unit workbook 4 calculus - wb 4.pdflet’s go on and see if we can find 𝑦𝑦 ′′ … 𝑦𝑦...

Unit WorkBook 4 – Level 4 ENG– U2 Engineering Maths © 2018 UniCourse Ltd. All Rights Reserved.

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Pearson BTEC Higher Nationals in Electrical and Electronic Engineering (RQF)

Unit 2: Engineering Maths (core)

Unit Workbook 4 in a series of 4 for this unit

Learning Outcome 4

Calculus

Sample


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3.1 Calculus There are a couple of great webpages for checking your answers to Calculus problems…

Check Differentiation answers

Check Integration answers

3.1.1 The Concept of the Limit, Continuity and the Derivative Calculus deals with functions which continually vary and is based upon the concept of a limit and continuity. Let’s refer to a diagram to understand the concept of a limit…

Here we have indicated a point P on part of a function (the red curve) with Cartesian co-ordinates (𝑥𝑥1,𝑦𝑦1). We require another point Q on the function to be a small increment away from P and will designate a small increment with the symbol 𝛿𝛿 (delta).

What we then have is…

𝑷𝑷 = (𝒙𝒙𝟏𝟏,𝒚𝒚𝟏𝟏)

𝑸𝑸 = (𝒙𝒙𝟏𝟏 + 𝜹𝜹𝒙𝒙, 𝒚𝒚𝟏𝟏 + 𝜹𝜹𝒚𝒚)

Look now at the chord PQ (drawn as the straight line in blue). If we can determine the slope of this chord and then make it infinitesimally short we will end up with a tangent to the function. It is the slope of this tangent which forms the basis of Differential Calculus (normally called differentiation).

By inspection, we see that the slope of the chord is given by…

𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 𝒔𝒔𝒐𝒐 𝒄𝒄𝒄𝒄𝒔𝒔𝒄𝒄𝒄𝒄 𝑷𝑷𝑸𝑸 =𝜹𝜹𝒚𝒚𝜹𝜹𝒙𝒙

If we deliberately make 𝛿𝛿𝑥𝑥 approach zero (i.e. make it as short as possible) then we shall reach a limit, which may expressed mathematically as…

Sample

http://www.derivative-calculator.net/

http://www.integral-calculator.com/


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Worked Example 1

𝒄𝒄𝒚𝒚𝒄𝒄𝒙𝒙

= 𝜹𝜹𝒙𝒙𝒔𝒔𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍�⎯⎯� 𝟎𝟎

𝜹𝜹𝒚𝒚𝜹𝜹𝒙𝒙

The term 𝑑𝑑𝑦𝑦 𝑑𝑑𝑥𝑥⁄ is written in Leibnitz notation and indicates the slope of the chord when the chord only touches the function at one single point. This is achieved by continual reduction of 𝛿𝛿𝑥𝑥.

What we can now say is that we are able to find the slope of any function by adopting this process. Hence, given a function 𝑓𝑓(𝑥𝑥) we are able to differentiate that function, meaning find its slope at all points. We can write…

The slope at any point of a function 𝑓𝑓(𝑥𝑥) is given by 𝒄𝒄(𝒐𝒐(𝒙𝒙))𝒄𝒄𝒙𝒙

This process is called finding the derivative of a function.

3.1.2 Derivatives of Standard Functions What we don’t want to be doing is to spend too much time drawing graphs of functions just to work out the derivative. Fortunately there are standard ways to determine the derivative of functions and some of the frequent ones which engineers meet are given in the table below.

Function Derivative 𝐴𝐴𝑥𝑥𝑛𝑛 𝑛𝑛𝐴𝐴𝑥𝑥𝑛𝑛−1

A sin (𝑥𝑥) A cos (𝑥𝑥)

A cos (𝑥𝑥) −𝐴𝐴 sin (𝑥𝑥)

𝐴𝐴 𝑒𝑒𝑘𝑘𝑘𝑘 𝑘𝑘𝐴𝐴𝑒𝑒𝑘𝑘𝑘𝑘

𝐴𝐴 𝑙𝑙𝑙𝑙𝑙𝑙𝑒𝑒(𝑥𝑥) 𝐴𝐴 𝑥𝑥�

A sinh (𝑥𝑥) A cosh (𝑥𝑥)

A cosh (𝑥𝑥) A sinh (𝑥𝑥)

Let’s look at some examples of using these standard derivatives…

Differentiate the function 𝒚𝒚 = 𝟑𝟑𝒙𝒙𝟒𝟒 with respect to 𝒙𝒙.

We see that this function is similar to that in row 1 of our table. We can see that 𝐴𝐴 = 3 and 𝑛𝑛 = 4. We may then write…

Sample


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Worked Example 2

Worked Example 3

Worked Example 4

𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

= 𝑛𝑛𝐴𝐴𝑥𝑥𝑛𝑛−1 = (4)(3)𝑥𝑥4−1 = 12𝑥𝑥3

Differentiate the function 𝒚𝒚 = 𝟔𝟔𝒙𝒙−𝟓𝟓 with respect to 𝒙𝒙.

We see that this function is similar to that in row 1 of our table. We can see that 𝐴𝐴 = 6 and 𝑛𝑛 = −5. We may then write…


= 𝑛𝑛𝐴𝐴𝑥𝑥𝑛𝑛−1 = (−5)(6)𝑥𝑥−5−1 = −30𝑥𝑥−6

Differentiate the function 𝒗𝒗 = 𝟏𝟏𝟏𝟏 𝒔𝒔𝒍𝒍𝒔𝒔(𝒍𝒍) with respect to 𝒍𝒍.

Don’t worry that 𝑦𝑦 and 𝑥𝑥 have disappeared here. We can use any letters we like. The letter 𝑣𝑣 means voltage and 𝑡𝑡 is time. All we want to do here then is to find 𝑑𝑑𝑣𝑣 𝑑𝑑𝑡𝑡� .

We see that the function is similar to that in row 2 of our table. We may write…

𝑑𝑑𝑣𝑣𝑑𝑑𝑡𝑡

= 𝐴𝐴 cos(𝑡𝑡) = 12 cos (𝑡𝑡)

Differentiate the function 𝒍𝒍 = 𝟓𝟓 𝒄𝒄𝒔𝒔𝒔𝒔(𝒍𝒍) with respect to 𝒍𝒍.

The letter 𝑖𝑖 means current and 𝑡𝑡 is time. All we want to do here then is to find 𝑑𝑑𝑖𝑖 𝑑𝑑𝑡𝑡� .

We see that the function is similar to that in row 3 of our table. We may write…

𝑑𝑑𝑖𝑖𝑑𝑑𝑡𝑡

= −𝐴𝐴 sin(𝑡𝑡) = −5 sin (𝑡𝑡)

Sample


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Worked Example 5

Worked Example 6

Worked Example 7

Worked Example 8

Questions

Differentiate the function 𝒚𝒚 = 𝟒𝟒𝒔𝒔𝟏𝟏𝒙𝒙 with respect to 𝒙𝒙.

From row 4 of our table we can write…


= 𝑘𝑘𝐴𝐴𝑒𝑒𝑘𝑘𝑘𝑘 = (2)(4)𝑒𝑒2𝑘𝑘 = 8𝑒𝑒2𝑘𝑘

Differentiate the function 𝒚𝒚 = 𝟏𝟏𝟔𝟔 𝒔𝒔𝒔𝒔𝒍𝒍𝒔𝒔(𝒙𝒙) with respect to 𝒙𝒙.

We refer to row 5 of our table and write…


=𝐴𝐴𝑥𝑥

=16𝑥𝑥

Differentiate the function 𝒚𝒚 = 𝟗𝟗 𝒔𝒔𝒍𝒍𝒔𝒔𝒄𝒄(𝒙𝒙) with respect to 𝒙𝒙.



= 𝐴𝐴 cosh(𝑥𝑥) = 9 cosh (𝑥𝑥)

Differentiate the function 𝒚𝒚 = −𝟏𝟏𝟒𝟒.𝟓𝟓 𝒄𝒄𝒔𝒔𝒔𝒔𝒄𝒄(𝒙𝒙) with respect to 𝒙𝒙.



= 𝐴𝐴 sinh(𝑥𝑥) = −14.5 sinh (𝑥𝑥)

Q3.1 Differentiate the function 𝒚𝒚 = 𝟖𝟖𝒙𝒙𝟓𝟓 with respect to 𝒙𝒙.

Sample


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Video

Q3.2 Differentiate the function 𝒚𝒚 = 𝟕𝟕𝒙𝒙−𝟒𝟒 with respect to 𝒙𝒙.

Q3.3 Differentiate the function 𝒗𝒗 = 𝟏𝟏𝟖𝟖 𝒔𝒔𝒍𝒍𝒔𝒔(𝒍𝒍) with respect to 𝒍𝒍.

Q3.4 Differentiate the function 𝒍𝒍 = 𝟏𝟏𝟏𝟏 𝒄𝒄𝒔𝒔𝒔𝒔(𝒍𝒍) with respect to 𝒍𝒍.

Q3.5 Differentiate the function 𝒚𝒚 = 𝟔𝟔𝒔𝒔𝟑𝟑𝒙𝒙 with respect to 𝒙𝒙.

Q3.6 Differentiate the function 𝒚𝒚 = 𝟏𝟏𝟎𝟎 𝒔𝒔𝒔𝒔𝒍𝒍𝒔𝒔(𝒙𝒙) with respect to 𝒙𝒙.

Q3.7 Differentiate the function 𝒚𝒚 = 𝟏𝟏𝟑𝟑 𝒔𝒔𝒍𝒍𝒔𝒔𝒄𝒄(𝒙𝒙) with respect to 𝒙𝒙.

Q3.8 Differentiate the function 𝒚𝒚 = −𝟏𝟏𝟗𝟗.𝟔𝟔𝟏𝟏 𝒄𝒄𝒔𝒔𝒔𝒔𝒄𝒄(𝒙𝒙) with respect to 𝒙𝒙.

ANSWERS

These videos will boost your knowledge of differentiation

3.1.3 Notion of the Derivative and Rates of Change The notion of taking the derivative of a function and examining rate of change can be readily explained with reference to the diagram below…

Q3.1 40𝑥𝑥4

Q3.2 −28𝑥𝑥−5

Q3.3 18 cos(𝑡𝑡)

Q3.4 −11 sin(𝑡𝑡)

Q3.5 18𝑒𝑒3𝑘𝑘

Q3.6 20 𝑥𝑥⁄

Q3.7 13 cosh(𝑥𝑥)

Q3.8 −19.62 sinh(𝑥𝑥) Sample

https://www.khanacademy.org/math/differential-calculus/taking-derivatives/derivative_intro/v/calculus-derivatives-1





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Challenge

Here we have plotted a section of a sine wave in red. Notice that we have used pink arrows to indicate the slope at various points on the sine wave. For example, at the start the sine wave is rising to a positive more rapidly than at any other point. The pink arrow indicates this positive growth by pointing upwards in the direction of the sine wave at that beginning point. The second arrow shows the slope of the sine wave at 𝜋𝜋 2⁄ radians (or 900 if you like). Here it is at a plateau and has zero slope (meaning that it just moves horizontally, with no vertical movement at all). The third arrow indicates a negative slope (its moving down very steeply). The picture continues in a similar way for the remaining arrows.

Here’s the really interesting bit. When we record all of the slopes on our sine wave we get the dotted waveform in blue. What does that look like? Of course, it is a cosine wave. So we have just proved graphically that finding the slope (differentiating) at all points on a sine wave leads to a cosine wave.

∴ 𝒄𝒄{𝐬𝐬𝐬𝐬𝐬𝐬 (𝒍𝒍)}

𝒄𝒄𝒍𝒍= 𝐜𝐜𝐜𝐜𝐬𝐬 (𝒍𝒍)

… as given in our table of standard derivatives in section 3.1.2.

If we had loads of time to kill we could prove all of the other derivatives graphically.

Determine a graphical proof (as above) for…

𝒄𝒄{𝐜𝐜𝐜𝐜𝐬𝐬 (𝒍𝒍)}𝒄𝒄𝒍𝒍

= −𝐬𝐬𝐬𝐬𝐬𝐬 (𝒍𝒍)

Sample


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3.2.3 Differentiation of Inverse Trigonometric Functions Suppose we have the equation…

𝑦𝑦 = 𝐴𝐴𝑖𝑖𝑛𝑛−1(𝑥𝑥)

How do we then go about finding 𝑑𝑑𝑦𝑦 𝑑𝑑𝑥𝑥⁄ ? Maybe we should write this in a different form…

𝑥𝑥 = sin(𝑦𝑦)

Now can differentiate…

𝑑𝑑𝑥𝑥𝑑𝑑𝑦𝑦

= cos(𝑦𝑦) ∴ 𝑓𝑓𝑙𝑙𝑖𝑖𝑓𝑓 𝑠𝑠𝑙𝑙𝑡𝑡ℎ 𝐴𝐴𝑖𝑖𝑑𝑑𝑒𝑒𝐴𝐴: 𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

=1

cos (𝑦𝑦)

Next we shall remember one of our handy trig. Identities…

𝑐𝑐𝑙𝑙𝐴𝐴2(𝑦𝑦) + 𝐴𝐴𝑖𝑖𝑛𝑛2(𝑦𝑦) = 1 ∴ 𝑐𝑐𝑙𝑙𝐴𝐴2(𝑦𝑦) = 1 − 𝐴𝐴𝑖𝑖𝑛𝑛2(𝑦𝑦) ∴ cos(𝑦𝑦) = �1 − 𝐴𝐴𝑖𝑖𝑛𝑛2(𝑦𝑦)

Since we know 𝑥𝑥 = sin(𝑦𝑦) then if we square both sides we shall have 𝑥𝑥2 = 𝐴𝐴𝑖𝑖𝑛𝑛2(𝑦𝑦)

So we may re-write our transposed trig. identity as…

cos(𝑦𝑦) = �1 − 𝑥𝑥2

Now we can put this latest result back into our differential…

𝒄𝒄𝒚𝒚𝒄𝒄𝒙𝒙

=𝟏𝟏

𝐜𝐜𝐜𝐜𝐬𝐬 (𝒚𝒚)=

𝟏𝟏√𝟏𝟏 − 𝒙𝒙𝟏𝟏

=𝟏𝟏

(𝟏𝟏 − 𝒙𝒙𝟏𝟏)𝟎𝟎.𝟓𝟓 = (𝟏𝟏 − 𝒙𝒙𝟏𝟏)−𝟎𝟎.𝟓𝟓

That was a bit tricky, but we managed to find 𝑦𝑦′.

Let’s go on and see if we can find 𝑦𝑦′′…

𝑦𝑦′ = (1 − 𝑥𝑥2)−0.5

This is where that shorthand notation for derivatives comes in really handy…

𝐿𝐿𝑒𝑒𝑡𝑡 𝑢𝑢 = 1 − 𝑥𝑥2 ∴ 𝑑𝑑𝑢𝑢𝑑𝑑𝑥𝑥

= −2𝑥𝑥

∴ 𝑦𝑦′ = 𝑢𝑢−0.5

∴ 𝑑𝑑𝑦𝑦′

𝑑𝑑𝑢𝑢= −0.5𝑢𝑢−1.5

∴ 𝑦𝑦′′ =𝑑𝑑𝑢𝑢𝑑𝑑𝑥𝑥

∙ 𝑑𝑑𝑦𝑦′

𝑑𝑑𝑢𝑢= −2𝑥𝑥 × −0.5𝑢𝑢−1.5

∴ 𝒚𝒚′′ = 𝒙𝒙𝒅𝒅−𝟏𝟏.𝟓𝟓 = 𝒙𝒙(𝟏𝟏 − 𝒙𝒙𝟏𝟏)−𝟏𝟏.𝟓𝟓

Sample


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Question

Question

Q3.23 Find 𝒚𝒚′′ for 𝒚𝒚 = 𝒄𝒄𝒔𝒔𝒔𝒔−𝟏𝟏(𝟏𝟏𝒙𝒙)

ANSWER

3.2.4 Differentiation of Inverse Hyperbolic Functions Consider the function…

𝑦𝑦 = 𝐴𝐴𝑖𝑖𝑛𝑛ℎ−1(𝑥𝑥)

Our task here is to find 𝑦𝑦′′. We proceed in a similar manner to the previous problem…

𝑥𝑥 = sinh (𝑦𝑦)

∴ 𝑑𝑑𝑥𝑥𝑑𝑑𝑦𝑦

= cosh(𝑦𝑦) ∴ 𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

=1

cosh (𝑦𝑦)

A useful identity is…

𝑐𝑐𝑙𝑙𝐴𝐴ℎ2(𝑦𝑦)− 𝐴𝐴𝑖𝑖𝑛𝑛ℎ2(𝑦𝑦) = 1 ∴ 𝑐𝑐𝑙𝑙𝐴𝐴ℎ2(𝑦𝑦) = 1 + 𝐴𝐴𝑖𝑖𝑛𝑛ℎ2(𝑦𝑦)

∴ 𝑐𝑐𝑙𝑙𝐴𝐴ℎ2(𝑦𝑦) = 1 + 𝑥𝑥2 ∴ cosh(𝑦𝑦) = �1 + 𝑥𝑥2

∴ 𝒄𝒄𝒚𝒚𝒄𝒄𝒙𝒙

= 𝒚𝒚′ =𝟏𝟏

√𝟏𝟏 + 𝒙𝒙𝟏𝟏= (𝟏𝟏 + 𝒙𝒙𝟏𝟏)−𝟎𝟎.𝟓𝟓

Now to find that second derivative…

𝑦𝑦′ = (1 + 𝑥𝑥2)−0.5

𝐿𝐿𝑒𝑒𝑡𝑡 𝑢𝑢 = 1 + 𝑥𝑥2 ∴ 𝑑𝑑𝑢𝑢𝑑𝑑𝑥𝑥

= 2𝑥𝑥

∴ 𝑦𝑦′ = 𝑢𝑢−0.5 ∴ 𝑑𝑑𝑦𝑦′

𝑑𝑑𝑢𝑢= −0.5𝑢𝑢−1.5

∴ 𝑦𝑦′′ =𝑑𝑑𝑢𝑢𝑑𝑑𝑥𝑥

∙ 𝑑𝑑𝑦𝑦′

𝑑𝑑𝑢𝑢= 2𝑥𝑥 × −0.5𝑢𝑢−1.5 = −𝑥𝑥(1 + 𝑥𝑥2)−1.5

Q3.24 Find 𝒚𝒚′′ for 𝒚𝒚 = 𝒄𝒄𝒔𝒔𝒔𝒔𝒄𝒄−𝟏𝟏(𝟏𝟏𝒙𝒙)

ANSWER

Q3.23 −8𝑘𝑘(1−4𝑘𝑘2)1.5

Sample


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Worked Example 22

3.3 Further Integration

3.3.1 Integration by Parts This technique provides us with a way to integrate the product of two simple functions. We shall develop a formula to use as a framework for the technique.

The starting point for our formula lies with the Product Rule, which you mastered in Section 3.1.5…

𝑑𝑑𝑑𝑑𝑥𝑥

(𝑢𝑢𝑣𝑣) = 𝑢𝑢𝑑𝑑𝑣𝑣𝑑𝑑𝑥𝑥

+ 𝑣𝑣𝑑𝑑𝑢𝑢𝑑𝑑𝑥𝑥

Let’s rearrange this formula…

𝑢𝑢𝑑𝑑𝑣𝑣𝑑𝑑𝑥𝑥

=𝑑𝑑𝑑𝑑𝑥𝑥

(𝑢𝑢𝑣𝑣) − 𝑣𝑣𝑑𝑑𝑢𝑢𝑑𝑑𝑥𝑥

If we now integrate both sides with respect to 𝑥𝑥 then we need to place a ∫ sign before each term and a 𝑑𝑑𝑥𝑥 at the end of each term, like so…

�𝑢𝑢𝑑𝑑𝑣𝑣𝑑𝑑𝑥𝑥

𝑑𝑑𝑥𝑥 = �𝑑𝑑𝑑𝑑𝑥𝑥

(𝑢𝑢𝑣𝑣)𝑑𝑑𝑥𝑥 −�𝑣𝑣𝑑𝑑𝑢𝑢𝑑𝑑𝑥𝑥

𝑑𝑑𝑥𝑥

Some simplifications can be noticed here:

In the first integral the two 𝑑𝑑𝑥𝑥 terms cancel each other out In the second integral we are taking the ‘integral of the differential’ of 𝑢𝑢𝑣𝑣. Since integration is the

reverse of differentiation then the ’integral of a differential’ drops away, just leaving 𝑢𝑢𝑣𝑣 In the third integral the two 𝑑𝑑𝑥𝑥 terms cancel each other out

Performing these simplifications gives…

�𝒅𝒅 𝒄𝒄𝒗𝒗 = 𝒅𝒅𝒗𝒗 −�𝒗𝒗 𝒄𝒄𝒅𝒅

That is our formula for integration by parts.

What we do with the formula is to look at the left hand side and see the integral of a product. That product is composed of 𝑢𝑢 and 𝑑𝑑𝑣𝑣. When using integration by parts you have a choice of which part of the product to call 𝑢𝑢 and which to call 𝑑𝑑𝑣𝑣. Fortunately, your choice is guided…

The 𝑢𝑢 part will become a constant after taking multiple derivatives The 𝑑𝑑𝑣𝑣 part is readily integrated by using standard integrals.

Sample


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Worked Example 23

Determine: ∫𝒙𝒙 𝐬𝐬𝐬𝐬𝐬𝐬(𝒙𝒙)𝒄𝒄𝒙𝒙

One part of our product is 𝑥𝑥 and the other part is sin (𝑥𝑥). We need to make a choice as to which one to make 𝑢𝑢 and which to make 𝑑𝑑𝑣𝑣. Our guidance tells us to make the 𝑥𝑥 part equal to 𝑢𝑢 since the derivative of 𝑥𝑥 becomes 1 (i.e. a constant)…

𝐿𝐿𝑒𝑒𝑡𝑡 𝑢𝑢 = 𝑥𝑥

𝐿𝐿𝑒𝑒𝑡𝑡 𝑑𝑑𝑣𝑣 = sin (𝑥𝑥)

We now try to form the terms in our integration by parts formula…

𝐼𝐼𝑓𝑓 𝑢𝑢 = 𝑥𝑥 ∴ 𝑑𝑑𝑢𝑢𝑑𝑑𝑥𝑥

= 1 ∴ 𝑑𝑑𝑢𝑢 = 𝑑𝑑𝑥𝑥

𝐼𝐼𝑓𝑓 𝑑𝑑𝑣𝑣 = sin(𝑥𝑥) ∴ 𝑣𝑣 = − cos (𝑥𝑥)

We can now write…

�𝒅𝒅 𝒄𝒄𝒗𝒗 = 𝒅𝒅𝒗𝒗 −�𝒗𝒗 𝒄𝒄𝒅𝒅

�𝑥𝑥 sin(𝑥𝑥)𝑑𝑑𝑥𝑥 = 𝑥𝑥(− cos(𝑥𝑥)) −�(− cos(𝑥𝑥))𝑑𝑑𝑥𝑥

∴ �𝑥𝑥 sin(𝑥𝑥)𝑑𝑑𝑥𝑥 = −𝑥𝑥 cos(𝑥𝑥) + � cos(𝑥𝑥)𝑑𝑑𝑥𝑥

∴ �𝒙𝒙𝐬𝐬𝐬𝐬𝐬𝐬(𝒙𝒙)𝒄𝒄𝒙𝒙 = −𝒙𝒙𝐜𝐜𝐜𝐜𝐬𝐬(𝒙𝒙) + 𝐬𝐬𝐬𝐬𝐬𝐬(𝒙𝒙) + 𝑪𝑪

That was fairly straightforward. Always remember the constant of integration 𝐶𝐶 at the end.

Let’s now try to integrate the product of an exponential and a cosine.

Determine: ∫𝒔𝒔𝟏𝟏𝒙𝒙 𝐜𝐜𝐜𝐜𝐬𝐬(𝒙𝒙)𝒄𝒄𝒙𝒙

Our two products here are 𝑒𝑒2𝑘𝑘 and cos (𝑥𝑥). Neither of them will reduce to a constant when differentiated so it doesn’t matter which one we call 𝑢𝑢 and which we call 𝑑𝑑𝑣𝑣. You will see here that we need to do integration by parts twice to arrive at our answer.

Just as in the previous worked example we need to decide on assignments for 𝑢𝑢 and 𝑑𝑑𝑣𝑣, then find 𝑣𝑣 and 𝑑𝑑𝑢𝑢. Let’s do that…

𝐿𝐿𝑒𝑒𝑡𝑡 𝑢𝑢 = 𝑒𝑒2𝑘𝑘 ∴ 𝑑𝑑𝑢𝑢𝑑𝑑𝑥𝑥

= 2𝑒𝑒2𝑘𝑘 ∴ 𝑑𝑑𝑢𝑢 = 2𝑒𝑒2𝑘𝑘𝑑𝑑𝑥𝑥

Sample


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Question

𝐿𝐿𝑒𝑒𝑡𝑡 𝑑𝑑𝑣𝑣 = cos(𝑥𝑥) ∴ 𝑣𝑣 = sin (𝑥𝑥)

To aid our working here (and save some ink) we normally assign the capital letter 𝐼𝐼 to the integral in our question.

∴ 𝐼𝐼 = �𝑒𝑒2𝑘𝑘 cos(𝑥𝑥)𝑑𝑑𝑥𝑥

∴ 𝐼𝐼 = 𝑒𝑒2𝑘𝑘 sin(𝑥𝑥) −� sin(𝑥𝑥) 2𝑒𝑒2𝑘𝑘𝑑𝑑𝑥𝑥

This needs tidying…

∴ 𝐼𝐼 = 𝑒𝑒2𝑘𝑘 sin(𝑥𝑥) − 2�𝑒𝑒2𝑘𝑘sin(𝑥𝑥)𝑑𝑑𝑥𝑥

We were rather hoping for an answer, but what we seem to have is another ‘integration by parts’ problem embedded into our development. Don’t worry, that’s quite normal for this type of problem. All we do is attack that last part with ‘integration by parts’ once more, using square brackets. Our desired result will then appear…

𝐼𝐼 = 𝑒𝑒2𝑘𝑘 sin(𝑥𝑥) − 2 �𝑒𝑒2𝑘𝑘(−cos (𝑥𝑥)) −�−cos (𝑥𝑥)2𝑒𝑒2𝑘𝑘𝑑𝑑𝑥𝑥�

Expanding these terms gives…

𝐼𝐼 = 𝑒𝑒2𝑘𝑘 sin(𝑥𝑥) + 2𝑒𝑒2𝑘𝑘 cos(𝑥𝑥) − 4�𝑒𝑒2𝑘𝑘 cos(𝑥𝑥)𝑑𝑑𝑥𝑥

That last integral is the same as 𝐼𝐼, which is really handy, so we may now write…

𝐼𝐼 = 𝑒𝑒2𝑘𝑘 sin(𝑥𝑥) + 2𝑒𝑒2𝑘𝑘 cos(𝑥𝑥) − 4𝐼𝐼

∴ 5𝐼𝐼 = 𝑒𝑒2𝑘𝑘 sin(𝑥𝑥) + 2𝑒𝑒2𝑘𝑘cos (𝑥𝑥)

If we divide both sides by 5 then we have our final answer…

𝑰𝑰 =𝒔𝒔𝟏𝟏𝒙𝒙 𝐬𝐬𝐬𝐬𝐬𝐬(𝒙𝒙) + 𝟏𝟏𝒔𝒔𝟏𝟏𝒙𝒙𝐜𝐜𝐜𝐜𝐬𝐬 (𝒙𝒙)

𝟓𝟓

Q3.25 Determine: ∫𝒔𝒔𝟒𝟒𝒙𝒙𝒔𝒔𝒍𝒍𝒔𝒔(𝟏𝟏𝒙𝒙)𝒄𝒄𝒙𝒙

ANSWER

Q3.25 4𝑒𝑒4𝑥𝑥 sin(2𝑘𝑘)−2𝑒𝑒4𝑥𝑥cos(2𝑘𝑘)

20

Sample


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3.4 Solution of Engineering Problems with Calculus

3.4.1 Charging RC Circuit For the circuit below, use Calculus to find the charge stored in the capacitor 2 seconds after the switch closes. Assume zero charge on the capacitor before this event.

For a capacitor charging via a DC source voltage, through a series resistor, the formula for instantaneous current is…

𝑖𝑖 =𝐸𝐸𝑅𝑅∙ 𝑒𝑒�−𝑡𝑡 𝑅𝑅𝑅𝑅� � 𝐴𝐴𝑠𝑠𝑓𝑓𝐴𝐴

Since current is the defined as the rate of change of charge (𝑠𝑠, carried by passing electrons) we may say…

𝑖𝑖 =𝑑𝑑𝑠𝑠𝑑𝑑𝑡𝑡

If we integrate both sides with respect to 𝑡𝑡 we will have…

� 𝑖𝑖 𝑑𝑑𝑡𝑡 = �𝑑𝑑𝑠𝑠𝑑𝑑𝑡𝑡

𝑑𝑑𝑡𝑡 = �𝑑𝑑𝑠𝑠 = 𝑠𝑠

So we may say that…

𝑠𝑠 = � 𝑖𝑖 𝑑𝑑𝑡𝑡

If we wish to find the amount of charge store stored within a certain time period (i.e. 2 seconds in this example) we write…

𝑠𝑠(0−2) = � 𝑖𝑖 𝑑𝑑𝑡𝑡2

0

= �𝐸𝐸𝑅𝑅∙ 𝑒𝑒�−𝑡𝑡 𝑅𝑅𝑅𝑅� �

2

0

𝑑𝑑𝑡𝑡

=𝐸𝐸𝑅𝑅�𝑒𝑒�−1𝑅𝑅𝑅𝑅�𝑡𝑡

−1𝑅𝑅𝐶𝐶�

�

0

2

=−𝐸𝐸𝑅𝑅𝐶𝐶𝑅𝑅

�𝑒𝑒�−𝑡𝑡 𝑅𝑅𝑅𝑅� ��0

2= −𝐸𝐸𝐶𝐶 �𝑒𝑒�−𝑡𝑡 𝑅𝑅𝑅𝑅� ��

0

2

Since we know that 𝑅𝑅 = 1𝑀𝑀Ω and 𝐶𝐶 = 1𝜇𝜇𝐹𝐹 then 𝑅𝑅𝐶𝐶 = 1 × 106 × 1 × 10−6 = 106−6 = 100 = 1 we may write…

Sample


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𝑠𝑠(0−2) = −𝐸𝐸𝐶𝐶 �𝑒𝑒�−𝑡𝑡 1� ��0

2= −𝐸𝐸𝐶𝐶[𝑒𝑒−𝑡𝑡]02 = −𝐸𝐸𝐶𝐶[𝑒𝑒−2 − 𝑒𝑒−0] = −𝐸𝐸𝐶𝐶[0.135 − 1]

We also know that the DC supply voltage (𝐸𝐸) is 10V…

∴ 𝒒𝒒(𝟎𝟎−𝟏𝟏) = −𝟏𝟏𝟎𝟎 × 𝟏𝟏 × 𝟏𝟏𝟎𝟎−𝟔𝟔[−𝟎𝟎.𝟖𝟖𝟔𝟔𝟓𝟓] = 𝟖𝟖.𝟔𝟔𝟓𝟓 × 𝟏𝟏𝟎𝟎−𝟔𝟔 𝑪𝑪𝒔𝒔𝒅𝒅𝒔𝒔𝒔𝒔𝒍𝒍𝑪𝑪𝒔𝒔 = 𝟖𝟖.𝟔𝟔𝟓𝟓 𝝁𝝁𝑪𝑪

The plot below illustrates the integration we have just performed…

Current is on the vertical axis and time on the horizontal. When we multiply current by time we get charge, which is shown as the shaded area between 0 and 2 seconds. You may like to experiment with the Graph simulator to produce similar results.

3.4.2 Energising Inductor Consider the series RL circuit below…

If 𝑬𝑬 = 𝟏𝟏𝟎𝟎 𝑽𝑽, 𝑹𝑹 = 𝟏𝟏𝟏𝟏𝛀𝛀 and 𝑳𝑳 = 𝟏𝟏𝟎𝟎𝟎𝟎𝒍𝒍𝟐𝟐, use Calculus to determine the voltage across the inductor after 𝟎𝟎.𝟏𝟏𝝁𝝁𝒔𝒔. Use the following formulae in your solution development…

𝒍𝒍 =𝑬𝑬𝑹𝑹�𝟏𝟏 − 𝒔𝒔−𝑹𝑹𝒍𝒍 𝑳𝑳� � 𝒂𝒂𝒔𝒔𝒄𝒄 𝒗𝒗𝑳𝑳 = 𝑳𝑳

𝒄𝒄𝒍𝒍𝒄𝒄𝒍𝒍

Sample


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We start with…

𝑣𝑣𝐿𝐿 = 𝐿𝐿𝑑𝑑𝑖𝑖𝑑𝑑𝑡𝑡

= 𝐿𝐿𝑑𝑑 �𝐸𝐸𝑅𝑅 �1 − 𝑒𝑒−𝑅𝑅𝑡𝑡 𝐿𝐿� ��

𝑑𝑑𝑡𝑡

=𝐿𝐿𝐸𝐸𝑅𝑅∙𝑑𝑑 �1 − 𝑒𝑒−𝑅𝑅𝑡𝑡 𝐿𝐿� �

𝑑𝑑𝑡𝑡=𝐿𝐿𝐸𝐸𝑅𝑅�𝑅𝑅𝐿𝐿∙ 𝑒𝑒−𝑅𝑅𝑡𝑡 𝐿𝐿� � = 𝐸𝐸𝑒𝑒−𝑅𝑅𝑡𝑡 𝐿𝐿�

Putting in the values for E, R, t and L gives…

𝒗𝒗𝑳𝑳 = 𝟏𝟏𝟎𝟎𝒔𝒔�−𝟏𝟏𝟎𝟎

𝟔𝟔×𝟎𝟎.𝟏𝟏×𝟏𝟏𝟎𝟎−𝟔𝟔𝟎𝟎.𝟏𝟏 �

= 𝟔𝟔.𝟎𝟎𝟔𝟔𝟓𝟓 𝒗𝒗𝒔𝒔𝒔𝒔𝒍𝒍𝒔𝒔

Sample

unit workbook 4 calculus - wb 4.pdflet’s go on and see if we can find 𝑦𝑦 ′′ … 𝑦𝑦...

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