Purpose

The purpose of this week's lab is to provide additional experience evaluating antiderivatives, formulating differential equations, and solving differential equations with Maple. While Maple will be used to evaluate most indefinite integrals in this worksheet, you should always check that Maple's results are correct. This check can always be done by differentiation but you should also attempt to identify the problems that you could solve by hand.

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Example 1: Evaluating Indefinite Integrals

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restart;

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The Maple commands Int( f, x );  and   int( f, x );  can be used to evaluate the indefinite integral .   For example,

>	f := 3*x^2;

>	F := Int( f, x );

>	value( F );

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Note that Maple does not automatically include the arbitrary constant in its result. We need to include an appropriate constant in the input to Maple (or remember to include the constant in an appropriate place when an answer is copied to another program).

>	F := Int( f, x ) + C;

>	value( F );

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The int  and Int  commands perform essentially the same operations.

>	F := int( f, x ) + C;

The difference is that the Int  command does not evaluate the integral until it is explicitly told to do so with the value  command. I highly recommend using the Int  command because it allows you to visually verify that you correctly entered the integrand.

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Regardless of how an indefinite integral is evaluated, the result can - and should - be checked for consistency.

>	CHECK := diff( F, x ) = f;

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Exercise 1 (# 49, page 215)

For each of these three indefinite integrals, note that the integrand is more complicated than anything we have seen. Later in this course we will learn to evaluate integrals like the one found in (a); integrals like the ones in (b) and (c) can be evaluated using techniques learned in Math 142. Visually verify that you enter the correct integrand and check that the answer returned by Maple is correct.

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Exercise 2 (# 50, page 215)

Replace the %?  in the first command below with an appropriate expression for this problem. The second command computes (with int) the next function in this sequence. Use similar commands to compute , , and .

>	F[0] := %?;

>	F[1] := int( F[0], x ) + C_0;

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Example 2: Modeling and Solving Differential Equations

Exercise #25 (page 220) will be presented by Jay and Wally. The following Maple commands could be used to solve this problem -- note that the individual operations involved in this problem are easier to do by hand. This Maple solution is presented as a guideline for more complicated problems, possibly even # 35 (page 221).

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The geometric condition that is always satisfied is the differential equation:

>	model := diff( V(t), t ) = -k*S(t);

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In this case, the volume and surface area are each expressible in terms of the radius of the snowball:

>	volume := V(t) = Pi*r(t)^3;

>	surface_area := S(t) = 4*Pi*r(t)^2;

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When the formulae for volume and surface area are inserted into the differential equation, we obtain:

>	ode := eval( model, [volume,surface_area] );

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Solving this equation for  yields:

>	ode := isolate( ode, diff(r(t),t) );

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Observe that this differential equation can be solved by direct integration (and we cannot forget the constant of integration):

>	general_solution := int( lhs(ode), t ) = int( rhs(ode), t ) + C;

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When the differential equation cannot be solved by direct integration, a solution method for differential equations must be used. We will learn about Separable Equations in Section 5.2. Maple's differential equation solver, dsolve , is very powerful. Here is how it could be used for this problem:

>	dsolve( ode, r(t) );

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The information about the radius at times  and  can be used to determine values for the constants of proporationality ( ) and integration ( ):

>	condition1 := [ t=0, r(t)=2 ];

>	condition2 := [ t=10, r(t)=1/2 ];

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Each of these conditions, when substituted into the general solution, creates a linear equation for  and :

>	eq1 := eval( general_solution, condition1 );

>	eq2 := eval( general_solution, condition2 );

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The solution to these two equations is found to be:

>	constants := solve( {eq1,eq2}, {k,C} );

>	solution := eval( general_solution, constants );

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This solution exactly matches the one given in the text. If we wanted to check that this solution satisfies all of the conditions of the problem we could compute the following:

>	eval( solution, t=0 );

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>	eval( solution, t=10 );

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>	vol := eval( volume, solution );

>	surf := eval( surface_area, solution );

>	eval( model, [vol,surf,op(constants)] );

>	simplify( % );

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Because everything agrees, our solution is correct. Notice also that we now have the explicit formulae for the volume and surface area of this snowball for any time . (You should be able to answer questions such as: At what time does the snowball have half its original surface area? At what time does the snowball disappear?)

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Exercise 3 (# 36, page 221)

The modeling in this problem is very easy. The differential equation cannot be solved by direct integration. Read about Separable Equations in Section 5.2. If you use dsolve , be sure you also know how to solve this problem by hand.

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Exercise 4 (# 35, page 221)

The modeling in this problem is similar to that presented in Example 2. The differential equation is separable; if you use dsolve , be sure you also know how to solve this problem by hand.

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