slope_fields_Jensen_handout

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9. SLOPE FIELPS ANn EULER TANGENTSpIf you are using a pre-reform textbook. there is a good chance that these two topics do not appear orare treated in a cursory manner. The fact is. these areas could not be meaningfully addressed beforethe graphing calculator because the work to produce them is so tedious. With technology. theyprovide streng support to both the field of differential equations and the notion of linearapproximation. Slope fields can be introduced when stude nts first encoun te r the generalantiderivative of a function to reinforce the idea that the solution is a family of functions.There are some very good calculator programs that can be used with these topics; however. somecare needs to be taken in using them. The program for slope fields will generate a flow field solutionfor a differential equation in x and/or y. However. when the window is notjudiciously chosen. theresults can be horrendous. especially with exponential functions. It is helpful to try to predict thesize of slopes and to adjust the window so that they do not become too large or too small. Forinstance,if dy/dx = xy. the window [-2.2] x [-2.2] gives a nice slope field with about 10 intervals forboth x and y; however, if we change the window to [-10.10] x [-10.10]. the slope field looks nearlyvertical. inasmuch as the slopes at the extremities have absolute value near 100. Similarly. oneneeds to be careful in deciding the number of x and y values selected. For instance. if dy/dx = x/yand the number of y values selected is 10. then it is very likely that the program will attempt toevaluate the slope when y = O. I have found it useful to choose a number of values such as 11 or 13to avoid having the slope evaluated at a window value of zero.The programs that produce Euler tangent values and graph Euler tangent solutions are very helpful.especially the latter. When using the Euler tangent graphic program. Ell/erg. try using a differentialequation whose solution can be found analytically so that they can see how making L1x smallprovides for closer solution approximations and how propagated error grows as the number of. values increases. [Note: Since this programs turns off functions. it will be necessary to go into theY= stack to tum on the solution to the D.E.]None of these pn:;>grall1s. however. is likely to be a direct help to students on the AP exam. There. has been much speculation about how the items would be tested; we discovered at least one answerto this speculation in the 1998 Be exam. where students ~ e r e asked to sketch a slope field and tocompute a few Euler tangent values with a given set of conditions. Students must show their work.so it will be necessary to know that ~ y '" d y / d x ' ~ x and how to apply it. In the multiple choice area.it is still likely that students will be aSked to do something such as matching a slope fie'ld with adifferential equation or finding an approximation to f ( x O + ~ x ) given an initial condition (xo.f(Xo)) anda value for ~ x .

, A final note. Slope fields and Euler tangents are accessible enough as a topic that it would notsurprise me if they eventually end up in the AB syllabus. Therefore. it is probably not a bad idea tointegrate this subject matter as time allows in an AB course.,

JJENSEN APCALC


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SI,OPE FIEI,PSA slope field is a graphical display which shows the flow of tangent lines to the family of solutioncurves of a differential equation. Often, this flow diagram provides valuable information about thenature of the solution curve, i.e., whether it is polynomial, exponential. circular, uigonomerric, etc.Slope fields can be.consllUcted. over a region .of the plane by direct substitution into the differentialequation, or they can be generated using the graphing calculator and an appropriate program.There are generally two types of problems that involve slope fields. In the first, we are given adifferential equation and asked to produce its slope field diagram. In the second, we are given aslope field and asked to match it to a paJ.ticular differential equation which has those characteristics .CONSTRUCTING A SLOPE FIELD.When constructing slope fields, it is helpful to first create a table of values for a specific intelval of xand y. For example, if we know that dy/dx = xy/4 for x in [-2,2] and yin [-2,2], we can consuuct agrid to evaluate the slopes at each integer pair of values. Complete the grid of slope values for thisdifferential equation: .

x values1. -2 1 o 1 2-( .. J:. 02.. 2

1 -\;:: -(.. 61-. .!'l .y values o ( ) 0 0

- 1 '-r.1 - l-,.-'-f 0t 1.... 6,1,.. ... " " ' ,0 - 2

Once the grid is complete, the slope segments can be drawn tlrrough the lattice points of a graph.Draw the slope field for the values obtained above: .2.

\ --. \Y " ~ .' ..., ..

/ ..--- ' , . " .../ / "/ " -.'

3. What kind of function seems to be pictured by the slope field you constructed?

JJENSEN APCALC

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READING A SLOPE FIELD.When reading a slope field, it is important to look for clues the slope segments give us about thebehavior of the differential equation and, by extension, its family of solutions.It is possible to read a slope field one segment at a time: however, this can be exhausting for verylarge fields. It is probably easier in many cases to spot trends in the slope field that tell us somethingabout how x and y are related in the differential equation. Here are some approaches you can use: Examine slope field segments along vertical lines. If the segments along each vertical linehave the same slope. then the differential equation does not depend on y, because, as y varies, theslope does not. Examine slope field segments along horizontal lines. If the segments along each horizontalline have the same slope. then the differential equation does not depend on x. Examine slope field segments in the first quadrant. If the segments have positive slope, thenthere are likely 110 negatives in the expression of the differential equation. If the slopes becomelarger as x gets larger. then dy/dx varies directly with x; likewise for y. Otherwise. we candetelmine that the slope is inversely related to one or both vaIiables. If the slope field evinces a curve which looks familiar, check by differentiating that curve tosee if its slope field fits the graphical d(i(d.NOTE: There are occasional anomalies in the appeaI'ance of slope fields. due to the way they aI'egenerated on a calculator. Small discrepancies in slope can usually be dismissed.Consider the following slope field: y' / ,. ,. / ' I... ../ / ,. ,. / ' /.. ...'/ / , / ' / , . . '.. . . ." / / '

/ / '/ / i / I. / / .. ,. / /.. ..I / ,. .. / /... ..

x [-,4 . What can you deduce from reading the slope segments?_____________5. , Which of these is most likely the differential equation:

(A ) dy/dx = .5xy (B) dy/dx =x2/Y (C ) dy/dx = .5x2

(3 )

JJENSEN APeALe


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Here are some more slope fields to practice on.equation.

6. Which of the following differentialequations has the ,solution slope fieldpictured at right?(A) dy/dx = .5y(B) dy/dx = .2x/y(C) dy/dx = xy(D) dy/dx = x + y(E) dy/dx = l/x

7. Which of the following differentialequations has the solution slope fieldpictured at light?

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(A ) dy/dx = x2(B) dy/dx = y/x

, '.(C) dy/dx =-'1 '(D) dy/dx = -x/y

(E ) dy/dx = x2 + y2

Which of the following differentialequations has the solution slope fieldpictured at right?(A ) dy/dx = x + Y(B)dy/dx=x-y(C) dy/dx = x2(D ) dy/dx = 2y(E ) dy/dx = y/x

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HENSEN APCALC

In each, match the slope field with its differential

.. .. ." ..". .... ....


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EULER TANGENTSImagine trying to measure a curve using a rigid ruler. Because the contours of the curve cannot befollowed by the IUler. you would have to make small interval measurements as an estimate. Thisprocess underlies a calculus procedure known as Euler Tanjients.Let dy/dx = f(x.y). that is. a diffe re ntial equation in x and /or y. and let (xO. yO) be an in itialcondition of its solution. Further. let's suppose that the differen tial equation cannot be solved by ourknown method of separating variables. Then. we can only estimate points on the solution curveusing the evidence provided.Illustration: dy/dx = x + y and (xO. yO) = (1.1)

1.xYdy/dxfly

2.

xydy/dxfly

HE NSEN APCALC

From previous work with differentials. we know that. when flx is "sufficientlysmall" fly:::: dy/dx Llx, which is to say that the curve is approximated by itstangent line over a small interval. Let's choose flx = .1: since dy/dx = x + y.at (1,1). dy/dx = 2 and fly "" 2(.1) = .2. This brings us to an estimate of anearby point on the solution curve, (1.1. 1.2). Call this (xl. yl).Now we can repeat the process. At (1.1, 1.2). dy/dx = 2.3 and fly "" (2.3 )(.1)= .23. Thus. our next point is (x2. y2) = (1.2. 1.43).As we continue. the y-value estimates become more complicated: however. the x-values continue to increment by .1 each time.Use the method of Euler tangents to complete the table below for this example.Give estimates accW'ate to three decimal places.1 . 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0, \1 1.2 1.43 \ \ \ \, \ \, \ \,,. .,. \ \ ,2 ' 2:3 2.63 \ \ \,, \ "\\ , \ \, , \.2 .23 .263 , , \ ,\ ,\., Given that dy/dx = 2y with initial condition (xO. yO) = (OJ). use Eulertangents to estimate the solution using LlX = .2 for five steps. i.e . from x = 0to x = I.

,,\

0 .2 .4 .6 .8 1.0j ( ' {

t> " 1'\ "', "\' \ . \ '\1-, '" " ' ~ '''.,,,'\, ",.

\ \ \

~ ~ .. V "-"

(5)

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(a) Solve the differential equation in example 2. Use the initial value to resolve theconstant. '

a. _________________________

4.

(b) In the table below, compare the values of y obtained in example 2 with thevalues of your solution in part (a). Compute the difference in each case.x 0 .2 A .6 .8 l.0" Euler y I I y L( "-\ \"", \ \ . "-t.b Ii q1., ..Actual v I '\,"\Difference 0 0 '1 '2v _ ~ (c) In the windows below, plot: (I ) the curve you found in pan (a) and,(2) the points you found in pan (b). The window dimensions are [0.1] x [0,8].

. .. , .. .-

For each of the following situations. indi


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. POSTSCRIPT ON EULER TANGENTS.The Fundamental Theorem with Euler Tangents .

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Given a function f. find an approximation to f with theinitial condi tion (xO. f(xO)).For small ~ x . t:.y/t:.x == f(xo)t:.y =:: f(xO) 6xf(x l) == f(xO) + f(xo) t:.xf(x2) == f(x 1) + f(x 1) 6x

= f(xO) + f(xO) t:.x + f(x 1) t:.x f(xn) =:: f(xO) + f(xO) 6x + f(x 1) t:.x + ... + f( -1) ~ x By the left endpoint method, the area bounded by f from xO to xn can beapproximated as:A =:: f(xO) ~ x + f( x1) ~ x + ... + f(xn-l) ~ x But, f(xn) =:: f(xO) + f(xO) t:.x + rex 1) t:.x + ... + f(xn-l) t:.x

" ... . . .or.f(xn) - f(xo) =:: f(xO) t:.x + f(x 1) Ax + ... + f(xll-l) t:.x

Thus. A = f(xn) - f(xO) as t:.x->O. which is the FTC!

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lJENSEN Arc-ALe

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