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Student[Calculus1]

  

ShowSolution

  

show all the steps in the solution of a specified problem

 

Calling Sequence

Parameters

Description

Examples

Compatibility

Calling Sequence

ShowSolution(p, opts)

Parameters

p

-

(optional) posint or a calculus1 problem; the problem to solve

opts

-

(optional) options of the form keyword=value, where keyword is one of maxsteps, searchoptimal, showrules, output, displaystyle.

Description

• 

The ShowSolution command is used to show the solution steps for a Calculus1 problem, that is, a limit, differentiation or integration problem such as can be expected to be encountered in a single-variable calculus course.

• 

If p is omitted, the current (most recently referenced) problem is solved.  Otherwise, the problem referenced by p is solved.  A problem can be referenced either by its problem number (see GetProblem and ShowIncomplete) or by the problem itself, for example via a label.

• 

These options can be used to control how the problem is solved:

– 

maxsteps = posint (default: 25)

This puts a limit on the number of rules which can be applied to solve a problem.

– 

searchoptimal = truefalse (default: true)

The default behavior is to try to determine the shortest solution sequence. If this option is given as searchoptimal = false, the first solution discovered will be displayed.

• 

These options can be used to control how the problem is displayed:

– 

showrules = truefalse (default: true)

Normally, the rule applied at each step of the solution is displayed; if this option is given as showrules=false, the rules are not shown, and the displaystyle is brief.

– 

output = canvas,script,record,list,print,printf,typeset,link (default: typeset)

The output options are described in Student:-Basics:-OutputStepsRecord

– 

displaystyle= columns,compact,linear,brief (default: linear)

The displaystyle options are described in Student:-Basics:-OutputStepsRecord.  Setting displaystyle = brief shows the steps in a compact form with one or two words at each step to describe the rule being applied.

• 

If you set infolevel[Student] := 1 or infolevel[Student[Calculus1]] := 1 (see infolevel), Maple may display some additional, useful information about the state of the problem and its solution.

Examples

withStudentCalculus1:

Diffx2sinx,x

ⅆⅆxx2sinx

(1)

ShowSolution

Differentiation Stepsⅆⅆxx2sinx1. Apply theproductruleRecall the definition of theproductruleⅆⅆxfxgx=ⅆⅆxfxgx+fxⅆⅆxgxfx=x2gx=sinxThis gives:ⅆⅆxx2sinx+x2ⅆⅆxsinx2. Apply thepowerrule to the termⅆⅆxx2Recall the definition of thepowerruleⅆⅆxx=x1This means:ⅆⅆxx2=So,ⅆⅆxx2=We can rewrite the derivative as:2xsinx+x2ⅆⅆxsinx3. Evaluate the derivative ofsin(x)Recall the definition of thesinruleⅆⅆxsinx=cosxThis gives:2xsinx+x2cosx

(2)

Intsinx2,x

sinx2ⅆx

(3)

Hint

rewrite,sinx2=12cos2x2

(4)

Rule

sinx2ⅆx=12cos2x2ⅆx

(5)

ShowSolution

Integration Stepssinx2ⅆx1. RewriteEquivalent expressionsinx2=12cos2x2This gives:12cos2x2ⅆx2. Apply thesumruleRecall the definition of thesumrulefx+gxⅆx=fxⅆx+gxⅆxfx=12gx=cos2x2This gives:12ⅆx+cos2x2ⅆx3. Apply theconstantrule to the term12ⅆxRecall the definition of theconstantruleCⅆx=CxThis means12ⅆx=x2We can now rewrite the integral as:x2+cos2x2ⅆx4. Apply theconstant multiplerule to the termcos2x2ⅆxRecall the definition of theconstant multiplerulefxⅆx=fxⅆxThis means:cos2x2ⅆx=cos2xⅆx2We can rewrite the integral as:x2cos2xⅆx25. Apply a change of variables to rewrite the integral in terms ofuLetubeu=2xIsolate equation forxx=u2Differentiate both sides=2Substitute the values for x and dx back into the originalcos2xⅆx=cosu2ⅆuThis gives:x2cosu2ⅆu26. Apply theconstant multiplerule to the termcosu2ⅆuRecall the definition of theconstant multiplerulefuⅆu=fuⅆuThis means:cosu2ⅆu=cosuⅆu2We can rewrite the integral as:x2cosuⅆu47. Evaluate the integral ofcos(u)Recall the definition of thecosrulecosuⅆu=sinuThis gives:x2sinu48. Revert change of variableVariable we defined in step5u=2xThis gives:x2sin2x4

(6)

UnderstandInt,`c*`,revert

Int=constantmultiple,revert

(7)

ShowSolutionInt14x2+4x+1,x

Integration Steps14x2+4x+1ⅆx1. Apply a change of variables to rewrite the integral in terms ofuLetubeu=x+12Isolate equation forxx=12+uDifferentiate both sides=Substitute the values for x and dx back into the original14x2+4x+1ⅆx=1u2ⅆu4This gives:1u2ⅆu42. Apply thepowerrule to the term1u2ⅆuRecall the definition of thepowerrule, for n-1uⅆu=This means:1u2ⅆu=So,1u2ⅆu=1uWe can rewrite the integral as:14x+2

(8)

Compatibility

• 

The Student[Calculus1][ShowSolution] command was updated in Maple 2021.

• 

The output and displaystyle options were introduced in Maple 2021.

• 

For more information on Maple 2021 changes, see Updates in Maple 2021.

See Also

Calculus1

GetProblem

infolevel

label

ShowIncomplete

ShowSteps