Math: Using constants on the Casio fx-260Solar

Tutoring math, calculator use is perennial.  The math tutor introduces a nifty trick for the Casio fx-260Solar.

In physics, you often have a constant in a formula.  An obvious example is

F=mg, where g=9.8m/s^2

High school physics students use the above formula hundreds of times.

There are a couple of ways to use the Casio fx-260Solar so that you don’t have to enter 9.8 each time. Here’s one:

Enter 9.8 then X X; doing so defines 9.8 as a multiplication constant. You’ll see “K” next to the “DEG” around top middle of the screen, meaning you’ve defined a constant. Now, pressing any number, then =, will give that number multiplied by 9.8.

As soon as you enter one of the arithmetic keys (let’s say, for instance, you enter 6 X 7), the constant is erased; the “K” next to “DEG” disappears.

In the Casio manual, this feature is referred to as “Constant Calculations.”

I’ll be saying more about calculator usage in future posts:)

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

Urine Regulation: Aldosterone

Tutoring Biology 12, you realize that with most organ systems, the hormonal control is the most difficult to retain.  The Biology tutor continues about urine regulation with this discussion of aldosterone.

Of course, urine is produced by the kidneys.  If you missed it, you can read how a kidney works in my post here.

In my previous post I talked about ADH and how the body uses it to regulate urine volume. There is another hormone – called aldosterone – that the body uses to control how much water is reclaimed from the filtrate. (Recall that the filtrate is the mix of water, ions, and small molecules first removed from the blood by the kidneys.)

Aldosterone is released by the adrenal cortex. However, the adrenal cortex needs to be informed to do so by the presence of renin in the blood. Renin is secreted by the cells of the juxtaglomerular apparatus, which are adjacent to the glomerulus and sense the blood pressure within. Specifically, when the cells of the juxtaglomerular apparatus sense that the blood pressure is too low, they respond by secreting renin into the bloodstream.

The renin circulates through the body to the adrenal cortex. Detecting the renin, the adrenal cortex responds by secreting aldosterone.

Aldosterone targets the cells of the distal convoluted tubule, telling them to let go of more K+ (K+ means potassium ions), but reclaim more Na+ (sodium ions) in compensation. The effect is that more water is reabsorbed from the filtrate, increasing blood volume and decreasing urine volume.

Unlike ADH, aldosterone does not result in blood dilution, since more ions are reclaimed alongside the extra water that is reabsorbed. Someone might ask, “If aldosterone increases the reclamation of sodium ions, how does that mean increased water reabsorption?” The answer is that sodium ions have a powerful pull on water – more powerful than potassium ions. So if you reabsorb sodium ions instead of potassium ions, more water will be drawn back into the blood as well.

Ultimately, the kidneys release renin – which leads to the release of aldosterone – in order to defend their own function.  If blood pressure is too low, the kidneys cannot filter the blood properly.  By increasing water reabsorption and therefore blood volume, aldosterone helps maintain the necessary blood pressure for proper filtration.

Source: Biology 12, Module 4: Human Biology 2. Open School BC, 2007.

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

Urine regulation: ADH

Tutoring biology 12, you cover kidney function.  The biology tutor introduces ADH, which is a hormone used to regulate urine volume.

For explanation of how a kidney works, see my post here.

Today, we focus on the fine tuning of urine volume. The hypothalamus monitors the concentration of the blood. It may decide, for instance, that the blood risks dehydration. How can the hypothalamus respond to help prevent dehydration?

The hypothalamus has the option of ordering the posterior pituitary to release ADH (anti diuretic hormone). ADH acts on the cells of the distal convoluted tubule and the collecting duct, causing them to be more permeable to water. The result is that more water will be reabsorbed back into the blood. Subsequently, blood volume will stay higher, while urine volume will decrease.

Let’s imagine the other situation: the person has just drunk lots of water to flush themselves out. In such a case, the hypothalamus will detect the surplus of water in the blood, so won’t order the secretion of ADH. The cells of the distal convoluted tubule and collecting duct will allow less water to be recollected, so more will be left in the urine. Urine volume will increase, while blood volume will decrease.

At night, the hypothalamus may order the secretion of ADH to keep urine acculumation low during sleep. The benefit: the person will not have to get up as often to urinate – or maybe not at all until morning.

Another hormone – aldosterone – can also be used to influence urine volume. It will be discussed in a future post:)

Source: Biology 12, Module 4, Open School BC, 2007.

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

Math: A counting problem

Tutoring math 12 and some university courses, you get asked about counting. The tutor opens the discussion with a couple of examples.

Questions that ask, “How many ways can people be lined up for a photograph” or “how many five-card hands have two aces” or “how many ways can you fill out a multiple choice test” are all counting problems.

When an item, once used, cannot be used again, we call that without replacement. Counting without replacement can often be done using permute or choose, but there are other options.  For a quick summary of permute and choose, see my article here.

When an item can be used over again, we call that with replacement. A question with replacement is often done using an exponential expression, as we shall see.

Example 1: How many ways can you arrange the letters of the word radio?

Solution: The answer is 5P5, or 120. You use P – which means Permute – because the order of the letters is what matters, and because the process is without replacement: that is, in a given arrangement, each letter can only appear once.

Example 2: How many five card hands from a standard deck of 52 cards have exactly 2 aces?

Solution: Here, the order doesn’t matter, but the process is still without replacement: if you get the queen of hearts, you cannot draw it again in the same hand. Without replacement, and where order doesn’t matter, points to Choose (aka Combination): Specifically, (4C2)(48C3) or 103776. The reasoning is as follows: from the 4 aces, choose 2: 4C2. Then, from the other 48 non-ace cards, choose 3: 48C3. Multiply the two results together since they both happen yet are independent of each other. Independent means that the choosing of the two aces has no influence on which non-ace cards will be chosen.

Example 3: How many possible ways can a 20-question multiple choice test be filled out, if each question has five options?

Solution: Order matters here, but it is with replacement: for instance, you can choose answer A over and over again. The number of ways you can fill out the test is 5^20. The reasoning is that you can fill out the first question five possible ways, then the second five possible ways, and so on. Your possibilities are (5)(5)(5)(5)……(5)=5^20. The number is absurdly large, so we’ll just refer to it as 5^20.

I’ll say more about counting in future posts. Hope your exams are going well:)

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

Physics: Impulse

Tutoring physics, you get asked about impulse now and again.  The tutor introduces the topic, explaining its connection with momentum.

Impulse has many definitions.  Fundamentally, it means change in momentum.  Momentum is a vector quantity, which means it has a direction as well as a magnitude.  Impulse, being the change in momentum, is a vector as well.

A formula definition for impulse is

I=Ft

where I=impulse, F=force, and t=time.

Generally, impulse is thought to mean a large force being applied over a very short time duration. A club hitting a golf ball is such a situation.

Example: A golf ball of mass 80.0g is struck by a golf club with a force of 90.0N. The duration of contact is 120.ms. Find

a) the impulse;

b) the velocity of the golf ball after being struck.

Solution:

a) the impulse is given by I=Ft. The time is 120.ms, where ms means milliseconds. Therefore, the time is 0.120s.

I=(90.0)(0.120)=10.8kgm/s

So the impulse is 10.8kgm/s. Note that the unit of impulse is the same as momentum: Newtons times seconds (Ns), which equates to kgm/s.

b) the velocity of the golf ball can be found using the momentum formula

p=mv

Here, p (strangely, but universally) stands for momentum,m for mass, and v for velocity.

Assuming the golf ball is at rest before it was struck, the impulse applied to it becomes its momentum, since its momentum was zero before. Therefore, its momentum after being struck is 10.8kgm/s. From there, we can find the velocity using p=mv, as follows:

10.8=0.0800v

Note that the mass, being 80.0g, is 0.0800kg. You need base units – i.e., kilograms, metres, and seconds – in the momentum and impulse formulas.

We arrive at

10.8/0.0800=135=v

So, the golf ball’s velocity is 135m/s as it leaves. In real life, air friction quickly takes effect, giving the ball the graceful arc golfers know.

I’ve never been golfing myself:)

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

Math: Systems of equations: a word problem

Tutoring math, you revisit this type of problem often.  The math tutor gives a quick reminder just ahead of exams.

Here’s the problem:

A saleswoman gets paid a base monthly salary plus commission.  One month she sells $50 000 and gets paid a total of $4200.  Another month she sells $75 000 and gets a total of $5700.  Find her base salary and commission rate.

Like any word problem, we begin with let statements:

let x=base salary
let y=commission rate

We set up the equations that reflect the two months’ earnings:

x+50000y=4200
x+75000y=5700

In a situation like this, substitution is easy. Looking at the first equation above, we notice that if x+50000y=4200, then x=4200 – 50000y. We can then substitute 4200 – 50000y for x in the other equation:

(4200 – 50000y) + 75000y = 5700

Continuing, we simplify the left side:

4200 – 50000y + 75000y=5700

4200+25000y=5700

Now we subtract 4200 from both sides:

25000y=1500

Finally we divide both sides by 25000:

y=1500/25000=0.06

Substitute the known value of y into either equation to find x:

x+50000(0.06)=4200

Simplify:

x+3000=4200

Subtract 3000 from both sides:

x=1200

To answer the problem: the saleswoman’s base monthly salary is $1200, while her commission rate is 0.06. Note, of course, that 0.06 is the same as 6 percent.

Hope your exam prep is going well:)

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

Math: a little more on factoring

Tutoring high school math, you might be in review mode now.  The math tutor brings up a couple of factoring instances.

Instance 1:  Suppose you need to factor -2x^2 + 4x + 70.  How would you proceed?

Solution: First, you would remove the common factor (for more explanation, see my post here). However, you are best to factor out the negative 2, rather than just the 2. In general, when factoring an expression with the lead term negative, you should factor out the negative as a common factor.

We proceed:

-2x^2+4x+70=-2(x^2-2x-35)

Next, we factor the inside expression using the easy trinomial method.

-2(x^2-2x-35)=-2(x-7)(x+5)

Sometimes you encounter an expression with just a negative in front.

Instance 2: Factor -x^2 +10x – 24

Solution: First, factor out the negative:

-x^2+10x-24=-(x^2-10x+24)

Now, factor inside, using the easy trinomial technique:

-(x^2-10x+24)=-(x-6)(x-4)

I’ll be mentioning more little tips and tricks the next few posts, with the hope of refreshing the skills towards first semester exams:)

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

Physics: back emf

Tutoring physics 12, you encounter this concept.  The tutor offers the formula with an explanation.

Induction refers to the creation of voltage in a wire that is not connected to a power source.  It’s the premise that enables generation of electricity.  To induce voltage in a loop of wire, you need to spin it in a magnetic field.  The axis of rotation must be at right angles to the field.  You’ll get induced voltage according to

Emf=-NΔ(BA)/Δt

I’ll be covering the above formula in more detail in future posts, but today the focus is back emf.  Back emf evolves during the operation of an electric motor due to the fact that, once again, a loop of wire is spinning in a magnetic field.  Back emf is so-called because it opposes the line voltage powering the motor.  However, back emf is good, not bad.

Consider the following question:

The resistance of an electric motor is 10.0Ω, while the line voltage is 120V.  At operating speed, the motor draws 2.5A.  What is the back emf?

Solution:

We know that, without considering back emf, I=V/R.  Therefore, the current drawn should be I=120/10.0=12A.  The fact that, at operating speed, the motor draws much less than 12A, is due to back emf.  Specifically,

Back emf=Vline-IR

In our specific case,

Back emf=120 – (2.5)(10.0)=95V

Essentially, as the coils of wire in the motor start to spin in the motor’s magnetic field, they become a generator. The generator opposes the line voltage. In our example, with Back emf=95V, the line voltage is reduced from 120V to 25V in the motor. However, that’s good, because it drastically reduces the motor’s power consumption. Recalling the equation for power, P=I^2R, we see that without back emf, our motor would consume (12)^2(10.0) or 1440 Watts. Because of back emf, it consumes only (2.5)^2(10.0)=62.5W.

The efficiency of an electric motor can be related as

eff=(Back emf)/Vline

By that relationship, our motor above is 95/120 or 79% efficient.

In future posts I’ll be revisiting the concepts touched upon herein:)

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

Math: factoring difference of cubes

The math tutor continues with factoring cubes.  Tutoring calculus, this formula is another standby.

Let’s turn to factoring x^3 – 64. Realizing that 64=4^3, you actually are facing x^3-4^3

You need the formula

a^3 – b^3=(a-b)(a^2 + ab + b^2)

For x^3-64, we substitute x for a and 4 for b. We arrive at

x^3-64=(x-4)(x^2+x(4)+16)=(x-4)(x^2+4x+16)

What about factoring 343x^6-8y^3? Since 343=7^3 and 8=2^3, we can rewrite the expression as
(7x^2)^3-(2y)^3

Carefully substituting 7x^2 for a and 2y for b, we arrive at

343x^6-8y^3=(7x^2-2y)((7x^2)^2+(7x^2)(2y)+(2y)^2))

Finally we simplify to

(7x^2-2y)(49x^4+14x^2y+4y^2)

The key with using these formulas is knowing how to rewrite your own expression so the proper substitution can be made.

Cheers,

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

Math: factoring sum of cubes

Tutoring high school math, you rarely see this; it comes up often, however, in calculus.  The tutor offers a light treatment.

How do you factor x^3 + 125? You need the following formula:

a^3 + b^3 = (a + b)(a^2 – ab + b^2)

Of course, 125=5^3. Carefully substituting x for a and 5 for b, we arrive at

x^3 + 125=(x+5)(x^2-5x+25)

What about factoring 8x^3+1?

We need to realize that, really,

8x^3+1=(2x)^3 + 1^3

Substituting 2x for a and 1 for b in the formula, we arrive at

8x^3+1=(2x+1)((2x)^2 – 2x(1) + 1^2)=(2x+1)(4x^2 – 2x + 1)

For a hint about factoring the difference of cubes, please come back soon:)

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