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 (h, k) 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 > b (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) = 9 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

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!

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 x 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)

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

Week of May 13th: SAT Reading and Writing Tips from Across the Web

You can get some great tips from websites that aren't so well known. For example, check out these tips from the SAT Ninja!

http://satninja.com/sat-reasoning-test-general/7-myths-about-the-sat-that-prevent-you-from-a-2400

http://satninja.com/sat-critical-reading/7-strategies-to-ace-the-sat-sentence-completion-problems

http://satninja.com/sat-writing/how-to-ace-the-sat-writing-sections-even-if-you-suck-at-writing

http://satninja.com/sat-critical-reading/how-i-improved-my-critical-reading-score-by-150-points

http://satninja.com/sat-critical-reading/how-to-answer-sat-critical-reading-questions-without-reading-the-passages

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.

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

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!

Go to Virtual Tutor Center (http://www.virtualtutorcenter.com) for all of your tutoring needs!

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

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.