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Karl Castleton

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:

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Karl Castleton

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  1. Karl Castleton Research Scientist Pacific Northwest National Laboratory

  2. 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

  3. 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?

  4. 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.

  5. F 1/2 E 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? 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 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 t A B C

  6. t A B C Diesel Tank continued C-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.

  7. 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?

  8. 120,100 65,100 75,70 garden 75,25 10,10 65,100 A1 10,10 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) 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.

  9. 120,100 65,100 75,70 garden 75,25 10,10 Wacky Gardener Continued 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 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

  10. Barn x x 100-2x Getting the most inside a piece of fence. 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 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?

  11. 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. 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 b/day

  12. 45 45 40 40 b/day b/day 25 25 12 12 1 2 3 4 1 2 3 4 day day Bottle Counting continued 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. 115 b This should start to look like the wacky gardener again. You could integrate and find the total number of bottles. 77 65 25 1 2 3 4 day

  13. Ci Q l C V 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.

  14. Ci Q l C V What’s going to happen continued 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. 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)

  15. 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

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