Exponential Decay and Time Constants

The other day, a colleague showed me a diagram about current vs. time in an RC circuit. The diagram helped me realize that there is another way to think about time constants and exponential decay, a way I had completely overlooked.  I had all the right pieces floating around but I had never put them together.

WHAT’S A TIME CONSTANT?

Here are three graphs showing a quantity that is decaying exponentially toward zero:

Each of these is a graph of a function: N(t) = Noe-kt

The only difference is the value of the constant, k.  Higher values of k lead, in a sense, to faster decay.

To help emphasize this, we can define a constant: τ = 1/k

Then we can  re-write the function this way: N(t) = Noe-t/τ

We call τ the “time constant” for this decay.  It has the units of time.  And it gives us an intuitive feeling for how fast a function is decaying.

For every time constant that passes, our decaying quantity gets reduced by another factor of e.

So after one time constant has passed, the function’s value is No/e.  After two time constants, it’s No/e2.  After three, No/e3 …and so on.

But there is another way to think about the time constant.

The time constant tells us how long it would take to reach the asymptote at the current rate of change.

Maybe this should have been obvious, but I never thought about that way until just a few days ago. Here’s the diagram that got me started:

The claim here is made in the context of a charging capacitor, but it is actually true for any function that is approaching an asymptote exponentially. Here is an animation I made in Desmos to illustrate the point:

In the still picture above, you can see the tangent line at t=0.  The dotted green segment on the t-axis represents the time it would take for the tangent line to reach the asymptote.  When you run the video, you will see the function decrease and the tangent line get shallower.  But that green segment stays the same length.

(If you want to play with the Desmos file, you can get it here🙂

WAIT – WHAT’S A TIME CONSTANT NOW?

A quantity that decays exponentially approaches its final value asymptotically. And as it decays, its rate of decay decreases as well.  (That’s the defining characteristic of exponential decay: the rate of decay toward the final value is proportional to the current distance from that value.  In fact, we can write that as a differential equation. I went on and on about that here.)

But now suppose that at some moment, the decay rate were to become constant. Instead of taking forever, the function would now be able to reach its asymptote in a finite amount of time. Not in just any old amount of time, but in fact in an amount that equals the “time constant” of the decay. It doesn’t matter what moment you pick.  From any starting point during exponential decay, if the rate were to stay constant (which it doesn’t, but still…) the time to reach the asymptote would always comes out to τ = 1/k.

Or to say the same thing geometrically: the tangent line from any point on the curve will intercept the asymptote after one time constant has elapsed. That’s what the animation is trying to emphasize.  I chose a k value of .2 so my time constant was 1/.2 = 5 seconds.  If you open the Desmos file, you can change the constants to see what changes and what doesn’t.

SO WHY DID THIS SURPRISE ME?

I wasn’t making the connection between the exponential decay and the differential equation lurking in the background (even though I teach that connection and have blogged about it).

That the function decreases with a rate of change that is proportional to the value of the function is actually a first-order differential equation.  The exponential decay is its solution. (Again, gory detail here.)

And if slope is rise over run, then run is rise over slope!  When the function is starting from a higher value, the tangent has a slope that is PROPORTIONALLY higher!  So that proportionally higher slope will get you back to the asymptote in the same amount of time every time.

To find how much time we are talking about, divide the “rise” (or in this case, a “fall”) by the slope:

I’ll leave a more rigorous, math-y proof as an exercise for those of you in your first calculus class.

James Tanton is Still Rolling

And now he asks:

“If I roll a die five times, how many distinct values should I expect to see?”

I was feeling pretty good about myself having solved the first one. (Here) And my Excel simulation told me I was at least close.  But when I attacked this new one, I hit some trouble.  I know I need the expected values, but to get them, I need the probabilities.  And I was bogging down in the calculations.  I could find the probability of getting one distinct value, but two was harder, three harder still…

Simulate first and calculate later

I already had the spreadsheet with the random integers.  But Excel does not have a “number of discrete values on the list” function.  So even there, I was stuck.

[Side note: back in the day when I did know how to code, I briefly knew an obscure programming language called APL.  I believe that in APL, a problem like this can be solved in a single line of code, dense with obscure symbols.  Document your code, campers, or you will never remember what you did!]

But since JT only rolled the dice 5 times, I did eventually come up with a way to have Excel do this for me. Let’s call it the “Go Fish” procedure, naming it after the simple card game.  It’s not very elegant but it works and, as you will see, it pays extra dividends.

For each possible dice value, 1 through 6, I made a column that answered the question: did this value appear on the list. For example, the formula in my first column answers the Go Fish question: got any ones?

=1×or((a2=1),(b2=1),(c2=1),(d2=1),(e2=1))

This generates a value of 1 if any of the dice came up as a 1 and zero if none of them did.

I made a total of 6 columns like this. Then, the sum of those columns tells me how many distinct values appeared in my original 5 rolls of the dice.

From there, it was just a matter of doing this in every row and taking the average. Again I did 10 years worth of rows.

You can see in the top row that the dice came up: 1, 3, 2, 1, 1

So the answer was yes if the question was got any 1’s, 2’s or 3’s and no for any 4’s, 5’s or 6’s.  That gave a total of 3 distinct values this time.

OK, so now I have a rough idea of the answer. But aren’t I just stalling?  I should get back to work calculating the probabilities and expected values.

[Really, I am stalling.  I should be grading lab reports.]

Then I realized that the procedure I used to generate the answer in the simulation can be used to calculate the answer directly:

Say you want to know the probability that your list contains a 1. That is more easily calculated as 1 minus the probability that it contains no 1’s.

P(got any ones) = 1 – (5/6)5

But that is also the probability for any of the single Go Fish questions:

P(got any ones) = P(got any twos) = P(got any threes)…and so on.

So the expected value of the total of the “Go Fish” questions = 6 times the value for any one of them!

That means we expect 6×(1 – (5/6)5) = 3.589 different results every time.

That’s not far from what the simulation told me to expect. I ran it 5 more times and got:

3.584, 3.591, 3.586, 3.587, 3.595

So I am feeling like this is a promising answer. I would still like to go back and finish calculating the probabilities the hard way.  But the point to notice here is that taking the time to simulate the problem also provided the key to a solution path.

James Tanton is on a Roll

A good puzzle is a torment.  And there is something interesting about probability puzzles in particular:  you can’t always tell how tricky they are until you dig into them for a while.  Sometimes a question that seems quite tractable ends up eating up more time and more note paper than you were expecting.

James Tanton has been dropping puzzle after tormenting puzzle on Twitter over these past weeks.  The first one was interesting – and I think I have it solved.  But he was just getting started.

I roll 5 dice every day for a year, recording my high score each day.  At the end of the year, I average those high scores.  What average should I expect?

What we need here is the expected value of the high scores. So first we need the probability of each high score.  Then we can multiply each of those probabilities by its corresponding score.  The sum of those products will be our expected average.

What is the probability of a high score of 1?

Well, you would have to get a 1 every time.  That probability is (1/6)5.  So we can write:

P(high score is 1) = (1/6)5.

What is the probability of a high score of 2?

Now you need a 1 or a 2 every time. That probability is (2/6)5.  But wait!  That includes the cases that turn out to be only 1’s.  But we can subtract the probability we just calculated above to get the probability we want:

P(high score is 2) = (2/6)5  – (1/6)5.

What is the probability of a high score of 3?

We use the same plan: find the probability of getting a high score of as much as 3, subtract the probability of getting a high score of as much as 2  and we will be left with the probability of getting a high score of exactly 3:

P(high score is 3) = (3/6)5  – (2/6)5.

And now we have a pattern to follow:

P(high score is 4) = (4/6)5  – (3/6)5.

P(high score is 5) = (5/6)5  – (4/6)5.

P(high score is 3) = (6/6)5  – (5/6)5.

OK, then we go ahead and multiply each probability by its score and add them up!

When we do, we get an expected value of:

6 – (5/6)5 – (4/6)5 – (3/6)5 – (2/6)5 – (1/6)5 = 5.431

BUT IS THAT RIGHT? Hmm…

I suppose a short computer program could test this. It’s been years since I have written code.  But there’s this program called Excel. And it has a random-integer-generating function. [I used Randbetween(1,6) ]

As you can see, I actually simulated 10 years of dice rolls. And then I did it again.  And again. And again…The last 5 times I did this, I got 5.4055, 5.4127, 5.4389, 5.4477 and 5.4211.

So I believe I am on the right track. Flush with success, I see on Twitter that Mr. Tanton has posted again:

“If I roll a die five times, how many distinct values should I expect to see?”

Well, how much harder can this be? Ha!  Stay tuned…here it is!

Alternative Solutions: Pratice Test #7

DITW = Do it their way

Trial and Error = pick an answer, play with it, see if it fits what they said in the problem…

Back Door = make up numbers for the variables in the problem, work out an answer that is based on the numbers you made up, then put those made up numbers back into the answer choices, ruling out any that don’t produce a matching answer.

More detailed explanations available in the book…

PRACTICE TEST #7, SECTION 3 – NO CALCULATOR

1. DITW

2. There way is to distribute and then group terms, which is fine.  But you can also make up a number for x (say x=5), see what you get, and then see which answer choice gives the same value.  In other words, this is actually an entry-level back door problem.

3. If you don’t feel liked doing algebra, just try each answer choice in both equations.

4. DITW

5. DITW

6. You don’t need to do nearly as much algebra as the official answer suggests.  You do need to think about linear functions:  if a linear function takes you from your current value down to zero in 12 years, then in 4 years it will take you 1/3 of the way to zero!  So you will still have 2/3 of the initial value.  [Shouldn’t we be encouraging this kind of thinking?]

7. Another back door:  make up a number for x, get an answer, put that made up number into the answers.

8. Their way is fine, but they say it the slow way.  Also, like many SAT questions, they intentionally give you the info in an inconvenient order.  Start with the fact that 90% of what he earns is \$270.  So without doing algebra, think: what number when reduced by 10% lands you on \$270?  I would not be surprised if \$300 was your first guess.  Then, since he already earned \$80, he needs another \$220.  At \$10/hr, that’s 22 hours.

Also, just for the record, you could also do this one by trying each answer choice one at a time.

9. DITW

10.  This looks like it is going to be much harder than it is!  But it turns out that when they gave you the roots, they also gave you the factors: (x+1), (x+3) and (x-5).  Only one of those is an answer choice.

11. Well, if this were in the calculator section, you could use the back door.  Alas, it isn’t, so you do need to know your exponent laws…DITW

12. Really useful parabola factoid: the vertex lies midway between the zeroes (which they gave you).  Amazingly, that is the way they did it too!  DITW

13. Seriously?  Yet another back door problem: make up a number for x….

14. Still no algebra required.  The lower limit is obviously zero – as it is in each answer choice.  The upper limit can be found by trial and error with the answer choices.  For example, is the width  20?   Well,  2.5 times 20 give a length of 50.  This gives a perimeter of 140 and we have not even added in the height of 60 yet! So 20 is too big…I’d try b next.  Can you see why?

15. OK, start by making up a number for x and getting your value for the expression.  Then, you can either solve for the k value that gives the same value – or use trial and error.  So, if you are keeping score, this problem lets you show off both of the two major algebra-avoidance techniques.  Nice.

16.  When they give you one algebraic expression  and they ask you for another, you should be on the lookout for simple ways to convert one into the other.  In this case, dividing both sides by 2 does the trick.

17. DITW (and who says there isn’t geometry on the new SAT!)

18.  If you only know one thing about radians, know that π radians is 180 degrees.  If you know a second thing about radians, know how to take that fact and scale it up or down.  So if π radians is 180 degrees, 2π radians is 360 degrees, and 4π radians is 720 degrees.

19. All of my students are laughing at this question.  They know that if you just make a really neat drawing, the answer will be staring right at you, no algebra required.

20. DITW

PRACTICE TEST #7, SECTION 4 – WITH CALCULATOR

1. DITW

2. Back door – make up a number…

3. DITW

4. DITW – and this is a common question-type on the new SAT.  Don’t assume more than what the problem actually states.

5. DITW

6. Stick in x = -3. Then, do trial and error with the answer choices.

7. DITW

8. Just read the graph!  At 1.2 on the x axis…

9. As we said earlier, look to transform what they give you into what they ask for.  In this case, you can add 6 to both sides, then divide by 9.  But if you don’t like that, here’s another way: mess around with numbers until you find an a, x and b that make the equation true.  For example, a=2, x=1, b=1…it works!  Then, use those numbers to answer the question!

10. DITW

11. DITW

12. You should learn the fast way to do percent changes: A 6% increase is the same as multiplying by 1.06.  If you knew that, you could just try each answer choice with ease!  Or take the \$53 and divide it by 1.06 to get the initial price.

13. DITW

14. The relevant geometry fact is that the angles of a quadrilateral add up to 360 degrees.  Then, while the algebra is pretty easy this time, you could also do trial and error if you prefer.

15. DITW (Actually, they give two solutions, both fine.)

16.  If you approach this by making up numbers that fit, you will find that you are done in moments.

17. And here it is: another example of y = b + mx.  The b is the flat fee and the m is the rate per mile.  This is a very common question on the new SAT.

18. Once you see what they are asking, this is not that bad.  I’ll translate for you: find the point with the highest y-value.  Then, tell me how far that point is vertically from the line of best fit.

19. They want you to solve for w in terms of A.  But if that gives you trouble, this is also a classic back door problem.  Make up w and A – if you are lazy, try w = 26, A = 1 – then go see which answer works.

20. DITW

21. This is another linear modeling question.  Make sure you know that if y=mx + b (or y=b+mx) the b value is the starting value and the m value is the rate of change.  This time, they are asking for the rate of change, so pick two points and find the slope.

22. This is really just asking if you know how to find a median.  But be careful!  Many students will go right to the middle of the chart and say that the median is 19.5.  That’s a trap: you have to list the data in ascending or descending order and only then can you go to the middle of the list.

23. DITW

24. Well, you can debate whether this is fair or not.  It certainly favors the kids who have had physics.  They would know right away that 72 is the starting height.  You can think of it as the height you get when you plug in t=0.

25. This one is silly.  If you could make up values for x  and k that go together, you could just see which answer works with those numbers. But wait!  You don’t have to make up the numbers – you can pick them right from the chart.  For example, use  x=9 and k=37.7 and see which answer is true.

26.  If you see the algebraic method, that’s fine.  But if not, use the values in the chart to make up numbers for p, f and c so that the total of the calories comes out to 180.  It will take you just a few moments to find a possible set.  Then you can stick those numbers into each answer choice.

27. You should know the quick way to do percent increases and how successive equal percent increases leads to exponential growth.  (See pages 181-183) This is a frequently occurring SAT item. You want to be able to do this about as fast as you can read it.

28. Make up numbers for a point on the line. A line has only one slope.  So far, we know it has risen 6 and run 3.  So keep doing that to find another point.  You could use (6,12).  For that matter, you can even use (3,6). Then, check out the ratio of t to s using your point.

29. DITW

30. DITW

31. DITW

32. They are hoping to make you use the slope formula.  And they know that some of you will mess up subtracting fractions (even with a calculator).  But you can also carefully find two point that have integer coordinates and then get rise over run by just counting boxes.  Of course, you first check to make sure the boxes are the same scale – and they are.

33. I suppose you could do algebra for this one.  But why not take a few attempts at trial and error first.  I’ll get you started: could it be 35 right and 5 wrong?  That would be 70 – 5 = 65 points…too high, try again.  I promise you will land on the right answer in just another guess or two.

34. DITW

35. Again, a little trial and error before you wade into the algebra. This time, it took me three tries.  But that’s because I didn’t notice that zero works! Also, you TI89 fans are probably laughing about this one.

36. DITW

37. It’s nice to see this old classic. They put some data in a chart and they ask you to find the mean.  (In other versions of this question, they ask for the median.) One way to get the answer is to pull the data out of the chart, listing out the individual results:

5,5,4,4,4,2,2,2,2,2,2,1,1,0,0,0

If you are wondering how I knew how many of each to list, you need to examine the chart a little more carefully.  I went by what the chart said!

Once you have the 20 data points listed, you can easily add them up and divide by 20 to get the average.

38. DITW

Eratosthenes with Modern Tools

It came up in class recently that the Ancient Greeks knew the Earth was round and that in fact, Eratosthenes even measured the Earth’s radius.  We can revisit his method using some fun web-based tools.

REVIEW OF THE METHOD OF ERATOSTHENES

1. If you happen to be standing on the Tropic of Cancer at noon on the day of the summer solstice, you will notice that you have no shadow.  The sun is directly over your head.
2. On that same day, if you are some distance north (or south) of the Tropic of Cancer, you WILL have a shadow.  You can use your height and the length of your shadow to calculate the angle of the sun overhead.  That angle is also the central angle from the center of the earth to your current location.
3. There are 360 degrees in a full circle.  And the corresponding distance is the full circumference of the Earth.

You can find tons of info about this on the web.  For example, here is a helpful image from Jochen Albrecht’s website at Hunter College:

I thought it might be fun to see if we can reproduce this result using some helpful sites on the web.

SUNCALC.ORG

This website lets you choose any location on the planet, choose the date you want to see, and then watch how an object’s shadow changes throughout the day.  So you can have your students look for a place and time when the object’s shadow disappears!  (You can provide more or less guidance, depending on how much time you want to spend on this activity.)

Here is a screenshot showing the shadow length of a 1 m tall object in Sombrerete, Mexico on June 21st:

You can see that the shadow is gone!  Depending on how much you have revealed to your students, it might be helpful to note the latitude (in the window on the lower left).

The next step is to choose another location north or south of our first location so that we can check out the shadow there.  I went north to the little town of Olney Springs in Colorado.  I chose this pretty much at random but nearly due north from Sombrerete.

You can see that on the same time and date as before, the shadow of the 1-m tall object is now .26 meters.  You can use the inverse tangent to get the angle of the sun.  Or, if you feel lazy, you can read the angle right from the website — but that is the complement of the angle we need.  Either way, you get 14.7 degrees.

Now we just need to know how far it is from Sombrete to Olney Springs.  Google Maps to the rescue:

That’s 1,614 km, nearly due north.  And it corresponds to a central angle of 14.7 degrees.  We are almost home free:

This solves to give as the circumference of the Earth: 39,526 km.

From there, we get a radius of 6,290 km — a result that is less than 2% away from the “official” value.

Of course, it’s good to remind students that Eratosthenes worked this out more than two millennia ago!

Alternative Solutions: Practice Test #6

DITW = Do it their way

Trial and Error = pick an answer, play with it, see if it fits what they said in the problem…

Back Door = make up numbers for the variables in the problem, work out an answer that is based on the numbers you made up, then put those made up numbers back into the answer choices, ruling out any that don’t produce a matching answer.

More detailed explanations available in the book…

PRACTICE TEST #6, SECTION 3 – NO CALCULATOR

1. Linear modeling again!  When y=mx + b (or b + mx) the b is the starting value.  In this case, that’s the service fee.
2. DITW
3. DITW
4. You don’t have to FOIL this out all the way.  Just notice that the x^2 term has to be p^2…
5. Yes, you should be able to match a graph and its equation.  But if you can’t…pick any x value,,,say x=2.  Find the y-value that goes with that x value.  (You get y=-1) .  Then, check which graph contains the point you have just found, (2,-1).
6. DITW
7. If you need to, use the back door: make up numbers, then put your numbers into the answer choices to see which one works.
8. No real trick to this one once you understand what they are asking.   In other words, when you add entry in the second column to the one in third column,  do you get the value in the first column?
9. Just try the answers…
10. DITW
11. Know your basic parabola facts and your shifts…see pages 165-166
12. They want you to do polynomial division.  But seriously, this one cries out for the back door.  Try letting x=1.  Work out the answer.  Then put x=1 into each answer choice.  It’s much easier my way.
13. This is actually pretty subtle.  You could try each value and see if it factors (or see if the discriminant is negative). That’s kind of a pain.  There is another way to get the answer but I admit it is sneaky.  Consider the quadratic function y=2x^2-4x-t.  It opens upward.  So if it has no real solutions, if you shift it upward it will still have no real solutions.  But the problem can only have one answer!  So it has to be A.  (If this doesn’t make sense to you, post a question and I will go into it in more detail.  Also, I predict that very few test-takers will get this one right.)
14. DITW
15. Try numbers!  I used a=1 and b=4 so I got 9.  But oh, no!  A and D both come out to 9.  No worries – just do it again with new numbers. I tried a=1, b=6 and got 16.  This time, only D worked.
16. Other than playing around with numbers, how else are you going to do this?
17. DITW
18. This is a Pythagorean Theorem question.  But first, when you see this shape you should automatically be on the alert for similar triangles and ratios…
19. DITW – even I admit that algebra is the fastest way.  But if you are comfortable with the idea of weighted averages and inverse proportions, there is another way to think about this one.

The goal is a 15% solution. Since that is closer to the 10% stuff than it is to the 25% stuff, I expect to need less of the 25% solution than the 3 liters I already have.  In fact, if you look at how far each solution is from the goa, the 10% stuff is 5% away, the 25% stuff is 10% away.  Like weights balancing on a see-saw, the farther away, the greater the effect.  In fact, it is an inverse proportion.  So I know I’ll need half as much of the 25% solution.

OK, I admit that if I were taking this test, I would never have come up with that method on time.  So in this case, do the algebra if you can.  If not, take a guess and move on.

1. DITW

PRACTICE TEST #6, SECTION 4 – WITH CALCULATOR

1. Combine like terms if you want.  Or try a number!
2. DITW
3. You are looking for shallow, then level, then steep.  Slope is rate of change.  Also, the snow was always getting deeper so C and D are silly.
4. DITW
5. They are just checking if you know that you can divide both sides of an inequality, in this case by 3.
6. DITW
7. It seems from the released tests that the SAT would really like you to know that to make accurate predictions, a survey must involve RA NDOM sampling.
8. If this is your sixth practice test, then you have seen this before.  But read carefully so you know if they are asking about everybody or just a subset.  This time, they only want to know about vanilla lovers.
9. DITW
10. By algebra, if x is the second voyage, then x + x + 43 = 1003.  But why not just try the answer choices?
11. DITW
12. You don’t have to calculate the average growth rates to see which is greatest.  Average growth rates are also slopes of the segments that connect the two points.  So draw them and then pick the steepest one.
13. And here we have yet another linear model: y= mx + b but with some of the letters changed.  The a value is still the rate of change, or in this case, the amount of daily growth.
14. Pick a value from the graph or chart . For instance, on day 28 the height was 98.  Try t = 28 in each answer choice.  You will see that only one answer is even close.
15. Really?  The very next question?  I tried x = 4…
16. These are similar triangles.  It might help you to see that better if you redraw the one on the left, turning it and flipping it so that it is oriented the same way as its neighbor…
17. Make up a set of numbers so that 2h + d = 25.  Then go to the answers…by now you should know that I call that “the back door play”.  In basketball, the back door play gets you an easy layup.
18. DITW
19. Trial and error!  For example, if d=7.2 inches then 2h = 25-7.2=17.8 inches so h=8.9 inches.  No good.  If it had fallen in the right range, we would then check if the number of steps came out odd.  Keep going until you find the one that works…
20. DITW
21. DITW
22. Many people find it helpful to list out the data:

10,10,10,10, 11,11,11,11,12,13,15,15,15,17,20,21,21,27,31,34,55

Once you do that, finding the median is just a matter of finding the middle number.  Also, once you have done a few this way, you won’t need to any more.  You can just think: there are 21 states on the list so 10 of them will be greater than or equal to the median, 10 will be less than or equal to the median and the 11th one will be the median.  Start at the top of the chart and keep a running tally of how many states you have accounted for.  By the fourth row, you will have counted 10 states.  The median will be in the next row.

1. DITW
2. You can DITW or you can use trial and error with each answer choice.
3. Linear functions are “evenly spaced”.  So to start,  figure out what goes between the first row and the second row.  Then you will know how much you have to go up by to get to f(3).
4. DITW
5. DITW
6. This is yet another example where you are better off finding your own answers first and then checking which equation matches them.  Just draw a number line and you can easily see that the two points they are talking about are -1 and -7.  So check each answer to see which one works with both of those values.
7. Let’s make up numbers again.  For example, if t= 25, then s = 80 and the average speed is 80/25=3.2 inches per second.  Now, it’s off to the answer choices!
8. Hmm..if we drew a curve through these points it would be a downward parabola.  So you know it is B or D.  When x=4, it looks like y= 800 or so.  Plug in x=4 into B and D.  It won’t be a close call.
9. Just try numbers…easy numbers.
10. DITW
11. DITW…remember the formulas are given in the front of the section
12. If you don’t want to do the algebra, you can graph both functions on a graphing calculator and then trace along until you come to the intersection point.
13. DITW
14. If the average is .1 greater and there are 6 items then the total is .6 greater…
15. Make a little chart, fill in the 480 dollars at the bottom,  then work your way back up, cutting the value in half each time.
16. Here’s a hint: the 6 students represent 15% of the committee.  You can set up ratios to find the rest of the committee members.  For example, to find the administrators:  6/15 = x/25

Alternative Solutions: Practice Test #5

DITW = Do it their way

Trial and Error = pick an answer, play with it, see if it fits what they said in the problem…

Back Door = make up numbers for the variables in the problem, work out an answer that is based on the numbers you made up, then put those made up numbers back into the answer choices, ruling out any that don’t produce a matching answer.

More detailed explanations available in the book…

PRACTICE TEST #5, SECTION 3 – NO CALCULATOR

1. While it would be best if you can immediately make the connection between an equation in slope-intercept form and its graph, you do have a backup plan here: pick some point on the graph, say (2,3) and check which answer choice fits.

2. Look at the central angle to see what fraction of the circle we are talking about.

3. If you are quick about it, factoring is fine.  But trial and error with the answer choices works pretty quickly too.

4. DITW

5. Trial and error

6. OK, algebra is pretty quick.  But this is also a classic “back door” problem.  Make up a value for ‘a’, use it to get a^2 – 1 and a + 1 and their sum.  Then put that same ‘a’ value into each answer choice…

7. DITW

8. This is an example (yet another) of y=mx+b where they want to see if you know what the y-intercept means…it’s the “starting value” or in this case, the speed at 0 degrees.

9. This LOOKS like you are going to have to do algebra – substitution and whatnot.  But before you jump in and do that, it’s worth playing around with some easy numbers to see if you can find a solution that way…I won’t ruin the fun by giving away numbers that work.  I’ll just say that you will be surprised to find them with ease!

10.  Again, if you don’t want to do algebra, use the back door: make up a and b, calculate z, y and 4z+8y.  Then go check the answers to find the match.

11. DITW

12. DITW

13. Make up numbers!  Say n=5…that makes t = 7.  For an additional cup, n=6 and now t=8. So one more cup meant one more tea bag.

14.  You should recognize a graph of an exponential function (see page 182 ) and you should know what happens when you shift it up 1 unit (See page 129) and then what happens when you flip it vertically (see page 130).  Update 3/18/17  — even if you don’t know what the graph should look like, you know that the function is f(x)=-(2^x +1).  You can make up an x value and see what why value you get. So for instance, if x=0 you get y=-2.  Then look at the graphs to see which one has (0,-2) as a point on the graph.

15.  I don’t like this kind of question.  It attempts to force you not just to think algebraically but to do it in precisely the way they expect you to.  But there is an alternative:  solve for m using any method you can come up with.  Then plug your m into each answer.  Only one equation will be true.  Here’s one way to do it:  gas costs \$4 per gallon.  So to save \$5, I need to use a little more than 1 fewer gallon per week – how much more?  The “extra” dollar is ¼ of \$4 so the extra amount of gas  is ¼ of a gallon.  So I need to use 1.25 gallons fewer gallons.  (We could have set up a ratio too.)  At 25 mpg, that makes m= 25 times 1.25 = 31.25 miles.  Now put that value into each answer…

16.  DITW

17. DITW

18. This time, with the ugly fraction, I could not find numbers just by playing…oh, well – sometimes you have to DITW

19.  Well, they WANT you to put the fractions over common denominators and clean up.  That will work.  But there is a neat alternative:  they said the expressions are equivalent for ANY value of x other than -2.  So pick a lazy value!  I went with x=0…a little bit of algebra (easier than what they wanted you to do) leads to a = 2.  If you don’t believe me, try it yourself with some other x value.  It will still lead you to a=2.

20.  DITW – and I admit, I am not sure how this question is worthy of being the last one in a section.

PRACTICE TEST #5, SECTION 4 – WITH CALCULATOR

1. DITW

2. Just pick a row of the chart and check which answer fits.

Pro tip: don’t start with the first row – they cleverly make it so that multiple answers work.  Try the second or third row…

3. Simple ratio problem, but don’t forget the units conversion.

4. Well, the algebra is literally one step: divide both sides by 3.  That’s faster than making up numbers.  In fact, to make up numbers, you would probably end up doing that same algebra anyway.

5. DITW

6. No need to set up a system of equations!  You can just try each answer one at a time.  But there is also a “think-about-it” method (see page 159 ). Suppose all she bought were magazines. That would cost \$11, leaving \$9 unaccounted for.  Novels cost \$3 MORE than magazines.  So how many must she have purchased?

7. No kidding, walk into the SAT knowing about models for linear growth (see pages 146-147).

8. They want you to FOIL the term on the left and then regroup, combining like terms.  If you are comfortable with that, go ahead.  But you do have a calculator and this is actually a classic back door problem.  Make up an x value and take it from there…

9.  If you are not sure how to convert the 2 kilometers back to miles, you could also use trial and error, multiplying each answer choice by 1.6 to see when you get closest to 2 miles.

10. You could make up numbers for this one too…

11. DITW

12. Algebra is easiest this time: you just add the equations.  The y terms drop out and you are done!

13.  Trial and error! But another pro tip: check the easier inequality first.  You will see that just the one inequality rules out all but one of the answer choices.

14. DITW – and if you have already worked through tests 1 – 4, this should look like a rerun.

15. DITW

16. DITW

17. They expect you to draw the line of best fit and then find its slope.  That will work, but there is an alternative:  pick a diameter, say 14.  Read the age: it’s about 105 years.  Now go to the chart with the growth factors and check which one has a factor where 14 times that factor is in the neighborhood of 105.  There’s only one that is even close…

18. DITW

19.  So many ways to do this…fastest is to recognize that the outer triangle is equilateral and that the little triangles cut it in half.  (Often very useful to see a 30-60-90 triangle as half of an equilateral triangle.)

20.  An odd little question.  But you can rule out A and C – they stay constant.  And you can rule out B – the distance from the starting point increases steadily.

21.  Just make up numbers and see what happens.  But be careful to make a negative and b positive.  And if you are shaky on your arithmetic with negative numbers, let your calculator take care of it.

22. DITW – but if you don’t feel like doing percents, notice that the answer choices are pretty widely spaced and that 34.6 % is a little more than a third.  But they want the kids with FEWER than 2 siblings, so roughly 2/3 of the class.  There are 1800 classes of that size.  So you could multiply 1800 by 26 and then take 2/3 of that answer.  That will give you a close enough result.  (But you wouldn’t use this rough estimate if the answer choices were tightly spaced.

23.  Trial and error – pick a purchase price (from column one, in thousands) and see which formula gives the right answer…

24.  It helps if you know the quick way to find a 40% discount followed by a 20% discount: multiply by .6 and then by .8.  Then you have to notice that the purchase price is the price AFTER the discounts.  So to “undo” these discounts, take \$140000 and divide by .6 and by .8 – or start from each answer choice and take the discounts, picking the answer that leads you to \$140,000.

25.  Again, if we can find the answer our own way, we can check which of their answers agrees with ours.  Since 20% of 150 is 30, the first batch of people gave us 6 more than we need.  So the second batch can give us 6 less than we need, which is 24.  So 24 is the lower bound for p.  Stick that into each answer…only d makes sense.

26.  DITW, though making up a number for a might help you to see what’s going on quicker.

27. DITW

28.  You don’t need to write equations but you do need to understand slopes.  The function f has slope = ½ (as you can see directly from the graph) so the function g has slope = ½ × 4 = 2.  You have been given the point (0,-4).  You could actually draw a picture, neatly counting as you rise 2, run 1 all the way to (9, 14).  Or you could notice that rise = slope times run so you have to rise 18 units.  Starting from y=-4, that takes you to y=14.

29. DITW

30.  You can think of this as the difference of two squares and then factor it.  But you can also just make up numbers for x and a, calculate y and then go check each answer choice.  In other words, this is another classic back door problem.

31. DITW – it’s a classic ratio problem.

32. DITW

33. DITW

34. Another “Case of the Missing Constant (see page 126).  They tell us that (2,5) is a point on the graph.  That means that when you use 2 as the x-value, the function output is a 5….

35.  Jump in and play with numbers.  Make the length 5 more than the width and keep trying until you find a pair that multiplies out to 104…it won’t take long.

36.  This is a very old trick that still shows up on the SAT.  Draw the segment from A to P.  You have just made two little triangles, each of which is ISOCELES.  How do we know that?  It’s because P is the center of the circle!  All of the radii are equal.  From there, finding the angles is much easier.

37. DITW

38. DITW

The Perfect is the Enemy of the Good – An SAT Attitude Adjustment

This is one of the “Attitude Adjustments” in The New Math SAT Game Plan: For 2016 and Beyond!

The Perfect is the Enemy of the Good  (With Apologies to Debbie Stier)

The SAT, and really the entire college application process, is notoriously stressful.  You may have already noticed this.  And it may be that your parents are adding to your stress, not quite able to maintain their cool as they guide, cajole, push and pull you, trying to help you navigate these waters.  I am well-known for being a calm voice of reason (with other people’s children) but as a parent, I can tell you it is hard for us too.  We are eager to see you succeed and we are older than you.  With our advanced years and wisdom, we can see some of the challenges that lie ahead with a clarity we are convinced that you lack, but we have less control of the situation than you do.  So we feel, and attempt to communicate, a sense of urgency.

Different parents deal with this stress in different ways.  The author and parent, Debbie Stier decided to immerse herself in the SAT experience.  Over a period of a year, she tried a variety of prep methods and then took the SAT every time it was offered.  She then wrote a book about it: The Perfect Score Project.  Though I never actually worked with Debbie, she did interview me, and my book was one of her resources.  She improved spectacularly in reading and writing and helped her own kids to great success.  But her math improvement was actually quite small.  When her project was finished, we emailed about what she could have done differently.  I’m going to tell you what I told her: the goal of “perfection” got in her way – and lowered her score!

I am not telling you to abandon that goal.  I have not met you!  If you are already scoring in the high 600s – low 700s and you have just begun to prepare, well then maybe a perfect 800 is in your future.  But for most students, obsessing over perfection makes it harder to achieve their personal best.  Look at your PSAT score or your last SAT score. Ask yourself: how you would feel about raising that math score to the next level? Let’s start by turning a 440 into a 570.  A 540 into a 650 or even a 700.  Take that 650 and turn it into a 720.  I’m not asking you to give up on long-term dreams, but let’s start by making incremental progress.  I am telling you that you are more likely to raise your score on the very next SAT you take if you approach it in a slower, more low-key and playful way.  You need to give yourself time to think and time to breathe.

As for Debbie Stier, though she never attained the perfect math score, I’d still say that her project was a tremendous success.  When I first heard about it, I admit that I thought it was a crazy idea and that she would drive her own kids nuts!  But that didn’t happen (much) and I really enjoyed reading her book.  One thing that comes across very clearly is a steady respect for how difficult this all can be.  There’s no sense of “hey-you-lazy-kid-why-can’t-you-be-perfect-like-me?”   It’s more like: “Wow. This is very challenging.  How can we find a path to help you succeed?”

So when I recommend that most of you take the SAT with a strategy that causes you to run out of time, please be open-minded.  If you do this, you probably will NOT get a perfect score.  That’s OK.  It’s also the single easiest thing you can do to raise your score to your own personal best.  And if when you walk away from this test (and this whole college application process), you have achieved your personal best, isn’t that the perfect score?

More Polar Graphs!

Lot’s more!

Since it does not look like I will be learning to write my own applets any time soon (or getting any papers graded!) I have continued using the Interactive Physics work-around to generate a gallery of these polar graphs.  If you are learning (or teaching) graphing in polar coordinates, I  hope you find these helpful.

NOTE: I am not claiming that these are the fastest ways to get these graphs.  There are many graphing programs that can sketch these diagrams more quickly.  But I think these might help you to build an intuitive understanding of why the graphs come out the way they do.

Reminder: I am using Anton’s idea about an ant that walks along a rotating clock arm.  The ant’s position along the clock arm is give by the polar function r=f(θ).  These animations show how the polar curve is constructed as the ant’s position changes while the clock arm rotates.

Anyway, here they are.  Click on a video to play it.  And it’s easier to see what’s happening if you expand the video to full-screen mode.

A CIRCLE on the x-axis

********************************

A CIRCLE on the y-axis

********************************

A CARDIOID — symmetrical w/ y-axis

********************************

ANOTHER CARDIOID — symmetrical w/ x-axis

********************************

A CARDIOID with an inner loop

********************************

A ROSE PETAL with 3 loops

********************************

ANOTHER ROSE PETAL with 3 loops — what’s different about it?

********************************

A ROSE PETAL WITH 5 LOOPS

********************************

OK, that’s all for now.  If you had another one of these that you were eager to see, post a comment and I’ll try to add the one you want to the gallery.

Alternative Solutions: Practice Test #4

DITW = Do it their way

Trial and Error = pick an answer, play with it, see if it fits what they said in the problem…

Back Door = make up numbers for the variables in the problem, work out an answer that is based on the numbers you made up, then put those made up numbers back into the answer choices, ruling out any that don’t produce a matching answer.

More detailed explanations available in the book…

PRACTICE TEST #4, SECTION 3 – NO CALCULATOR

1. Try x=0 and see what happens.

2. This is a Case of the Missing Constant.  See page 126.

3. Let’s use trial and error: if y=2 then x has to be 12.  Put them into the second equation and see what happens.

4. Classic back door – make up a number for x…I used x=2. So -3x = -6 and f(-6)=17.  Now put x=2 into each answer choice.

5. Another back door.  Try x=0…see what you get and see which answers match. Then try x=1…it’s magic.

6. And yet another back door.  I made b=7 and a=10 because they were the laziest numbers I could find that fit the equation.  Then it’s off to the answer choices…

7. DITW

8. DITW

9. Use trial and error to show that 6 works and 3 does not.

10.  Hmm…if your arithmetic is good enough, you can use trial and error.  But otherwise, go ahead and DITW.

11.  One way to avoid the algebra this time is to make a really neat graph.  You have to solve the first equation for y and you do have to know one of the ways for finding the vertex of a parabola.   But your reward is that you can look at your picture and see that the line crosses the parabola twice.  You may think that my way is no fun.  Look at their way too and see what you prefer.

12. Classic back door.  I started with x=5…

13. DITW

14. DITW – across the country, students are now going to be really motivated to learn how to divide complex numbers!

15.  You can back door this one too!  Make up a k and a p…then you can use the quadratic formula or you can complete the square.  But it’s easier with concrete numbers.

16. DITW

17.  You can make up numbers for the lengths of the triangle.  But the sine of x has to be .6.  So I used 6 for the vertical leg, 10 for the hypotenuse.  Then, Pythagoras says the horizontal leg is 8 – it’s a 6-8-10 right triangle.  But that makes the cosine of y 6/10 or .6 as well.  THEY REALLY WANT YOU TO KNOW THE COMPLEMENT RULE FOR SINES AND COSINES!

18.  This is obnoxious.  In general, there is no way to factor a cubic equation.  But there must be a way to factor THIS one or the problem can’t be solved.  Group the first two terms together and the second two together.  You can factor out (x-5) from each of them.  So the whole thing becomes:

(x2 + 2)(x – 5)=0.  The solution is x=5.

19. DITW

20. I found it helpful to draw a diagram: a number line labeled below with the numbers 50, 60, 70, and 80.  I wrote a -5 above the 50 and a -80 above the 80.  So what I had looked like this:

-5……………………-80

50        60        70        80

Then you can see that in 3 10-kilometer jumps, the temperature went down by 75 degrees so it must be 25 degrees per 10-kilometer jump.

PRACTICE TEST #4, SECTION 4 – WITH CALCULATOR

1. Well, this is easy either way, but you COULD use trial and error if you felt like it.

2. And this could be done by the back door. But it is quicker if you recognize  y = b +mx when you see it.

3. DITW

4. First of all, this is a probability scale-up.  And second of all, you should notice that the percentage is close to 25% or ¼.  Only one answer choice is in the neighborhood of ¼ of 225.

5. Algebra or trial and error…

6. Algebra, trial and error or just play around with numbers until you find a pair where one is 11 more than the other and they add up to 59.

7. Just read the chart to get 4/50.  Also, I believe this question has a minor math vocabulary error.  In regular English usage, the word “proportion”  is sometimes used interchangeably with the word “fraction”.  But mathematically, a proportion is a statement that two (or more) fractions are equivalent.  I wish they had used the word “fraction” in this question.  Nitpicking, I know, and yet…

8. Draw it! You will see that the only way to avoid quadrant I is to slope down hill.

9. DITW

10. DITW (just read the graph)

11. You don’t actually have to calculate the ratios or even estimate them.  The ratio is also the rise over run, starting from the origin.  So imagine lines (or actually draw them if you like) starting from the origin and going to each of the four labeled points.  Pick the line with the steepest slope.

12. DITW

13. Classic example of exponential growth: population increases by the same factor (x10 in this case) in equal time increments (5 years in this case).

14. DITW

15. Exponential decay is one of the graphs you want to recognize just by looking at it.  (Exponential growth is another. See #20 below)  See pages 181-188.

16. To start, this is yet another example where the key is to understand the form y = b + mx.  In this case, the b (or starting amount) is the material cost and the m is the rate of change.  But as the given formula tells you, you have to add the two daily rental costs to get that rate of change.

So at store A you get y = 750 + 80x and at store B you gety = 600 + 105x.

But then what?  You could set up and solve an inequality (DITW) or you can just try numbers.  If you try x = 7 you will see that it is already too many days.

17. See above – the slope is the daily cost.

18. DITW

19. Trial and error: choice A works. Choice B doesn’t.  So A is the smallest…but you can also do some very quick algebra.  Remember that you don’t have to solve for the variable.  As the official key does point out, you can just add 4 to both sides.

20. See the note for #15 above

21. DITW

22. It drops from 2.25 to 2, a decrease of .25.  And that’s a little more than 10% of the starting number.  (10% is one that you want to know how to do without a calculator – you just divide by 10.)

23. You are not expected to know how to calculate standard deviation.  They just want you to eyeball the data and notice that for City A, the data is clustered closer to the mean and for City B there are more data points out on the margins.

24. DITW

25. OK, I admit that I never would have come up with their solution to this one! I approached this in a completely different way.  Here goes:

If the polynomial is divisible by (2x+3) then -1.5 must be a root.  That means that when you put -1.5 into the function, you get zero out.  So I put -1.5 into the two given functions…

f(-1.5)=.75

g(-1.5)=-.25

Then look at those numbers: if you triple the second one and add it to the first one, you get zero!

And the polynomial in choice B does exactly that: p(x)=f(x) + 3g(x)

26. Make up numbers and draw it on a number line.  You will see that y has to be bigger than x but that x can be positive or negative…it just has to stay between y and –y.

27. DITW – and it’s y = b + mx again!

28. While foiling is easy and completing the square is fun, you do have another option: graph the given function on your calculator and then use your calculator to find the minimum value – you should get -25.  That only appears in choice D.  But wait – if they had been nastier, they could have had two answers with -25 in them but one of them not even the right parabola!  If that happens, just graph the answer choices as well to see which one gives the same graph as the original function.  But like I said…foiling is easy and completing the square is fun.

29.  A classic back door!  Make up a value for m, work out the answer, plug your m-value back into the answer choices.

30.  This one is supposed to be hard.  But I think the hardest part is figuring out what they are asking you!  Try reading it one more time…ok, now I will translate.  The question is essentially asking this: At what y-value can we draw a horizontal line that will cross this graph three times?

Only one answer works.

31. DITW – and note: you can think of this one as y = b + mx too.

32. Make up an x, use it to find P.  Add 1 to your x, find P again.  You will see that P goes up by ½ or .5.  Or re-write the equation as P =(1/2) x + 110.  Then you know that 110 is the starting value and ½ is the rate of change.

33. DITW (but if you have taken chemistry, this would be a good time to use factor-label unit conversion techniques.)

34. If 3/5 of the bats are male, then 2/5 are female.  So the male: female ration is 3:2.  And we have 260 female bats.  Set up the proportion, cross multiply and divide and you will see that we need 390 male bats…but we already have 240, so subtract to get 150.

35. Make up numbers for n and v.  Calculate q.  Then multiply to get 1.5v and recalculate q.  No matter what numbers you make up, the ratio of your new q to your old q will be 2.25.

By the way, if you have taken physics, you have probably done problems just like this but involving kinetic energy.  That formula, K = ½ mv2 works the same way…

36. The full circumference is 20pi.  You have to multiply that by the fraction x/360  and have the answer come out between 5 and 6.  Why not use trial and error?

37.  The quick way to find a 28 percent decrease (see page 74) is to multiply by .72.

38.  And the quick way to find successive percent decreases: \$360 × .72 × .72 × .72 or if you prefer: \$360 × (.72)3

And this time, we found alternative solutions to about 40 out of the 58!