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.

C

BA t

A diesel truck driver needs to know how much fuel is in the tank

The fuel gage is broken. He wants to use a dip stick. The mark at full, 1/2 and empty are easy. But where does the 1/4 and 3/4 go?

F

1/2

E

The area above the 1/4 mark should be

equal to the area below. Only 1/4 of

the tank need be considered.The Area

of an angular segment will be

useful.

Area A+B =C -B

Lets name the angle t

C

BA t

Diesel Tank continuedC-B=B+A

tr2 - ½ r2cos(t)sin(t)=

½ r2cos(t)sin(t)+(pi/2-t)r2

assume r=1 (for simplicity)

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

has coordinates of the “corners”. Clearly this is just integration (for those who know what

it is) in hiding. But the strange shape might

initially make it seem difficult. Trapezoid

summation of the areas “under” the line

segments. Area of a trapezoid

A=1/2h(b1+b2)

garden

10,10

65,100 120,100

75,25

75,7010,10

65,100

A1

h=65-10, b1=10,b2=10

h will be negative for the lines that go

towards the Y axis. So the line (120,100)

- (75,25) will have h=-45

Our gardener likes non rectangular shapes. Each

sprinkler head can covers 2000ft2.

How many does he need.

Wacky Gardener Continued

garden

10,10

65,100 120,100

75,25

75,70

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.

r

Getting the most inside a piece of fence.

This one does appear in many calculus texts. A farmer has

100 feet of fence and he wants to enclose the largest

rectangular area along side his barn. What should the

dimensions of the area be?

Students should get used to the idea that if you are maximizing or

minimizing something you are going to be taking the derivative

and setting it equal to 0. The most important part of this question is

setting it up properly. Assume the small side length is x then the large has to be 100-2x. Students may try to call this distance y and be stuck.

A=x * (100-2x)=100x-2x2

dA/dx=100-2*2x=0

100=4x or x=25

Barn

x x

100-2x

Measure the total with sampled rates.Your company produces

pop/soda. Estimate the total number of bottles leaving the plant without adding equipment to count every bottle.(cheap boss) You

know that the plants production rate in

bottles/day does not change instaneously but slowly increases or decreases.

b/day

We should assume that we can every once in a while measure the number of bottles that left over a

short period of time or monitor how long it takes for a certain amount of

bottles to leave the plant. Either would give you b/day estimates

(slopes) at given points in time. Lets assume we got the following

measures.

Day 1 25 b/day, Day 2 40 b/day

Day 3 12 b/day, Day 4 45 b/day

Bottle Counting continued

1 2 3 4

12

25

4540

day

b/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.

1 2 3 4

12

25

4540

day

b/day

1 2 3 4

25

65

115

77

day

b

What’s going to happen in the future This is a Constantly Stirred Tank

Reactor (CSTR) model and is the 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)

du=-(1+lT)dC

-(1+lT)dC/C(1+lCT)=-(1+lT)dt/T

du/u|utuo

ln(ut/uo)=-(1+lT)t/T

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.

l C

V

QCi

What’s going to happen continued

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)

l C

V

QCi

Now this equation can be 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/calculus