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Work and Kinetic Energy

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New Maple Operators

  • int   The integration operator computes both indefinite and definite integrals.  It also has an inert form, Int, for the pretty display of an integral in standard mathematical form.  The constant term that arises in indefinite integral calculations is not displayed.  Some examples:

    > int( x^2, x );

    (1/3) x^3

    > Int( x^2, x);

    int( x^2, x)

    > value( % );

    (1/3) x^3

    > int( %, x=a..b );

    (1/12) b^4 - (1/12) a^4

    > Int( %%, x=a..b );

    int(  (1/3) x^3, x=a..b )

    > a := -2; b := 3;

    a:= -2

    b := 3

    > value( %%% );

    65/12

    > Int( sin(t), t=0..Pi/2 );

    int( sin(t), t=0..(1/2) Pi )

    > value( % );

    1

  • diff   The expression differentiation operator.  The inverse of the int function.  It also comes with an inert form.

    > diff( int(x^3,x), x );

    x^3

    > diff( sin(t), t );

    cos(t)

    > int( %, t );

    sin(t)

    > Diff( 4*s^3, s);

    d/ds (4 s^3)

    Unlike the D function, the diff function does not return or expect functions as arguments.

    > f := p -> ln( p^2 );

    f := p -> ln(p^2)

    > diff( f );

    Error, wrong number (or type) of parameters in function diff

    The diff function does accept functions as arguments when you present them in the f(x)  symbol form.

    > diff( f(p), p );

    2 (1/p)

    > diff( f(t), t );

    2 (1/t)

    In the context of the diff function, an unbound variable is considered a constant function, and the symbol f is a variable, not the function f.

    > f; f(q);

    f

    ln(q^2)

    > diff( f, p );

    0

    Maple help describes the relationship between D and diff as D(f) = unapply(diff(f(x), x), x).  Work with this definition until you understand it, it's a good exercise in understanding Maple's interpretation of the functions involved.  Keep in mind that functions/operators take certain types of objects as arguments, and return other types of objects as results.  Work from the inside out.

  • with   The complete library of Maple functions is quite large (check out ?index,function), and many computers would slow down and operate very poorly, or not at all, if all of them were initiated/defined each time the software was started.  For this reason the complete library of Maple functions is broken up into packages, most of which are not activated at start up.  A standard package of the most commonly used functions is activated at start up.  Use the with operator to activate the functions of a particular package.  The packages divide up the functions in the Maple library by mathematical topic.  To see a complete list of the provided packages, check out ?packages and/or ?index,packages.  Be sure to check out ?with, also.

    The student package provides an excellent set of calculus tutorial functions.   I encourage you to use the Maple help pages to investigate these functions.

    > with(student):

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Encapsulated Clown

Back on the movie set they are preparing a scene in which the clown is launched into flight inside a small capsule.  It's the last scene of the movie, and where the clown thinks he's going to go in this capsule is unknown, but it certainly makes a nice set up for a sequel, which is what it's all about anyway. 

The capsule is launched towards the sea from a ramp on the beach.  The ramp is at a 45 degree angle with the surface of the sea, and the capsule is accelerated a distance of 40 m along the ramp by a large compressed spring (The spring is compressed 40 m from equilibrium, and when not compressed the capsule end of the spring rests at the top edge of the ramp).  Given that there is still no air on the set, that the coefficient of kinetic friction between the capsule and the surface of the ramp is .25, that the combined mass of the capsule, clown, and captive duck is 500 kg, and that the linear constant of the spring is 10,000 N/m, how fast is the capsule moving when it leaves the top end of the ramp?

Solution

Remember to use Maple help or the Maple Index if you encounter a function or operator you do not remember.

ramp.gif (3755 bytes)

 

Let the x-axis of the coordinate system lies along the surface of the ramp.

> restart;

Define the capsule as a system, and let the spring, the ramp, and the Earth be outside of the system. With these assumptions we can use the Work-Kinetic Energy Theorem to solve this problem.  The Work-Kinetic Energy Theorem: The work done by the net external force on a body/system during a change in the body's position is equal to the change of kinetic energy of the body.

> WKEThm := Work = Delta(KE);

WKEThm := Work = delta (KE)

The Greek letter Delta is used to indicate a change in a quantity, and typically expressions like the change in kinetic energy would look like delta KE , not delta (KE).  But, to achieve the typical look in Maple output requires the use of multiplication, and in our case this causes a problem.  We want to use it as a variable name, and Maple does not allow such assignments.

> Delta * KE := puppy_love;

Error, invalid lhs of assignment

So we'll stick with delta (KE).

The definition of work is

> Work := force * distance;

Work := force distance

A spring generates a force that varies, and for situations of varying force the useful form of the Work equation is

> Work := Int( net_force, r = r[i]..r[f] );

Work := int( net_force, r= r[i]..r[f] )

A change in kinetic energy is expressed mathematically as

> Delta(KE) := (1/2)*mass*(v[f]^2-v[i]^2);

delta (KE) := (1/2) mass (v[f]^2-v[i]^2)

Thus, a useable form of the work kinetic energy theorem is

> WKEThm;

int( net_force, r=r[i]..r[f] ) = (1/2) mass (v[f]^2 - v[i]^2)

Every vector equation resolves into a set of equations that expresses the situation in terms of component vectors which lie along the coordinate axes.  There are two coordinate axes in our system and the two following equations express the above information in set form.

> { Int(F[x], x=x[i]..x[f]) = (1/2)*mass*(v[x,f]^2-v[x,i]^2), Int(F[y], y=y[i]..y[f]) = (1/2)*mass*(v[y,f]^2-v[y,i]^2) };

{ pretty display of command line }

Our choice of where to place the axes saves us from a lot of extra mathematics.  Although the equation on the left correctly models its part of the situation, all we need is the equation on the right.  Why?  

To simplify notation, make the following change to the Work equation.

> Work := Int( F[x], x = x[i]..x[f] );

Work := int( F[x], x=x[i]..x[f] )

The WKEThm equation becomes

> WKEThm;

int( F[x] , x=x[i]..x[f] ) = (1/2) mass (v[f]^2 - v[i]^2 )

Note that we are now also making the substitutions v[x,i] = v[i] and v[x,f] = v[f]

Rotating the picture above -45 degrees produces the following picture, and using the information it provides, we can construct a useful expression for F[x].  Three forces contribute to F[x]: the friction force, the spring force, and the weight force.  Let theta = Pi/4.

axes.gif (2340 bytes)

> fric_force := mu*norm_force;

fric-force := mu norm_force

Where

> norm_force := mass*10*cos(theta);

norm_force := 10 mass cos(theta)

Thus,

> 'fric_force' = fric_force;

fric_force = 10 mu mass cos(theta)

> grav_force := mass*10*sin(theta);

grav_force := 10 mass sin(theta)

> spr_force := -k*(x-x[equalib]);

spr_force := -k (x - x[equalib])

The term x[equalib] is the position along the axis at which the spring is in equilibrium.  For our situation, a more useful version of the equation results if we place the origin of the x-axis at the top end of the ramp, which is the spring's equilibrium position.  Thus,

> x[equalib] := 0;

x[equalib] := 0

and

> 'spr_force' = spr_force;

spr_force = -k x

Note that at this point we have made two choices about our coordinate axes which together have greatly simplified the mathematical representation of the problem.

The net force along the x-axis is now represented as

> F[x] := spr_force - grav_force - fric_force;

F[x] := - k x - 10 mass sin(theta) - 10 mu mass cos(theta)

And with the following substitutions,

> k := 10000; mass := 500; theta := Pi/4; mu := .25; x[i] := -40; x[f] := 0; v[i] := 0;

k := 10000

mass := 500

theta := (1/4) Pi

mu := .25

x[i] := -40

x[f] := 0

v[i] := 0

F[x] becomes

> 'F[x]' = F[x];

F[x] = -10000 x - 3125 sqrt(2)

and WKEThm becomes

> WKEThm;

int( -10000 x - 3125 sqrt(2), x=-40..0 ) = (1/2) mass (v[f]^2 - v[i]^2 )

> Work := value( Work );

Work := 7823223.305

> WKEThm;

7823223.305 = (1/2) mass (v[f]^2 - v[i]^2 )

> solve( WKEThm, v[f] );

-176.8979740, 176.8979740

As a check,

> subs( v[f] = %[2], WKEThm );

7823223.305 = (1/2) mass (31292.89321 - v[i]^2 )

> evalf( % );

7823223.305 = 7823223.305

As you are well aware, integration is a complicated and subtle process, and Maple employs many complicated and subtle algorithms to do integration.  Unless instructed specifically on which technique to use, Maple uses some standard techniques.   Blind trust of these algorithms without a good understanding of each of their assumptions and processes is a recipe for incorrect results. 

Until you know a lot more about different integration techniques and how to ask Maple to use them, never just accept any integration result that Maple produces.  Although we did not in the problem above, which involved a definite integral of a rather simple polynomial, you should always make some check of the result of an int call for an even moderately complicated expression.  In cases of indefinite integration use the diff function to differentiate the result and see if it produces the expression you originally integrated.  When differentiating in these situations, Maple might return results that look nothing like the expression originally integrated.  Don't immediately think that something is amiss; use the simplify function on the result first.  Check out ?simplify and the related functions listed at the bottom of this help page.

 

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