Single Variable Calculus

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Microsoft Word Free Math Add-In
Single Variable Calculus
Differential
The deriv function will yield the first derivative with respect to the variable indicated. The syntax is
deriv(function, variable). The deriv function may be omitted. There is an alternate way to find the first
derivative with respect to the particular variable, x. Enter the function as an expression in the variable x.
Right-click and select Differentiate on x from the menu.
The derivn function will yield a higher order derivative for a given function. The syntax is
derivn(function, variable, n) where n ∈ {2, 3, 4, …..}, the order. Three arguments are required. There is
an alternate way to find the nth derivative with respect to the particular variable, x. Enter the function
as an expression in the variable x. Right-click and select Differentiate on x from the menu n times using
the previous input lines.
Example 1: Find the first derivative for a given function.
Consider the input:
π‘‘π‘’π‘Ÿπ‘–π‘£(arccos(π‘₯ − 3) , π‘₯)
Right-click on the input line and select Simplify.
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The output is:
−
1
√−(π‘₯ − 3)2 + 1
An easier way to get the same answer does not involve the deriv command. Type in the expression as
follows:
arccos⁑(π‘₯ − 3)
Right-click and select Differentiate on x.
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The output is:
−
1
√−(π‘₯ − 3)2 + 1
Example 2: Find the third derivative with respect to x for a given function.
Consider the input:
sin−1 π‘₯
π‘π‘œπ‘ β‘π‘₯
Right-click and select Differentiate on x.
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The answer for the first derivative is:
sec(π‘₯)⁑tan(π‘₯)⁑sin−1(π‘₯) +
sec(π‘₯)
√−π‘₯ 2 + 1
Right click within the last output and select Differentiate on x . The answer is:
π‘₯ sec(π‘₯)
(−π‘₯ 2 +
3
1)2
+ 2 sec(π‘₯)⁑sin−1(π‘₯) tan(π‘₯)2 +
2 sec(π‘₯) tan(π‘₯)
√−π‘₯ 2 + 1
+ sec(π‘₯)⁑sin−1(π‘₯)
Right click within in the last output and select Differentiate on x a last time. The answer is:
3⁑π‘₯ 2 sec(π‘₯)
5+
(−π‘₯ 2 + 1)2
3⁑π‘₯ sec(π‘₯) tan(π‘₯)
(−π‘₯ 2 +
3
1)2
+ 6 sec(π‘₯)⁑sin−1(π‘₯) tan(π‘₯)3 +
+ 5 sec(π‘₯)⁑tan(π‘₯)⁑sin−1 (π‘₯) +
6 sec(π‘₯) tan(π‘₯)2
√−π‘₯ 2 + 1
+
sec(π‘₯)
3
(−π‘₯ 2 + 1)2
3 sec(π‘₯)
√−π‘₯ 2 + 1
The derivn command will allow the result to be displayed after one input line. Consider the input:
sin−1 π‘₯
π‘‘π‘’π‘Ÿπ‘–π‘£π‘›(
, π‘₯, 3)
π‘π‘œπ‘ β‘π‘₯
Right click and select Simplify.
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The output is:
3⁑π‘₯ 2 sec(π‘₯)
5+
(−π‘₯ 2 + 1)2
3⁑π‘₯ sec(π‘₯) tan(π‘₯)
(−π‘₯ 2 +
3
1)2
+ 6 sec(π‘₯)⁑sin−1(π‘₯) tan(π‘₯)3 +
+ 5 sec(π‘₯)⁑tan(π‘₯)⁑sin−1 (π‘₯) +
6 sec(π‘₯) tan(π‘₯)2
√−π‘₯ 2 + 1
+
sec(π‘₯)
3
(−π‘₯ 2 + 1)2
3 sec(π‘₯)
√−π‘₯ 2 + 1
Example 3: Given a polynomial, find the 3rd derivative with respect to x.
Consider the input:
π‘‘π‘’π‘Ÿπ‘–π‘£π‘›(2π‘₯ 5 − 2π‘₯, π‘₯, 3)
The output is:
120⁑π‘₯ 2
The same output can be obtained with an initial input of an expression:
2π‘₯ 5 − 2π‘₯
Right click and select Differentiate on x. Repeat two more times with right-clicking on the previous
outputs. The first and second and third derivatives are:
10⁑π‘₯ 4 − 2
40⁑π‘₯ 3
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120⁑π‘₯ 2
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Single Variable Calculus
Integrals
There are three ways to evaluate an indefinite integral.
ο‚·
ο‚·
ο‚·
Type an expression and select Integrate on x.
Use the integral symbol in the top ribbon and select Simplify.
Use the command, integral, and input integral(input, variable) and select Simplify.
I have seen that the third approach execute when involving a radical, here as the first two approaches
may not evaluate and simply follow a return of the original input. The software appears to be cautious
when working with a radical and requiring that the radical does not take on a value that is zero (or
negative). The first approach is the easiest to input.
There are two ways to evaluate a definite integral.
ο‚·
ο‚·
Use the integral symbol in the top ribbon and select Simplify.
Use the command, integral, and input integral(input, variable, lowervalue, uppervalue).
Example 1: Type an expression. Select Integrate on x from the drop-down menu to integrate on x with
the expression.
Consider the example:
√4 − π‘₯^2
Place the cursor in the middle and right click. Choose Integrate on x.
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The output is:
π‘₯⁑√−π‘₯ 2 + 4
π‘₯
+ 2 tan−1 (
)+Π‘
2
√−π‘₯ 2 + 4
Example 2: Evaluate a definite integral. Use the integral symbol from the top ribbon.
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Consider the input:
5
∫ √π‘₯ 2 − 4 𝑑π‘₯
4
The right arrow button was used after the square root not place the dx underneath the square root.
Right click and select Simplify from the menu.
The output is:
2⁑ ln(2⁑√3 + 4) − 2 ln(√21 + 5) +
5⁑√21
− 4⁑√3
2
Right click within this output and select Calculate.
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The output is:
4.0285533270189
Example 3: Evaluate an indefinite integral that would involve a substitution.
The input is:
∫ π‘₯ √π‘₯ − 1⁑𝑑π‘₯
Right click and select Simplify to yield the output:
3
2⁑π‘₯⁑(π‘₯−1)2
3
5
−
4⁑(π‘₯−1)2
15
The output does not include the integration constant. This will occur when evaluating some indefinite
integrals.
Consider another example where the input is only an expression:
π‘₯√π‘₯ 2 − 9
Right click and select Integrate on x to give:
3
(π‘₯ 2 − 9)2
+Π‘
3
The command integral would have yielded the same answer. The right arrow button should be used
before inputting the comma. This will ensure that the comma and dx are not under the radical. The
input is:
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2
π‘–π‘›π‘‘π‘’π‘”π‘Ÿπ‘Žπ‘™(π‘₯ √π‘₯ 2 − 9, π‘₯)
Select Simplify to give the answer:
3
(π‘₯ 2 − 9)2
+Π‘
3
Example 4: Evaluate the definite integral:
1.5
∫ √4 − π‘₯ 2 𝑑π‘₯
1
Using the integral symbol in the top ribbon and selecting Simplify gives the output:
1.5
∫ √−π‘₯ 2 + 4 𝑑π‘₯
1
The definite integral is not evaluated. Consider trying this another way using the integral command and
inputting:
π‘–π‘›π‘‘π‘’π‘”π‘Ÿπ‘Žπ‘™(√−π‘₯ 2 + 4, π‘₯, 1,1.5)
Again, the output is not simplified:
1.5
∫ √−π‘₯ 2 + 4 𝑑π‘₯
1
A work-around would be to consider an indefinite integral:
∫ √4 − π‘₯ 2 𝑑π‘₯
The output is simplified:
π‘₯⁑√– π‘₯ 2 + 4
π‘₯
+ 2 tan−1 (
)+Π‘
2
√−π‘₯ 2 + 4
Apply the Fundamental Theorem by evaluating F(1.5) – F(1) and cut and paste from the previous output.
Consider the new input:
1.5⁑√−1.52 + 4
1.5
1.0⁑√−1.02 + 4
1.0
(
+ 2 tan−1 (
) −⁑(
+ 2 tan−1 (
))
2
2
2
√−1.5 + 4
√−1.0 + 4
Select Calculate.
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The output is:
2⁑ tan−1 (
3⁑√7
3⁑√7 πœ‹ √3
− −
)+
7
8
3
2
Select Calculate again to give:
0.7750579446311
Mathematica gives the same answer and evaluates the definite integral directly as shown below:
Example 5: Evaluate a definite integral involving a trigonometric function.
Consider the input:
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5
∫ sinx 𝑑π‘₯
0
The Math Preferences needs to be set.
The angle measure should be in radian mode as shown below:
The output from our example is:
1 − cos(5)
Right-click and select Calculate from the menu to give:
0.7163378145368
Example 6: Evaluate an indefinite integral involving a trigonometric function.
Consider the input:
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(cos π‘₯)2
Right-click and select Simplify to give:
sin(2⁑π‘₯) π‘₯
+ +Π‘
4
2
Example 7: Take the derivative with respect to x for the following example involving an integral.
π‘₯
∫ 3𝑑⁑𝑑𝑑
2
Right-click and select Simplify to give:
3⁑π‘₯ 2
−6
2
Right-click and select Differentiate on x.
The Fundamental Theorem tells us the answer is:
3⁑π‘₯
Example 8: Evaluate the following definite integral by using the integral command.
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6
∫ π‘₯ √π‘₯ − 3⁑𝑑π‘₯
4
The input is:
π‘–π‘›π‘‘π‘’π‘”π‘Ÿπ‘Žπ‘™(π‘₯√π‘₯ − 3, π‘₯, 4,6)
Select Simplify to give:
5
4 · 32
12
−
+ 12⁑√3 −
15
5
Select Calculate to give:
14.2276877526612
Example 9: Integrate using trigonometric functions and the superscript feature in the top ribbon.
Consider the example:
sin⁑π‘₯⁑⁑cos⁑(π‘₯)^5
Trigonometric functions are recognized. After typing one in, press enter to key in the argument. A blue
box will appear as shown below:
Parentheses are optional. As long as what is typed is within the blue box, the add-in will understand this
as the argument. In the example below, typing cos (x) followed with a right arrow will ensure that the
cos(x) is the base.
sin π‘₯⁑ cos(π‘₯) ^5
Select Integrate on x to give:
−
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cos(π‘₯)6
+Π‘
6
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Single Variable Calculus
Graphing
The table below describes some graphing commands. In most instances, the command may be
absent. The exception is the plotparam command that is required. The commands are not case
sensitive. That is, the use of capitalization is optional and will not change whether the line executes.
Command
Example
Notation
Requirements
plot
π‘π‘™π‘œπ‘‘(3 ∗ π‘₯ 2 + 12 ∗ π‘₯)
plotDataSet
plotEq
plotIneq
π‘π‘™π‘œπ‘‘π·π‘Žπ‘‘π‘Žπ‘†π‘’π‘‘({{2, 3}, {5, −1}})
π‘π‘™π‘œπ‘‘πΈπ‘ž(π‘₯ 2 + 𝑦 2 = 9)
π‘π‘™π‘œπ‘‘πΌπ‘›π‘’π‘ž(π‘₯ ≤ 3 + 𝑦)
plotParam
π‘π‘™π‘œπ‘‘π‘ƒπ‘Žπ‘Ÿπ‘Žπ‘š(sin 𝑑, 𝑑^2)
plotPolar
π‘π‘™π‘œπ‘‘π‘ƒπ‘œπ‘™π‘Žπ‘Ÿ(3 × sin πœƒ)
πœ‹
πœ‹
π‘π‘™π‘œπ‘‘π‘ƒπ‘œπ‘™π‘Žπ‘Ÿπ·π‘Žπ‘‘π‘Žπ‘†π‘’π‘‘({{2, } , {8, − }})
3
2
Input function,
f(x).
Input point, {x, y}.
Input f(x, y) = c.
Input inequality
in x and y.
Input (f(t),
g(t))where x=f(t)
and y=g(t).
Input π‘Ÿ = 𝑓(πœƒ).
plotPolarDataSet
Input point {π‘Ÿ, πœƒ}.
DropDown
Menu
Option
to
Execute
Simplify
Calculate
Simplify
Simplify
Simplify
Simplify
Calculate
Graphing in two dimensions
Here are a couple of options of how to input multiple equations.
Method 1: Use the show command and this will allow the input of both equations with a single
application of the Insert New Equation command. Insert New Equation can be found here:
Input:
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π‘ β„Žπ‘œπ‘€(π‘π‘™π‘œπ‘‘(π‘Ž ∗ π‘₯ 2 + 𝑏 ∗ π‘₯), π‘π‘™π‘œπ‘‘(2π‘Žπ‘₯ + 𝑏))
Select Simplify from the right-click menu. This will bring up the two graphs.
The show command needs the following syntax: show(plot1example, plot2example)
Method 2: Here are the equations. Each equation is inserted separately with Insert New Equation.
𝑦 = π‘Ž ∗ π‘₯2 + 𝑏 ∗ π‘₯
𝑦 =2∗π‘Ž∗π‘₯+𝑏
Use the mouse (left click and drag over the top of these equations with keeping the mouse pressed
down) and highlight both simultaneously now. There cannot be any text between these equations.
You only want the equations highlighted.
You should see this image on your screen:
After both are highlighted (in blue), right click to bring up the option, Plot in 2D. You need to right
click and not left click. You are now working with both equations.
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Either method should return this graph:
The animate option appears by default when a variable is introduced such as a.
Example 1: Graph the following. Use the Trace command to find information about points on the curve.
π‘Ÿ = 1 − 2 cos 3πœƒ
Input the following using Insert New Equation.
π‘Ÿ = 1 − 2 cos 3πœƒ
Select Plot in 2D.
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The following will appear:
Select Trace.
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Example 2: Give an example using the plotparam command.
Consider the input:
π‘π‘™π‘œπ‘‘π‘π‘Žπ‘Ÿπ‘Žπ‘š(2 sin(2𝑑) , 3 sin(3𝑑))
Select Simplify to get:
Example 3: Give an example using the plotparam command and the Trace feature.
Consider the input:
π‘π‘™π‘œπ‘‘π‘π‘Žπ‘Ÿπ‘Žπ‘š(2 sin(2𝑑) , 3 sin(𝑑))
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Select Simplify and then apply the Trace feature.
Example 4: Plot the following equation.
π‘₯ 2 − π‘₯𝑦 − 5𝑦 − 1 = 0
Input the equation, right-click, and select Plot in 2D. Use the Zoom out feature to capture the
hyperbola.
Zoom out
using this
button.
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