
Assume the gardener
65,100
120,100
65,100
has coordinates of the
“corners”. Clearly this
A1
is
just
integration
(for
75,70
garden
those who know what
10,10
it is) in hiding. But the
75,25
h=65-10,
10,10
strange shape might
b1=10,b2=10
initially make it seem
Our gardener likes
h will be negative for
difficult. Trapezoid
non rectangular
the lines that go
summation of the areas
shapes. Each
towards the Y axis.
“under” the line
sprinkler head can
So the line (120,100)
segments. Area of a
covers 2000ft2.
- (75,25) will have
trapezoid
How many does he
h=-45
A=1/2h(b1+b2)
need.
Wacky Gardener Continued
65,100 120,100
garden
75,70
r
75,25
10,10
A1=1/2(65-10)(10+100)
A2=1/2(75-65)(100+70)
A3=1/2(120-75)(70+100)
A4=1/2(75-120)(100+25)
A5=1/2(10-75)(25+10)
A1+A2+A3+A4+A5=3750ft2
The math skill required can be kicked
up a notch by not giving the students
the coordinates of the vertices and
have them devise a technique for
measuring them. I would suggest the
you give them the picture of the plot
on “weird” shape paper. The
technique is simply to draw a line r
you do know the length of, then
measure the distance between the ends
of the line segment r and any corner.
From these two distances the X and Y
can be computed relative to ruler r. It
is just some algebra.
Getting the most inside a piece of
fence.
Students should get used to the idea
This one does appear in many
that if you are maximizing or
calculus texts. A farmer has
minimizing something you are
100 feet of fence and he wants going to be taking the derivative and
to enclose the largest
setting it equal to 0. The most
rectangular area along side his
important part of this question is
barn. What should the
setting it up properly. Assume the
dimensions of the area be?
small side length is x then the large
has to be 100-2x. Students may try
to call this distance y and be stuck.
Barn
A=x * (100-2x)=100x-2x2
x
x
dA/dx=100-2*2x=0
100-2x
100=4x or x=25
Measure the total with sampled rates.
Your company produces
We should assume that we can
pop/soda. Estimate the total every once in a while measure the
number of bottles leaving
number of bottles that left over a
the plant without adding
short period of time or monitor how
equipment to count every
long it takes for a certain amount of
bottle.(cheap boss) You
bottles to leave the plant. Either
know that the plants
would give you b/day estimates
production rate in
(slopes) at given points in time. Lets
bottles/day does not change
assume we got the following
instaneously but slowly
measures.
increases or decreases.
Day 1 25 b/day, Day 2 40 b/day
b/day
Day 3 12 b/day, Day 4 45 b/day
Bottle Counting continued
45
40
b/day
25
12
1 2 3 4
day
This should start to
look like the wacky
gardener again. You
could integrate and
find the total
number of bottles.
Bt=1/2(1)(25+40)+
1/2(1)(40+12)+
1/2(1)(12+45)=87
Check your calculus
intuition and see if you
can see what is wrong
with the two pictures
to the right?
Remember y=mx+b
would represent the
line between the points
on the rate graph.
Integrate with respect
to x.
45
40
b/day
25
12
115
b
77
65
1 2 3 4
day
25
1 2 3 4
day
What’s going to happen in the
This is a Constantly Stirred Tank
future
Reactor (CSTR) model and is the
How long do you need to put
clean water into your
swimming pool if you
accidentally put 10 times the
chlorine you should have.
The pool is already full.
Ci
Q
bread and butter of civil engineering.
VdC/dt=QCi-QC-lCV
dC/dt=Q/VCi-Q/VC-lC
V/Q=T
dC/dt=(Ci-C-lCT)/T
dC/dt= (Ci-C(1+lCT))/T
dC/C(1+lCT)=1/Tdt
u=Ci-C(1+lT)
lC
du=-(1+lT)dC
V
-(1+lT)dC/C(1+lCT)=-(1+lT)dt/T
du/u|utuo
ln(ut/uo)=-(1+lT)t/T
What’s going to happen
Now this equation can be
continued
Ci
Q
lC
V
ln(ut/uo)=-(1+lT)t/T
ln((Ci-C(1+lT))/(Ci-Co(1+lT)))=(-t/T)(1+lT)
Ci-C(1+lT)=exp((-t/T)(1+lT))*(Ci-Co(1+lT))
C(1+lT)= exp((-t/T)(1+lT))*(Ci-Co(1+lT))+Ci
C= exp((-t/T)(1+lT))*(Ci/(1+lT)Co)+Ci/(1+lT)
C= exp((-tQ/V)(1+lV/Q))*(Ci/(1+lV/Q)Co)+Ci/(1+lV/Q)
rearranged for t and assuming
.1*Co=C and Ci=0
With this result you could even
account for the fact that the
water you are putting into the
pool has chlorine as well. The
students need to realize that this
problem really does make a
prediction of the future based on
how the system works.
Environmental issues are
described and decided upon
using such equations.
Conclusions?
• Thanks for the time.
• I hope this gives you some ideas that
you can use to inspire students.
• More examples will be added to this set
and available at
http://home.mesastate.edu/~kcastlet/c
alculus