Analytical Methods - Calculus
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Transcript of Analytical Methods - Calculus
Steve Goddard
Contents
Topic PageDifferentiation of Common functions 1Differentiation of a Product 1Differentiation of a quotient 1Function of a Function 2Successive Differentiation 3Logarithmic Differentiation 4Differentiation of Inverse Trigonometry and Hyperbolic Functions
5
Integration of Common Functions 5Integration Using Algebraic Substitutions 6Integration Using Partial Fractions 7Integration by Parts 8Analyse engineering Situations and solve Engineering Problems Using Calculus
9
Maclaurin’s Series 16
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Steve Goddard
Analytical Methods – Assignment 1
Calculus
Differentiation of Common Functions
1.
If
Since , a = 3 and n = 2 thus
2.
Differentiation of a Product
3.
Differentiation of a Quotient
4.
Using the quotient rule:
Let U =
And
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U=
V=
Steve Goddard
Let V =
Putting these values into the equation:
=
Function of a Function
5.
Using the function of a function rule:
Rewriting U as gives:
Successive Differentiation
6. Find:
6.1
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Steve Goddard
6.2
Logarithmic Differentiation
7. Use logarithmic differentiation to differentiate the following:
First of all I took logs from each side:
Differentiation of Inverse Trigonometric and Hyperbolic Functions
Differentiate the following with respect to the variable:
8.
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9.
Integration of Common Functions
Determine the following indefinite integral:
10.
+c
Evaluate the following definite integrals correct to 4 significant figures:
11.
12.
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Integration Using Algebraic Substitutions
Integrate with respect to the variable:
13.
14.
Integration Using Partial Fractions
Integrate with respect to x:
15.
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Steve Goddard
Next I multiplied the numerators by the main denominator and cancelled out the relevant values
Next I will substitute a strategic value to make one side of the equation = 0. Firstly I will make x = -3
Therefore:
Doing the same again but for the other side I will use x = +3
From this I now know that: A= 2 and B = -2
Now that I now A and B I can put these into the original equation
To integrate this I split it into two parts
Therefore:
Integration by Parts
Determine the following integrals using integration by parts:
16.
Let u = Let du =
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Steve Goddard
Putting this into the by parts formula:
17.
From the integration by parts formula
Let from which i.e.
And let from which
Expressions for u, du and v are now substituted into the by parts formula
Analyse engineering Situations and Solve Engineering Problems Using Calculus
18. Find the turning points of:
And distinguish between them, showing your calculations and deductions
Given that
I determined that
Let Now I will solve the values for
If then:
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Using the quadratic formula,
if
Steve Goddard
Putting these values into the Y equation:
19. In an electrical circuit an alternating voltage is given by Volts.
Determine to 2 decimal places over the range t=0 to t = 10ms:
19.1 The mean and
From excel I have determined that the equation produces a sine wave:
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Because it is a sine wave I can use the equation
From the graph I can see the maximum value is 25 so:
19.2 The r.m.s.
This is very similar to the above equation
I already know that the maximum value is 25 so:
20. A lidless box with square ends is to be made from a thin sheet of metal. Determine the least area of the metal for which the volume of the box is
Area x =
Area y =
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-30
-20
-10
0
10
20
30
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41
Series1
Series2
XY
Steve Goddard
I already know that V = so:
To work out what is, using the total area equation:
Next I differentiate this answer:
Putting the value for x back into the original equation for total area:
21. The distance, x, moved by a body in t seconds is given by:
Distance =
Therefore:
Velocity = v =
Acceleration
Find:
21.1 The velocity and acceleration at the start
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Steve Goddard
Velocity =
Acceleration =
21.2 The velocity and acceleration when t = 3s
Velocity =
Acceleration =
21.3 The values of t when the body is at rest
This is a quadratic equation:
21.4 The values of t when the acceleration is
21.5 The distance travelled in the third second
For 3 seconds:
Distance =
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Steve Goddard
So if t = 3 then:
For 2 seconds:
So if t = 3 then:
99.5 - 24.33 = 75.17 metres
22. An alternating current i amps is given by:
Where: f is the frequency in Hz t is time in seconds
Determine the rate of change of current when t = 20ms, given that f=50Hz
23. The speed of a car, v, in metres per seconds is related to time, t, in seconds by the following:
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Determine the maximum speed of the car in kilometres per hour
Max Speed =
24. Determine the area enclosed by:
The lines on the graph to the right represent the equations below:
From these I will work out the x values at which the two lines intersect.
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-2
-1
0
1
2
3
4
5
1 2 3 4 5 6 7 8 9 10 11
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Maclaurin’s Series
25. Use Maclaurin’s series to find a power series for:
As far as the term
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26. Show, using Maclaurin’s series, that the first 4 terms of the power series for
is given by:
By definition;
Now use the known series for (which is ):
=
Note the even power terms cancel out and the odd powers appear twice:
=
(All summations go from n=0 to infinity).
So the series goes:
27. Find the first 4 terms of the series for by applying Maclaurin’s Theorem
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So the first four terms are:
28. Determine the following limiting values:
If you substitute x=1 directly into the expression, you obtain 0/0, which is undefined
Using l'hopital's rule:
Differentiate both the numerator and denominator with respect to x.
Thus:
When x=1 substituted into the above equation is definable (i.e doesn’t = 0/0), l’Hopitals rule doesn’t need to be used again. Therefore this expression is correct.
Then substitute the value x=1 into this new expression,
=1/9
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Steve Goddard
Bibliography
Higher Engineering Mathematics 5th Edition – John birdIn-class notes – Roger MaceyCourse Hand outs – Roger Macey
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