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Maple Solutions to Selected Review Problems
prepared by
Douglas B. Meade
(meade@math.sc.edu)
25 August 2003
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1.
> | restart; |
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> | f := sqrt(x); g := exp(-3*x); |
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> | plot( [f,g], x=0..1 ); |
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> | c := solve( f=g, x ); |
> | c := fsolve( f=g, x ); |
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(a) Area of R
> | A := Int( f-g, x=c..1 ); |
> | A = value( A ); |
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(b) Volume of solid formed when R is revolved around y=1
> | with( Student[Calculus1] ); |
> | VolumeOfRevolution( 1-g, 1-f, x=c..1, output=integral ); |
> | VolumeOfRevolution( 1-g, 1-f, x=c..1, output=plot, view=[0..1,-1..1,DEFAULT] ); |
> | VolumeOfRevolution( 1-g, 1-f, x=c..1, output=value ); |
Note: The VolumeOfRevolution command is available through a friendly graphical user interface (GUI) via the WWW. Links are provided on the course website. (You can use the Maplet link if you have Maple 8 on the computer you are using; otherwise, assuming you have an Internet connection, use the MapleNet link.)
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> | V := Int( Pi*(1-g)^2 - Pi*(1-f)^2, x=c..1 ); |
> | V = value( V ); |
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(b) Volume of solid with R as base and height = 5*base for all cross-sections.
> | plot3d( 5*(f-g), y=g..f, x=c..1, axes=boxed, shading=z ); |
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> | V2 := Int( (f-g)*5*(f-g), x=c..1 ); |
> | V2 = value( V2 ); |
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2.
> | restart; |
> | V := -(t+1)*sin(t^2/2); |
> | plot( V, t=0..3 ); |
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(a) Find a(2).
> | A := diff( V, t ); |
> | a(2) = eval( A,t=2 ); |
> | evalf( % ); |
At this time the velocity is negative and increasing. The speed is the absolute value of the velocity. At t=2, the speed is positive, and decreasing.
> | plot( [V,abs(V)], t=0..3, style=[point,line], color=[red,blue], legend=["velocity","speed"] ); |
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(b) Find times when particle changes direction
Direction change occurs when velocity changes sign. From graph, this occurs exactly once in [0,3], near t=2.5.
> | solve( V=0, t ); |
> | fsolve( V=0, t ); |
> | fsolve( V=0, t=2..3 ); |
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(c) Find total distance travelled.
> | TotDist := Int( abs(V), t=0..3 ); |
> | TotDist = value( TotDist ); |
> | TotDist = evalf( TotDist ); |
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(d) Find greatest distance from particle to origin during 0 <= t <= 3.
The position of the particle can be given as:
> | P := 1 + Int( V, t=0..T ); |
> | plot( P, T=0..3 ); |
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The greatest distance from the origin occurs at the critical point near t=2.5. Note that this critical point for the position occurs at the same time that the particle changes direction
> | Tcrit := fsolve( V=0, t=2..3 ); |
> | MaxSep := eval( P, T=Tcrit ); |
> | evalf( MaxSep ); |
Thus, the maximum separation from the origin is 2.265 (units).
6.
> | restart; |
> | f := piecewise( x<=3, sqrt(x+1), x<5, 5-x ); |
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(a)
> | plot( f, x=0..5 ); |
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(b)
> | AvgVal := 1/5 * Int( f, x=0..5 ); |
> | AvgVal = value( AvgVal ); |
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> | plot( [f, AvgVal], x=0..5 ); |
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(c)
> | g := piecewise( x<=3, k*sqrt(x+1), x<=5, m*x+2 ); |
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g differentiable at x=3 means that i) g is continuous at x=3 and ii) g' is continuous at x=3.
> | q1 := Limit( g, x=3, left ) = Limit( g, x=3, right ); |
> | eq1 := value( q1 ); |
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> | Dg := diff( g, x ); |
> | q2 := Limit( Dg, x=3, left ) = Limit( Dg, x=3, right ); |
> | eq2 := value( q2 ); |
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> | solve( {eq1,eq2}, {k,m} ); |
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> | G := eval( g, {k=8/5,m=2/5} ); |
> | plot( G, x=0..5 ); |
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