Thursday, May 16, 2013

Week of May 13th Math Problem: Finding the Lengths of the Major and Minor Axes for a Tricky Ellipse

For many of us, we begin our discussion of conic sections in high school with circles and ellipses. We first learn the equation for a circle is of the general form:

(x - h)2 + (y - k)2 = r2

where (hk) is the set of coordinates for the center of the circle and r is the radius of the circle. We then move on to ellipses, and find the general equation is

[(x - h)2] / a2 + [(y - k)2] / b2 = 1

when a > (horizontally elongated ellipse; switch the positions of the x and y terms for a vertically elongated ellipse). The length of the major axis is therefore 2a and the minor axis 2b. Here is a horizontally-elongated ellipse (image from Wikipedia):


How would one answer the following question?

What are the lengths of the major and minor axes for the following ellipse?

4x2 + 8x + 16y2 - 64y - 13 = 0

We would first complete the square by adding a total of 68 to both sides (split up as 4 and 64) to reveal two factorable trinomials:

 (4x2 + 8x + 4) + (16y2 - 64y + 64) - 13 = 68

Factoring out a 4 and 16 from the first and second trinomials respectively, and adding 13 to both sides, this becomes:

4(x2 + 2x + 1) + 16(y2 - 4y + 4) = 81

Factoring the trinomials 

4(x + 1)2 + 16(y - 2)2 = 81

and dividing both sides by 81

(4/81)(x + 1)2 + (16/81)(y - 2)2 = 1

This is nearly in the form requisite for an ellipse. How do we rewrite this to show a2 and b2 in the denominators? The key is to rewrite the equations first as 

 [(x + 1)2 / (81/4)] + [(y - 2)2 / (81/16)] = 1
(Note: 4/81 = 1 / (81/4)  and  16/81 = 1/ (81/16))

which equals: [(x + 1)2 / (9/2)2] + [(y - 2)2 / (9/4)2] = 1 


This ellipse, centered at (-1, 2), has major axis length (2a) of 2(9/2) = whereas the minor axis (2b) measures 2(9/4) = 9/2.

For an informative website on conic sections, see http://en.wikipedia.org/wiki/Conic_section

For online tutoring help, go to Virtual Tutor Center at http://www.virtualtutorcenter.com

Wednesday, May 15, 2013

Week of May 13th Test Taking Tip: Develop a Time Cushion to Answer More Questions


One of the most common ways people miss out on maximum scores on standardized exams is not necessarily from wrong answers. Rather, it is from lots of questions that remain unanswered once time expires! This is especially detrimental on exams where you do not lose points for incorrect responses on the multiple choice (such as AP exams). Take a typical SAT math section, where you must complete 18 questions in 25 minutes. This means you have an average of

(25 min x 60 sec) / 18 questions = 83 seconds per question

We also know that not every question is a question that is difficult for us. Some of these questions we know the answer to immediately! A conservative estimate is that you can answer 20% of these questions in 10 seconds or less. If this is the case, then let's say 4 questions of 18 are answerable in 10 seconds (total time = 40 seconds). This means you now have an average of 

(24.3 min x 60 sec) / 15 questions = 104 seconds per question

Thus, though it may seem a little disorganized, a viable strategy on the first pass through a section is to only answer questions to which you immediately know the answer. This means you should, on the second pass, have more than enough time to make an educated guess on every question. Don't wildly guess, but just remember that you have more time than you think. Make sure you keep this in mind for AP exams, where you are not penalized for incorrect responses.

Visit Virtual Tutor Center at http://www.virtualtutorcenter.com for your online tutoring needs!

Tuesday, May 14, 2013

Week of May 13th Biology Problem: Analyzing a Dihybrid Cross When One Trait Exhibits Incomplete Dominance

Students who study genetics are familiar with monohybrid crosses, that is, crosses between two parents of (usually) known genotype to produce offspring of various genotype and phenotype. Typically, the genotypic and phenotypic ratios are easily analyzed:


In this cross of pea plants, a yellow seed parent (heterozygous, Yy) is crossed with a homozygous recessive (green, yy) parent to produce offspring pea plants of known phenotypic and genotypic ratios, as shown in the Punnett square above. 

How about a more difficult problem? 

Predict the percentage of pink-flowered tall plants (F1 generation) that result from the cross between a red-flowered tall (heterozygous) plant and a white-flowered short plant. Assume tall is dominant to short.

Since the offspring (F1 generation) have pink flowers, we note red and white flower color exhibit incomplete
dominance to give pink-flowered offspring. These must have a heterozygous genotype with respect to flower color. Therefore, the cross is TtRR ttWW. This dihybrid cross is shown below:


The tall pink-flowered plants, all of which have the same genotype (TtRW), are highlighted in red. Statistically, 50% of the F1 generation is predicted to be tall and pink based on the parents of the offspring. The rest of the offspring are pink-flowered and short (ttRW).

For more on dihybrid crosses, see http://www.youtube.com/watch?v=FJKHC6wX1Hk.

Need online tutoring: try Virtual Tutor Center (http://www.virtualtutorcenter.com)

Monday, May 13, 2013

Week of May 13th Chemistry Problem: Analyzing a Galvanic Cell

One of the most potentially-intimidating questions you can encounter on the SAT Chemistry Subject Test is one which requires the analysis of a galvanic cell. For example, suppose you were given the following diagram for a galvanic cell and asked to provide analysis:


Where would we even begin?

We would start by noting that in a galvanic cell, a spontaneous reaction is observed (in this case, powering the light bulb). We recall that a galvanic cell always has an anode (where oxidation occurs) and a cathode (where reduction occurs). In this system, we have 4 metallic components: Zn(s), Zn(2+, aq), Cu(s), and Cu(2+, aq). As metal atoms typically are redox active, and since 2-electron oxidized species for each metal are present in solution, we suspect these are the atoms involved in the redox process. On any exam, we would be given the following reduction potentials:

Zn(2+, aq) + 2e-  → Zn(s)          E° = -0.76 V
Cu(2+, aq) + 2e-  → Cu(s)           E° = 0.34 V

and recalling that since a positive potential indicates a spontaneous redox process (which is what a galvanic cell is known for), then it must be the case that Zn should be oxidized at the anode and Cu(2+) should be reduced at the cathode. Therefore, the overall spontaneous process, with a potnetial difference of +1.10 V, should be

Zn(s) + Cu(2+, aq)  →  Zn(2+, aq) + Cu(s)  E° = 0.34 V + -(-0.76 V) = 1.10 V

The following diagram gives us a great visual of what occurs in this galvanic cell:


Keep in mind the salt bridge plays an important role here too, providing the ions necessary to balance excess positive or negative charge as Zn(2+) is formed and Cu(2+) is reduced. For a video on galvanic cells see: http://www.youtube.com/watch?v=0oSqPDD2rMA

Saturday, May 11, 2013

Week of May 6th Vocabulary Builder: Building Vocabulary by Association

Building an impressive vocabulary for the SAT and GRE requires you to memorize many words, but sometimes you can build your vocabulary by making an association. Here is an example of how this could work.

Hard work pays off, but how might one describes a person with a great work ethic and who never quits? We might describe this person as:

1. Diligent - done of pursued with persevering attention
2. Sedulous - diligent in application or attention; persevering
3. Assiduous - constant in application or effort; working diligently at a task; persevering; industrious
4. Tenacious - holding together; cohesive; not easily pulled asunder
5. Indefatigable  - incapable of being tired out; not yielding to fatigue
6. Industrious - working energetically and devotedly; hard-working

You can learn more about this technique here: http://www.literacyandnumeracyforadults.com/resources/355536.

Friday, May 10, 2013

Week of May 6th Physics Problem: Determining the Range and Maximum Height of a Projectile

In a student's first semester of college physics and after learning the fundamental equations governing the displacement, velocity, and acceleration of an object with respect to one dimension (e.g., thex direction), students oftentimes have a little more difficulty once two dimensional motion is considered. A typical problem which incorporates both x and y movement is one which involves the trajectory of a projectile. For example:

Suppose a cannonball is fired at an angle of 30° (relative to the horizontal) at an initial velocity of 100 m/s. What is the range and maximum height of the projectile? (Ignore air resistance)

To calculate the range (or horizontal distance traveled) of the cannonball, we need to calculate thex and y components of the initial velocity and the time the projectile is in the air. Thus,

velocity(x) = velocity(initial) (cosine θ) = 100 m/s (cosine 30) = 86.6 m/s
velocity(y) = velocity(initial) (sine θ) = 100 m/s (sine 30) = 50.0 m/s
time in air = [2 (velocity initial) (sine θ)] / g = [2 (100 m/s) (0.5)] / 9.81 m/s^2 = 10.2 s  

Using these data, we calculate the range as follows:

Range = (velocity x) (time) = (86.6 m/s) (10.2 s) = 883 m
Maximum height = (velocity y) (½time in air) - [½ g (½time in air)^2] = (50.0 m/s) (5.1 s) - [(½ (9.81 m/s^2) (5.1 m/s)^2] = 128 m

Note that only half the time in the air was used (5.1 s) because the projectile traces out a parabolic path, with its maximum height achieved after it has traveled halfway to its destination (due to the mirror symmetry of the parabola). For a great video: https://www.youtube.com/watch?v=Q53HHMMWtf0


Need help in physics? Go to Virtual Tutor Center at http://www.virtualtutorcenter.com

Thursday, May 9, 2013

Week of May 6th Math Problem: Using Geometry and Precalculus to Determine the Area Under a Curve

One of the best tips to scoring well on the AP Calculus AB or BC exams is to make sure you know your antiderivatives like the back of your hand. We are trained to easily recognize the antiderivatives of linear, quadratic, trigonometric, and other functions, and are taught a series of techniques to evaluate definite integrals. However, what happens when you see an integral you do not know how to evaluate. Ironically, one tip to solving this calculus problem would be to remember your geometry and precalculus!

How would you calculate area underneath the curve for the following integral?

ʃ √(1 - x^2) dx (evaluated from x = 0 to x = 1)

You take a quick glance and realize this integral is not quickly (if at all) solved by u substitution nor integration by parts, and is not in a form which invokes the use of an inverse trigonometric function as an antiderivative. What do we do here?

Recalling our studies from precalculus, we may remember the equation for a circle of radius 1, centered at (0, 0) is:

x^2 + y^2 = 1

Note that the expression inside the integral above (first bold expression) is what one would get if we solved the second bold expression for y.

y^2 = 1 - x^2
y = ±√(1 - x^2)

Evaluated from x = 0 to x = 1, we notice the integral above is simply the area of the unit circle in the first quadrant. Therefore, since we recall from geometry the area of a circle is equal to πr^2, this means the area of the unit circle is simply equal to π. One-fourth of this area, as the integral above asks for, would be π/4

Therefore, this calculus problem was solved using geometry and precalculus. The more tools in your repertoire, the better off you will be at problem solving!


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Wednesday, May 8, 2013

Week of May 6th Test Taking Tip: Use Your Knowledge of Other Subjects to Answer Difficult Questions!

We've all encountered seemingly impossible questions on our standardized tests that we choose to skip. However, there are cases where you can take advantage of the gamut of your academic repertoire to answer these questions! Here's one example:

None of the following cellular processes results in a net consumption of energy EXCEPT:

(a) Facilitated diffusion
(b) Osmosis
(c) Oxidation of glucose in eukaryotes
(d) Displacement of oxygen on hemoglobin by carbon monoxide in eukaryotes
(e) The removal of water by contractile vacuoles in Paramecium caudatum

This question may seem difficult at first glance, but let's use a combination of chemistry-, physics-, and biology-based knowledge to answer the question.

First of all, our biology courses teach us that diffusion along a concentration gradient is a spontaneous process that does not require energy (ATP in cells). Therefore, the answer is not (a). Also, we note that osmosis is a spontaneous process that results in net movement of water across a semi-permeable membrane. This occurs spontaneously, without any energetic input. This eliminates (b).

We next turn to chemistry-based reasoning. If you didn't recall that glucose is an energy source for eukaryotes, you should recall that the most oxidized form of carbon commonly found in biological systems is carbon dioxide. It is a molecule of high relative stability, and since we exhale it every second of the day, one must surmise glucose (sugar) is ultimately oxidized in our bodies to carbon dioxide, so this should yield much energy (which it does). This eliminates (c). In addition, hemoglobin is an iron-based globular protein, and oxygen binding occurs on the iron atom. Students of chemistry should recall that carbon monoxide is a much better ligand for iron (a Lewis acid) than oxygen (a poor ligand / Lewis base), thus displacement of oxygen by carbon monoxide should be energetically favorable (and thus (d) is eliminated).

This leaves (e). We may recall that Paramecium is a unicellular species that commonly lives in water. If we remember this, then using the information of choice (b) should tell us that water spontaneously enters Paramecium. This being said, it would take energy to reverse a spontaneous process (as physics teaches us that work, such as pumping water out of the Paramecium, requires energy). Thus, the correct answer is (e)!
A good video is shown here: http://www.youtube.com/watch?v=iG6Dd3COug4


For help on tests, go to Virtual Tutor Center at www.virtualtutorcenter.com

Tuesday, May 7, 2013

Week of May 6th Biology Problem: Interpreting Graphical Data Relevant to an Ecosystem

If you take the SAT Biology Subject Test, one of the decisions you will have to make is whether you will take the ecology-focused exam (Biology-E) or the molecular biology focused exam (Biology-M). Many students who take biology courses which thoroughly cover ecology in the curriculum find the Biology E exam a better choice.

A classic ecology problem for these exams traces the cyclical fluctuations in the population densities of two species intertwined in a predator-prey relationship, that is, the snowshoe hare (prey) and the Canadian lynx (graph copied from the Encyclopedia Brittanica Blog: http://www.britannica.com/blogs/2011/06/rise-fall-canada-lynx-snowshoe-hare/).



The data makes several suggestions. First of all, one can see that maxima in hare population densities are consistently simultaneous with, or slightly preempt, lynx population maxima. The strongly correlated data suggest the hare is a key source of food for the lynx, such that higher hare population densities provide lynx populations with the energy necessary to grow in size. However, the hare population minima which always follow maxima suggest that once lynx populations increase past a critical threshold, hares cannot maintain their populations due to the presumed ubiquity of the predator (the lynx), and hence hare populations crash. Clearly, the data seems to suggest that a key limitation to each species' carrying capacity in this ecosystem is each other!

For more info, please go to Virtual Tutor Center at http://www.virtualtutorcenter.com.

Monday, May 6, 2013

Week of May 6th Chemistry Problem: Analysis of a Common Redox Reaction

The thermite reaction is one of the most spectacular chemical reactions one can demonstrate in a classroom or laboratory (for a controlled example of this reaction, see http://www.youtube.com/watch?v=a8XSmSdvEK4). It is the reaction of aluminum metal with iron(III) oxide to yield iron metal and aluminum oxide.


How would one describe this reaction and calculate the change in free energy for an Introductory Chemistry course?

For starters, the reaction is clearly a single replacement reaction that is also described as a redox reaction. The reducing agent is clearly aluminum metal, which provides the electrons to reduce the iron in iron(III) oxide as shown below (the oxidation states are written above each of the metals)


Each aluminum atom provides 3 electrons to each iron atom. Note that since the reaction releases so much heat (see video above), this reaction is clearly exothermic (and therefore the change in enthalpy (G°) is negative, about -848 kJ/mol). Since the reaction does occur spontaneously, this must mean the free energy change (G°) is negative also. To calculate the change in free energy at room temperature (knowing the enthalpy change to be -848 kJ/mol) we must first calculate the entropy (S°) for the reaction, by summing the entropy of the products and subtracting the entropy of the reactants (values are listed in the Appendix of all good general chemistry textbooks).

Change in entropy = [2(27.3) + (50.9)] J/(mol K) - [2(28.3) + 87.4] J/(mol K) =  -38.5 J/(mol K)

Since the entropy change is negative, this is clearly an enthalpy-driven reaction according to the equation

G° = H° - TS°

Using our calculated values for enthalpy and entropy change for the reaction at room temperature (298 K), we calculate

G° = -837 kJ/mol

This very negative value indicates the thermite reaction is highly-favored energetically (under standard conditions), with products that are much more thermodynamically stable relative to reactants.


Need tutoring? Try Virtual Tutor Center! See http://www.virtualtutorcenter.com for more details.

Friday, May 3, 2013

Physics Problem of the Week: Calculating Magnetic Force of a Moving Charged Particle


Magnetic force calculations sometimes cause confusion for students because the calculation is different depending on the polarity of the moving particle (that is, whether the particle is a negatively charged electron or positively charged proton). For example:

Suppose a proton is moving at 8.0 · 10^6 m/s east, perpendicularly through an applied magnetic field (directed north) of 1.0 T. What is the magnitude of the magnetic force exerted on the proton, and in what direction? If an electron was moving through the same field at the same angle with the same velocity, what would be the direction of the magnetic force?

To begin with, physics students will recall that
F(magnetic) = qvB(sine θ)

where q is the charge of the proton, v is its velocity, and B is the magnitude of the magnetic field. Since the proton moves perpendicularly through the field, sine θ = 1. Substituting, we see
F(magnetic) = (1.6 · 10^-19 C)(8.0 · 10^6 m/s)(1.0 T) = 1.3 · 10^-12 N

Now that we have calculated the magnitude, what is the direction? Use the right hand rule. With flat hand, point your fingers in the direction of the particle’s motion (east). Curl your fingers towards the direction of the magnetic field (north). Your thumb points in the direction of the magnetic force (upwards), perpendicular to the xy-plane. For an electron, everything would be the same except the magnetic force would point downwards but perpendicular to the xy-plane (see http://www.youtube.com/watch?v=LK7hv4LX3ys for a helpful video).

Thursday, May 2, 2013

Math Problem of the Week: Finding the Surface Area to Volume Ratio of a Sphere

One of the tenets of cellular biology is that as the volume of a cell increases, its surface area to volume ratio decreases. After a certain volume has been attained, this spells trouble for a cell, because the surface area is too small to allow for adequate diffusion of nutrients and wastes per unit of time. How can we demonstrate this mathematically?

For starters, we note the following formulas:

Surface area of sphere = 4πr^2
Volume of sphere = (4/3)πr^3

This being the case, we note that the Surface Area to Volume ratio is:

SA / Vol = (4πr^2) / [(4/3)πr^3]

Therefore, the factors π and 4 in numerator and denominator cancel, as well as r^2 in both numerator and denominator, to give:

SA / Vol = 3 / r

Thus, the surface area to volume ratio is clearly inversely proportional to r. Therefore, a larger volume means there is less surface area in relation to the volume. No wonder why there are volume limitations in cells; beyond a certain size threshold, there is simply not enough surface area for nutrients and waste to diffuse through relative to the volume!

Wednesday, May 1, 2013

Test Taking Tip of the Week: Saving Time on the Math Section of Standardized Exams

Some of the standardized tests we encounter, such as the AP Chemistry exam, do not allow use of a calculator on some sections. However, most exams do, so you can rely on the calculator to save time, right? Not necessarily....here are some key tricks you should incorporate into your repertoire to save yourself lots of time on calculations.

1. Memorize the common constants
Some constants appear so often on calculations that you should memorize their approximate values so you do not have to waste precious seconds looking them up. Some examples are:


π ≈ 3.14
e ≈ 2.718
c (speed of light) ≈ 3.0 · 10^8 m/s
R (universal gas constant) ≈ 0.0821 L · atm / K · mol
h (Planck's constant) ≈ 6.63 · 10^-34 J ·s


2. Memorize common formulas
You should not have to look up some formulas. For example:

Area of a circle = πr^2
E = mc^2
PV = nRT
velocity · time = displacement


3. Estimate effectively
You should be able to look at answer choices, or an answer you have calculated, and have an intuitive sense about whether or not the answer "sounds right". For example, if the question is:

Calculate the area of a circle with radius 5 and your answer choices are

(a) 31.46 square units
(b) 15.71 square units
(c) 28.12 square units
(d) 78.54 square units
(e) 97.31 square units

You should be able to estimate the right answer choice is (d) because the area of a circle is πr^2
and 25π is roughly 75 (25 · 3).


4. Use the distributive property to aid in multiplication
Quick! What is 31 x 29? If you're not sure, use the distributive property to reveal:

31 x 29 = 31 x (30 - 1)

It is much easier to add and subtract quickly, so 31 x (30 - 1) = 930 - 31 = 899.


5. Become an expert with powers of 10
Any calculation involving powers of 10 should be automatic. For example, you should be able to quickly perform these calculations without the use of a calculator.

a. 1.17 x 1000 = 1170
b. 75.43 / 10000 = 0.007543
c. 10000 x 100000 = 10^9
d. 4.331 / 0.001 = 4331

If you consistently find yourself running out of time, these techniques should help you!