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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
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3.1 Calculus There are a couple of great webpages for checking your answers to Calculus problemsβ¦
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. Letβs refer to a diagram to understand the concept of a limitβ¦
Here we have indicated a point P on part of a function (the red curve) with Cartesian co-ordinates (π₯π₯1,π¦π¦1). 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 πΏπΏ (delta).
What we then have isβ¦
π·π· = (ππππ,ππππ)
πΈπΈ = (ππππ + πΉπΉππ, ππππ + πΉπΉππ)
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 byβ¦
ππππππππππ ππππ ππππππππππ π·π·πΈπΈ =πΉπΉπππΉπΉππ
If we deliberately make πΏπΏπ₯π₯ approach zero (i.e. make it as short as possible) then we shall reach a limit, which may expressed mathematically asβ¦
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Worked Example 1
ππππππππ
= πΉπΉπππππππππππποΏ½β―β―οΏ½ ππ
πΉπΉπππΉπΉππ
The term πππ¦π¦ πππ₯π₯β 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 πΏπΏπ₯π₯.
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 ππ(π₯π₯) we are able to differentiate that function, meaning find its slope at all points. We can writeβ¦
The slope at any point of a function ππ(π₯π₯) is given by ππ(ππ(ππ))ππππ
This process is called finding the derivative of a function.
3.1.2 Derivatives of Standard Functions What we donβt 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 π΄π΄π₯π₯ππ πππ΄π΄π₯π₯ππβ1
A sin (π₯π₯) A cos (π₯π₯)
A cos (π₯π₯) βπ΄π΄ sin (π₯π₯)
π΄π΄ ππππππ πππ΄π΄ππππππ
π΄π΄ ππππππππ(π₯π₯) π΄π΄ π₯π₯οΏ½
A sinh (π₯π₯) A cosh (π₯π₯)
A cosh (π₯π₯) A sinh (π₯π₯)
Letβs look at some examples of using these standard derivativesβ¦
Differentiate the function ππ = ππππππ with respect to ππ.
We see that this function is similar to that in row 1 of our table. We can see that π΄π΄ = 3 and ππ = 4. We may then writeβ¦
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Worked Example 2
Worked Example 3
Worked Example 4
πππ¦π¦πππ₯π₯
= πππ΄π΄π₯π₯ππβ1 = (4)(3)π₯π₯4β1 = 12π₯π₯3
Differentiate the function ππ = ππππβππ with respect to ππ.
We see that this function is similar to that in row 1 of our table. We can see that π΄π΄ = 6 and ππ = β5. We may then writeβ¦
πππ¦π¦πππ₯π₯
= πππ΄π΄π₯π₯ππβ1 = (β5)(6)π₯π₯β5β1 = β30π₯π₯β6
Differentiate the function ππ = ππππ ππππππ(ππ) with respect to ππ.
Donβt worry that π¦π¦ and π₯π₯ have disappeared here. We can use any letters we like. The letter π£π£ means voltage and π‘π‘ is time. All we want to do here then is to find πππ£π£ πππ‘π‘οΏ½ .
We see that the function is similar to that in row 2 of our table. We may writeβ¦
πππ£π£πππ‘π‘
= π΄π΄ cos(π‘π‘) = 12 cos (π‘π‘)
Differentiate the function ππ = ππ ππππππ(ππ) with respect to ππ.
The letter ππ means current and π‘π‘ is time. All we want to do here then is to find ππππ πππ‘π‘οΏ½ .
We see that the function is similar to that in row 3 of our table. We may writeβ¦
πππππππ‘π‘
= βπ΄π΄ sin(π‘π‘) = β5 sin (π‘π‘)
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Worked Example 5
Worked Example 6
Worked Example 7
Worked Example 8
Questions
Differentiate the function ππ = ππππππππ with respect to ππ.
From row 4 of our table we can writeβ¦
πππ¦π¦πππ₯π₯
= πππ΄π΄ππππππ = (2)(4)ππ2ππ = 8ππ2ππ
Differentiate the function ππ = ππππ ππππππππ(ππ) with respect to ππ.
We refer to row 5 of our table and writeβ¦
πππ¦π¦πππ₯π₯
=π΄π΄π₯π₯
=16π₯π₯
Differentiate the function ππ = ππ ππππππππ(ππ) with respect to ππ.
We refer to row 6 of our table and writeβ¦
πππ¦π¦πππ₯π₯
= π΄π΄ cosh(π₯π₯) = 9 cosh (π₯π₯)
Differentiate the function ππ = βππππ.ππ ππππππππ(ππ) with respect to ππ.
We refer to row 7 of our table and writeβ¦
πππ¦π¦πππ₯π₯
= π΄π΄ sinh(π₯π₯) = β14.5 sinh (π₯π₯)
Q3.1 Differentiate the function ππ = ππππππ with respect to ππ.
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Video
Q3.2 Differentiate the function ππ = ππππβππ with respect to ππ.
Q3.3 Differentiate the function ππ = ππππ ππππππ(ππ) with respect to ππ.
Q3.4 Differentiate the function ππ = ππππ ππππππ(ππ) with respect to ππ.
Q3.5 Differentiate the function ππ = ππππππππ with respect to ππ.
Q3.6 Differentiate the function ππ = ππππ ππππππππ(ππ) with respect to ππ.
Q3.7 Differentiate the function ππ = ππππ ππππππππ(ππ) with respect to ππ.
Q3.8 Differentiate the function ππ = βππππ.ππππ ππππππππ(ππ) with respect to ππ.
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 belowβ¦
Q3.1 40π₯π₯4
Q3.2 β28π₯π₯β5
Q3.3 18 cos(π‘π‘)
Q3.4 β11 sin(π‘π‘)
Q3.5 18ππ3ππ
Q3.6 20 π₯π₯β
Q3.7 13 cosh(π₯π₯)
Q3.8 β19.62 sinh(π₯π₯) Sample
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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 2β 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.
Hereβs 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.
β΄ ππ{π¬π¬π¬π¬π¬π¬ (ππ)}
ππππ= πππππ¬π¬ (ππ)
β¦ 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) forβ¦
ππ{πππππ¬π¬ (ππ)}ππππ
= βπ¬π¬π¬π¬π¬π¬ (ππ)
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3.2.3 Differentiation of Inverse Trigonometric Functions Suppose we have the equationβ¦
π¦π¦ = π΄π΄ππππβ1(π₯π₯)
How do we then go about finding πππ¦π¦ πππ₯π₯β ? Maybe we should write this in a different formβ¦
π₯π₯ = sin(π¦π¦)
Now can differentiateβ¦
πππ₯π₯πππ¦π¦
= cos(π¦π¦) β΄ ππππππππ π π πππ‘π‘β π΄π΄πππππππ΄π΄: πππ¦π¦πππ₯π₯
=1
cos (π¦π¦)
Next we shall remember one of our handy trig. Identitiesβ¦
πππππ΄π΄2(π¦π¦) + π΄π΄ππππ2(π¦π¦) = 1 β΄ πππππ΄π΄2(π¦π¦) = 1 β π΄π΄ππππ2(π¦π¦) β΄ cos(π¦π¦) = οΏ½1 β π΄π΄ππππ2(π¦π¦)
Since we know π₯π₯ = sin(π¦π¦) then if we square both sides we shall have π₯π₯2 = π΄π΄ππππ2(π¦π¦)
So we may re-write our transposed trig. identity asβ¦
cos(π¦π¦) = οΏ½1 β π₯π₯2
Now we can put this latest result back into our differentialβ¦
ππππππππ
=ππ
πππππ¬π¬ (ππ)=
ππβππ β ππππ
=ππ
(ππ β ππππ)ππ.ππ = (ππ β ππππ)βππ.ππ
That was a bit tricky, but we managed to find π¦π¦β².
Letβs go on and see if we can find π¦π¦β²β²β¦
π¦π¦β² = (1 β π₯π₯2)β0.5
This is where that shorthand notation for derivatives comes in really handyβ¦
πΏπΏπππ‘π‘ π’π’ = 1 β π₯π₯2 β΄ πππ’π’πππ₯π₯
= β2π₯π₯
β΄ π¦π¦β² = π’π’β0.5
β΄ πππ¦π¦β²
πππ’π’= β0.5π’π’β1.5
β΄ π¦π¦β²β² =πππ’π’πππ₯π₯
β πππ¦π¦β²
πππ’π’= β2π₯π₯ Γ β0.5π’π’β1.5
β΄ ππβ²β² = πππ π βππ.ππ = ππ(ππ β ππππ)βππ.ππ
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Question
Question
Q3.23 Find ππβ²β² for ππ = ππππππβππ(ππππ)
ANSWER
3.2.4 Differentiation of Inverse Hyperbolic Functions Consider the functionβ¦
π¦π¦ = π΄π΄ππππββ1(π₯π₯)
Our task here is to find π¦π¦β²β². We proceed in a similar manner to the previous problemβ¦
π₯π₯ = sinh (π¦π¦)
β΄ πππ₯π₯πππ¦π¦
= cosh(π¦π¦) β΄ πππ¦π¦πππ₯π₯
=1
cosh (π¦π¦)
A useful identity isβ¦
πππππ΄π΄β2(π¦π¦)β π΄π΄ππππβ2(π¦π¦) = 1 β΄ πππππ΄π΄β2(π¦π¦) = 1 + π΄π΄ππππβ2(π¦π¦)
β΄ πππππ΄π΄β2(π¦π¦) = 1 + π₯π₯2 β΄ cosh(π¦π¦) = οΏ½1 + π₯π₯2
β΄ ππππππππ
= ππβ² =ππ
βππ + ππππ= (ππ + ππππ)βππ.ππ
Now to find that second derivativeβ¦
π¦π¦β² = (1 + π₯π₯2)β0.5
πΏπΏπππ‘π‘ π’π’ = 1 + π₯π₯2 β΄ πππ’π’πππ₯π₯
= 2π₯π₯
β΄ π¦π¦β² = π’π’β0.5 β΄ πππ¦π¦β²
πππ’π’= β0.5π’π’β1.5
β΄ π¦π¦β²β² =πππ’π’πππ₯π₯
β πππ¦π¦β²
πππ’π’= 2π₯π₯ Γ β0.5π’π’β1.5 = βπ₯π₯(1 + π₯π₯2)β1.5
Q3.24 Find ππβ²β² for ππ = ππππππππβππ(ππππ)
ANSWER
Q3.23 β8ππ(1β4ππ2)1.5
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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 3.1.5β¦
πππππ₯π₯
(π’π’π£π£) = π’π’πππ£π£πππ₯π₯
+ π£π£πππ’π’πππ₯π₯
Letβs rearrange this formulaβ¦
π’π’πππ£π£πππ₯π₯
=πππππ₯π₯
(π’π’π£π£) β π£π£πππ’π’πππ₯π₯
If we now integrate both sides with respect to π₯π₯ then we need to place a β« sign before each term and a πππ₯π₯ at the end of each term, like soβ¦
οΏ½π’π’πππ£π£πππ₯π₯
πππ₯π₯ = οΏ½πππππ₯π₯
(π’π’π£π£)πππ₯π₯ βοΏ½π£π£πππ’π’πππ₯π₯
πππ₯π₯
Some simplifications can be noticed here:
In the first integral the two πππ₯π₯ terms cancel each other out In the second integral we are taking the βintegral of the differentialβ of π’π’π£π£. Since integration is the
reverse of differentiation then the βintegral of a differentialβ drops away, just leaving π’π’π£π£ In the third integral the two πππ₯π₯ terms cancel each other out
Performing these simplifications givesβ¦
οΏ½π π ππππ = π π ππ βοΏ½ππ πππ π
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 π’π’ and πππ£π£. When using integration by parts you have a choice of which part of the product to call π’π’ and which to call πππ£π£. Fortunately, your choice is guidedβ¦
The π’π’ part will become a constant after taking multiple derivatives The πππ£π£ part is readily integrated by using standard integrals.
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Worked Example 23
Determine: β«ππ π¬π¬π¬π¬π¬π¬(ππ)ππππ
One part of our product is π₯π₯ and the other part is sin (π₯π₯). We need to make a choice as to which one to make π’π’ and which to make πππ£π£. Our guidance tells us to make the π₯π₯ part equal to π’π’ since the derivative of π₯π₯ becomes 1 (i.e. a constant)β¦
πΏπΏπππ‘π‘ π’π’ = π₯π₯
πΏπΏπππ‘π‘ πππ£π£ = sin (π₯π₯)
We now try to form the terms in our integration by parts formulaβ¦
πΌπΌππ π’π’ = π₯π₯ β΄ πππ’π’πππ₯π₯
= 1 β΄ πππ’π’ = πππ₯π₯
πΌπΌππ πππ£π£ = sin(π₯π₯) β΄ π£π£ = β cos (π₯π₯)
We can now writeβ¦
οΏ½π π ππππ = π π ππ βοΏ½ππ πππ π
οΏ½π₯π₯ sin(π₯π₯)πππ₯π₯ = π₯π₯(β cos(π₯π₯)) βοΏ½(β cos(π₯π₯))πππ₯π₯
β΄ οΏ½π₯π₯ sin(π₯π₯)πππ₯π₯ = βπ₯π₯ cos(π₯π₯) + οΏ½ cos(π₯π₯)πππ₯π₯
β΄ οΏ½πππ¬π¬π¬π¬π¬π¬(ππ)ππππ = βπππππππ¬π¬(ππ) + π¬π¬π¬π¬π¬π¬(ππ) + πͺπͺ
That was fairly straightforward. Always remember the constant of integration πΆπΆ at the end.
Letβs now try to integrate the product of an exponential and a cosine.
Determine: β«ππππππ πππππ¬π¬(ππ)ππππ
Our two products here are ππ2ππ and cos (π₯π₯). Neither of them will reduce to a constant when differentiated so it doesnβt matter which one we call π’π’ and which we call πππ£π£. 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 π’π’ and πππ£π£, then find π£π£ and πππ’π’. Letβs do thatβ¦
πΏπΏπππ‘π‘ π’π’ = ππ2ππ β΄ πππ’π’πππ₯π₯
= 2ππ2ππ β΄ πππ’π’ = 2ππ2πππππ₯π₯
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Question
πΏπΏπππ‘π‘ πππ£π£ = cos(π₯π₯) β΄ π£π£ = sin (π₯π₯)
To aid our working here (and save some ink) we normally assign the capital letter πΌπΌ to the integral in our question.
β΄ πΌπΌ = οΏ½ππ2ππ cos(π₯π₯)πππ₯π₯
β΄ πΌπΌ = ππ2ππ sin(π₯π₯) βοΏ½ sin(π₯π₯) 2ππ2πππππ₯π₯
This needs tidyingβ¦
β΄ πΌπΌ = ππ2ππ sin(π₯π₯) β 2οΏ½ππ2ππsin(π₯π₯)πππ₯π₯
We were rather hoping for an answer, but what we seem to have is another βintegration by partsβ problem embedded into our development. Donβt worry, thatβs quite normal for this type of problem. All we do is attack that last part with βintegration by partsβ once more, using square brackets. Our desired result will then appearβ¦
πΌπΌ = ππ2ππ sin(π₯π₯) β 2 οΏ½ππ2ππ(βcos (π₯π₯)) βοΏ½βcos (π₯π₯)2ππ2πππππ₯π₯οΏ½
Expanding these terms givesβ¦
πΌπΌ = ππ2ππ sin(π₯π₯) + 2ππ2ππ cos(π₯π₯) β 4οΏ½ππ2ππ cos(π₯π₯)πππ₯π₯
That last integral is the same as πΌπΌ, which is really handy, so we may now writeβ¦
πΌπΌ = ππ2ππ sin(π₯π₯) + 2ππ2ππ cos(π₯π₯) β 4πΌπΌ
β΄ 5πΌπΌ = ππ2ππ sin(π₯π₯) + 2ππ2ππcos (π₯π₯)
If we divide both sides by 5 then we have our final answerβ¦
π°π° =ππππππ π¬π¬π¬π¬π¬π¬(ππ) + πππππππππππππ¬π¬ (ππ)
ππ
Q3.25 Determine: β«ππππππππππππ(ππππ)ππππ
ANSWER
Q3.25 4ππ4π₯π₯ sin(2ππ)β2ππ4π₯π₯cos(2ππ)
20
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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 isβ¦
ππ =πΈπΈπ π β πποΏ½βπ‘π‘ π π π π οΏ½ οΏ½ π΄π΄π π πππ΄π΄
Since current is the defined as the rate of change of charge (π π , carried by passing electrons) we may sayβ¦
ππ =πππ π πππ‘π‘
If we integrate both sides with respect to π‘π‘ we will haveβ¦
οΏ½ ππ πππ‘π‘ = οΏ½πππ π πππ‘π‘
πππ‘π‘ = οΏ½πππ π = π π
So we may say thatβ¦
π π = οΏ½ ππ πππ‘π‘
If we wish to find the amount of charge store stored within a certain time period (i.e. 2 seconds in this example) we writeβ¦
π π (0β2) = οΏ½ ππ πππ‘π‘2
0
= οΏ½πΈπΈπ π β πποΏ½βπ‘π‘ π π π π οΏ½ οΏ½
2
0
πππ‘π‘
=πΈπΈπ π οΏ½πποΏ½β1π π π π οΏ½π‘π‘
β1π π πΆπΆοΏ½
οΏ½
0
2
=βπΈπΈπ π πΆπΆπ π
οΏ½πποΏ½βπ‘π‘ π π π π οΏ½ οΏ½οΏ½0
2= βπΈπΈπΆπΆ οΏ½πποΏ½βπ‘π‘ π π π π οΏ½ οΏ½οΏ½
0
2
Since we know that π π = 1ππΞ© and πΆπΆ = 1πππΉπΉ then π π πΆπΆ = 1 Γ 106 Γ 1 Γ 10β6 = 106β6 = 100 = 1 we may writeβ¦
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π π (0β2) = βπΈπΈπΆπΆ οΏ½πποΏ½βπ‘π‘ 1οΏ½ οΏ½οΏ½0
2= βπΈπΈπΆπΆ[ππβπ‘π‘]02 = βπΈπΈπΆπΆ[ππβ2 β ππβ0] = βπΈπΈπΆπΆ[0.135 β 1]
We also know that the DC supply voltage (πΈπΈ) is 10Vβ¦
β΄ ππ(ππβππ) = βππππ Γ ππ Γ ππππβππ[βππ.ππππππ] = ππ.ππππ Γ ππππβππ πͺπͺπππ π πππππππͺπͺππ = ππ.ππππ πππͺπͺ
The plot below illustrates the integration we have just performedβ¦
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 belowβ¦
If π¬π¬ = ππππ π½π½, πΉπΉ = ππππππ and π³π³ = ππππππππππ, use Calculus to determine the voltage across the inductor after ππ.ππππππ. Use the following formulae in your solution developmentβ¦
ππ =π¬π¬πΉπΉοΏ½ππ β ππβπΉπΉππ π³π³οΏ½ οΏ½ ππππππ πππ³π³ = π³π³
ππππππππ
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We start withβ¦
π£π£πΏπΏ = πΏπΏπππππππ‘π‘
= πΏπΏππ οΏ½πΈπΈπ π οΏ½1 β ππβπ π π‘π‘ πΏπΏοΏ½ οΏ½οΏ½
πππ‘π‘
=πΏπΏπΈπΈπ π βππ οΏ½1 β ππβπ π π‘π‘ πΏπΏοΏ½ οΏ½
πππ‘π‘=πΏπΏπΈπΈπ π οΏ½π π πΏπΏβ ππβπ π π‘π‘ πΏπΏοΏ½ οΏ½ = πΈπΈππβπ π π‘π‘ πΏπΏοΏ½
Putting in the values for E, R, t and L givesβ¦
πππ³π³ = πππππποΏ½βππππ
ππΓππ.ππΓππππβππππ.ππ οΏ½
= ππ.ππππππ ππππππππππ
Sample
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