# The tutor gives an example of the disc method for finding volume of revolution.

Usually, the disc method is preferred when the graph is revolved about the x axis.

Example: Find the volume generated, x=7 to x=10, when y=-(x-9)2+5 is revolved about the x axis.

Solution:

Here we have an ideal case for the disc method. The radius of each disc is the height of the graph above the x axis; each disc’s area, therefore, is π(height)2. Of course, the height is y, or -(x-9)2+5. The “thickness” of the disc is dx. Using the concept that

volume=(area)(thickness)

we construct the integral:

V=∫710π(-(x-9)2+5)2dx

which becomes, from expanding the outer square,

V=∫710π((x-9)4-10(x-9)2+25)dx

We can factor out π, then integrate term by term:

π|107 (x-9)5/5 – 10(x-9)3/3 + 25x

Plugging in for the limits of integration we get

π((10-9)5/5-10(10-9)3/3+25(10)
-[(7-9)5/5-10(7-9)3/3+25(7)]

then

π(1/5-10/3+250 -[-32/5+80/3+175])

=π(33/5-90/3+75)=π(33/5+45)=π(258/5)

Apparently the volume arising from the revolution described above is 258π/5.

I’ll be discussing other methods of finding volumes of revolution in future posts:)

Source:

Larson, Roland and Robert Hostetler. Calculus, third edition. Toronto: D C Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor shows how to integrate exsinxdx by parts.

I’ve written a couple of posts on integration by parts: here and here. The method is used on products, and depends on choosing one function to integrate, then differentiating the other:

∫uv’=uv-∫vu’

With one of the functions xk for some integer k, that one will typically be differentiated, since it will eventually disappear (after k iterations). For k>2, the process will perhaps be long and messy, with lots of chances for error, but likely possible nonetheless – especially if the other function is integrable.

What if neither function will disappear, no matter how many times it’s differentiated? Today’s example is such a case:

∫exsinxdx

Happily, ex is easily integrable. Let’s imagine it as ∫(sinx)exdx: we’ll integrate exdx to ex, then differentiate sinx to cosx:

∫uv’=uv-∫vu’

becomes

∫(sinx)exdx=(sinx)ex -∫excosxdx

For the second round, we’ll integrate exdx again, while differentiating cosx:

∫excosxdx=(cosx)ex – ∫ex(-sinx)dx

⇒∫excosxdx=(cosx)ex + ∫ex(sinx)dx

Notice that our original integral, ∫exsinxdx, has emerged at the end of the second iteration. Believe it or not, this is exactly what we want. Let ∫exsinxdx = A. Substituting and carefully retracing the procedure, we arrive at

A = exsinx – [excosx + A]

or

A = exsinx – excosx – A

Adding A to both sides we get

2A = exsinx – excosx

Dividing both sides by 2, then resubstituting ∫exsinxdx for A, gives

A = ∫exsinxdx = (exsinx – excosx)/2 + C

I’ll be talking more about integral calculus in coming posts:)

Source:

Larson, Roland and Robert Hostetler. Calculus. Toronto: D C Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor explains how, on Excel, to make a grid useful for displaying a graph.

In these days of software availability, one would expect making squares on a page to be obvious. Here’s the easiest way I know:

• In Excel, in the Cells cluster (Home menu), click Format→Column Width. A diaologue box comes up in which you can enter the column width you want. In my case, the default column width is 8.43; I change it to 2.2, then the cell looks square.

Now, to copy the square shape to a range of other cells:

• Select the square cell, right click it, then click Copy. Next, select a row of cells you want to become the square shape. Right click, then click Paste Special→Column widths, then OK. Hopefully you’ll see the associated cells are now square.

Once you have a spacious region of square cells, you can capture it as an image:

1. Select the region of cells, right click it, then click Copy.
2. Open Paint, then click Paste.
3. Save the image (I use .png format).

I’ll be talking more about Excel, Paint, and other software in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor investigates a problem involving the remainder of a power.

On page 48 of his Elementary Number Theory, second edition, Underwood Dudley requests the remainder when 20012001 is divided by 26.

Solution:

2001 mod 26 = 25 ⇒ 2001 = 26k – 1 for some integer k.

Therefore,

20012001 = (26k-1)2001=(26k-1)(26k-1)(26k-1)…(26k-1)

where there are 2001 brackets of (26k-1) forming the product.

Expanding that product, every term is of the form (26k)m(-1)2001-m. Of those, the only term not divisible by 26 will be (26k)0(-1)2001 = -1. Therefore,

20012001 = 26q – 1 for some integer q.

⇒20012001 mod 26 = -1.

Now, -1 ≡ 25 mod 26 (because adding 26 to -1, you get 25).

So, 20012001 mod 26 = 25. Indeed, the remainder when 20012001 is divided by 26 is 25.

Source:

Dudley, Underwood. Elementary Number Theory, second ed. New York: W H Freeman and Company, 1978.

Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor investigates a problem involving composite numbers.

For problem 4b, page 19, of his Elementary Number Theory (second edition), Dudley invites the reader to prove there are infinite n such that both 6n-1 and 6n+1 are composite. (Composite means not prime.)

The first such n I find is 20: 6(20)-1=119=7(17), while 6(20)+1=121=11(11). Now, adding a product of 7 and 11 (such as 77) to 20, we arrive at

6(97)-1=6(20+77)-1=6(20)-1 + 6(77)=119+6(77), which must be divisible by 7, since 119 is. Therefore, it’s composite.

Similarly,

6(97)+1=6(20+77)+1=6(20)+1 + 6(77)=121 + 6(77), which must be divisible by 11, since 121 is.

So, n=20+77k, where k is any integer, gives 6n-1, 6n+1 both composite.

Source:

Dudley, Underwood. Elementary Number Theory, second ed. New York: W H Freeman and Company, 1978.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor shows an essential example of integration by parts.

Back in my July 23 post, I show the formula for integration by parts as

∫uv’=uv-∫vu’

Let’s consider the example

∫xsinxdx

Integration by parts is used to integrate a product of functions. One function needs to be identified as a derivative; it’s the one you integrate. The other function will be differentiated.

In the case of ∫xsinxdx, it’s best to integrate sinxdx, but differentiate x. The reason: differentiating x will get rid of it.

∫xsinxdx = -xcosx – ∫-cosxdx.

which becomes

∫xsinxdx = -xcosx + ∫cosxdx.

Since ∫cosxdx = sinx, we arrive at

∫xsinxdx = -xcosx + sinx + C

The + C part compensates for an unknown constant that was destroyed by the original differentiation.

I’ll be talking more about integral calculus:)

Source:

Larson, Roland and Robert Hostetler. Calculus, third edition.
Toronto: D C Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor solves a system of linear congruences.

Back in my post from March 25, 2014, I explain that “mod” means remainder: for instance, 7 mod 3 = 1.

Two numbers that, divided by a number n, give the same remainder, are called congruent mod n. For example, 24 mod 5 = 4, and also 29 mod 5 = 4. Therefore, 24 is congruent to 29 mod 5. This can be written

24 ≡ 29 mod 5

Example: question 10a, page 41, from Underwood Dudley:

Consider the system of linear congruences

x + 2y ≡ 3 (mod 9)
3x + y ≡ 2 (mod 9)

Solution: The first congruence suggests that

x + 2y = 9k + 3 for some integer k

⇒ x = 9k + 3 – 2y

Substituting into the second congruence, we get

3(9k + 3 – 2y) + y ≡ 2 (mod 9)

then

27k + 9 -6y + y ≡ 2 (mod 9)

which means

27k + 9 -5y = 9m + 2 for some integer m.

-5y = 9m + 2 – 27k – 9

Therefore,

-5y = 9(m -3k -1) + 2 ≡ 2 (mod 9)

From

-5y ≡ 2 (mod 9)

we can multiply both sides by -1 to arrive at

5y ≡ -2 (mod 9)

Since -2 + 9 = 7, we can rewrite the congruence as

5y ≡ 7 (mod 9)

Mod 9, there are only the integers 0 to 8. Trying each one, we realize y = 5:

5(5) ≡ 7 (mod 9) since 25 mod 9 = 7

If y=5, we can sub it into the first equation

x + 2(5) ≡ 3 (mod 9)

x + 10 ≡ 3 (mod 9)

Subtracting 10 from both sides gives

x ≡ -7 (mod 9)

meaning

x ≡ 2 (mod 9)

So we have x = 9p + 2 and y = 9q +5, for any integers p,q.

I’ll be discussing more about linear congruences in future posts:)

Source:

Dudley, Underwood. Elementary Number Theory,
second edition.  New York: W H Freeman and Company, 1978.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor tests graphics with the jsDraw2DX library.

jsDraw2DX is a graphics library for JavaScript. I’d call it object-oriented: for programmers used to that style, it may be very intuitive. The tilted text above was generated from the following code:

var gr0=new jxGraphics(document.getElementById(“graph0”));

var col0=new jxColor(“#770077″);

var pen0=new jxPen(col0, ‘2px’);

var font0=new jxFont(‘serif’);

var brush0=new jxBrush(col0);

font0.size=”28px”;

var text0= new jxText(new jxPoint(85,105), “jsDraw2DX”, font0, pen0, brush0,35);

text0.draw(gr0);

document.getElementById(“graph0″).style.margin=”auto”;

I’ll be talking more about JavaScript graphics in future posts:)

Source:

jsDraw2DX.js

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor shows the motivation behind the integration by parts formula.

Integration by parts is a reversal of the product rule (see my post here).

(uv)’=uv’ + u’v

then we integrate both sides, we get

∫(uv)’=∫uv’ + ∫u’v

Integral is opposite of derivative, so ∫(uv)’ = uv:

uv = ∫uv’ + ∫u’v

More conveniently, it can be written

∫uv’ + ∫u’v = uv

Next, subtracting ∫u’v from both sides, we have

∫uv’ = uv – ∫u’v

The last term is usually written ∫vu’:

∫uv’ = uv – ∫vu’

so that we have the familiar formula for integration by parts.

In a coming post I’ll show an example of how to use the integration by parts formula.

Source:

Larson, Roland E. and Robert P. Hostetler. Calculus, third edition. Toronto:
D C Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor brings up a curiosity from psychology.

Around 1860, a scientist named Gustav Fechner was interested in how physical input becomes mental experience – how sensation leads to perception. One of Fechner’s angles for studying that connection was the jnd – just noticeable difference.

A just noticeable difference is the quantity by which two stimuli must differ before the subject can tell them apart. An example is giving the subject two masses that are slightly different. By how much must they differ for the subject to know that one is heavier than the other?

Peoples’ typical jnd depends on what the stimulus is. However, it turns out that 1/30 of the first stimulus is a value for more than one jnd.

This is paragraph one of the two test paragraphs.

This is paragraph two; you may want to study them.

Can you notice the difference between the two test paragraphs?

I’ll be talking more about the fascinating jnd and other facets of psychology.

Source:

Weiten, Wyane. Psychology: Themes and Variations. Belmont: Wadsworth, Inc., 1992.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.