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

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.

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.

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., the x 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 the x 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