Precalculus 1 040120

Write down the full statement of the quadratic formula.

What three equations would you get if you plugged the points (2, 88), (5, 37) and (9, 11) into the model y = a t^2 + b t + c of a quadratic?

We get

4 a + 2 b + c = 88

25 a + 5 b + c = 37

81 a + 9 b + c = 11.

If you obtained the model y = .05 t^2 - 3 t + 72 what are the first two steps you would take to determine the clock time when depth is 15?

We first substitute 15 for y to get the equation 15 = .05 t^2 - 3 t + 72.

Then we subtract 15 from both sides to get into the standard form of the quadratic:

.05 t^2 - 3 t + 57 = 0.

From here we can factor, use the quadratic formula, or whatever to solve for t.

What are the forms of the basic functions we have looked at so far?

We've looked at least briefly at the following:

quadratic, linear, polynomial, power and exponential functions.

The basic forms are

y = a t^2 + b t + c (quadratic)

y = m x + b  (linear)

y = an x^n + a(n-1) x^(n-1) + a(n-2) x^(n-2) + ... + a1 x + a0. (polynomial)

y = A x^n (power)

y = A b^t (exponential)

We want to look at the how quickly depth is changing over different time intervals for the function y(t) =  .05 t^2 - 3 t + 72:

• At what average rate does the function y(t) =   .05 t^2 - 3 t + 72 change between t = 0 and t = 10?

t = 0 gives us y(0) = 72 and

t = 10 gives us y(10) = .05 * 10^2 - 3 * 10 + 72 = 47

so the graph goes from (0, 72) to (10, 47) between t = 0 and t = 10.

The average rate of change is the slope between the points, which is

slope = rise / run = (47 - 72) / (10 - 0) = -25 / 10 = -2.5.

• At what average rate does the function change between t = 10 and t = 20?

We follow the same process and get ave rate = -1.5.

• At what average rate does the function change between t = 20 and t = 30?

We again follow the same process and get ave rate = -.5.

• Is there a clear pattern to these average rates?
• If we looked at average rates over the interval between t = 0 and t = 5, then over the interval between t = 5 and t = 10, then between t = 10 and t = 15, etc., by how much do you think the average rate would change for each interval?

  What do you think would be the average rate on each interval?

• Looking more closely at average rates which are based at the t = 10 point:

  What was the average rate between t = 10 and t = 20, and interval of length 10?

  Slope was -1.5.

  What is the average rate between t = 10 and t = 11, and interval of length 1?

  t = 11 point is (11, 45.05), found by substituting t = 11.

  Slope is thus rise / run = (45.05 -47) / (11-10) = -1.95.

  What is the average rate between t = 10 and t = .1, and interval of length .1?

  Using the same process we get slope -1.995

  What do you conjecture the average rates will be for intervals of length .01, .001 and .0001, each interval starting at t = 10?

   We get -1.9995, -1.99995, -1.999995

  What do you think will happen as the interval length continues to decrease?

  Is there a limiting value to the average rate as the interval length approaches zero? 

  How would we interpret such a limiting value?

The temperature of that Coke increased from 2 Celsius to 4 Celsius in the first 10 minutes. 

• Let's start our clock when the temperature is 2 Celsius, so we have the data points (0, 2) and (10, 4) on a graph of temperature T vs. clock time t. 
• Starting an hour later the temperature increased from 10.2 Celsius to 11.2 Celsius.  Note that this would give us the data points (60, 10.2) and (70, 11.2).
• Sketch all these data points on a graph of T vs. t.
• The room temperature was 22 Celsius.  Where are all the point on your graph at which the temperature is equal to the room temperature 22 Celsius?
• What do you think the graph of the temperature of the Coke looked like for the next three hours?
• Why did the temperature of the Coke change more rapidly during the first 10-minute period than during the 10-minute period starting at t = 60?
• At what average rate did the temperature of the Coke change for the first 10 minutes, and at what average rate did it change for the 10 minutes starting at t = 60?
• In mathematical terms, how do you think the rate of change of the temperature of the Coke is related to the temperature of the Coke and the room temperature?
• Let Tdiff be the difference in temperature between the Coke and the room.  What are the values of Tdiff at t = 0, 10, 60 and 70?
• Sketch a quick graph of Tdiff vs. t.  Draw a smooth curve indicating how you think Tdiff behaves betwee t = 0 and t = 70.
• Pick two points on your smooth curve for Tdiff vs. t.
• Plug the coordinates of those two points into the form y = A b^t.  What two equations do you get?

y(t) := .05 t^2 - 3 t + 72

VECTOR(y(t), t, 0, 30, 10)

[72, 47, 32, 27]

vector(y(t+10)-y(t),t,0,20,10)

[-25, -15, -5]

vector((y(t+10)-y(t))/10,t,0,20,10)

[-2.5, -1.5, -0.5]

dt :=

aveRate(t, dt) := (y(t + dt) - y(t))/dt

aveRate(0, 10) -2.5

aveRate(10, 10)

-1.5

aveRate(10, 1)

-1.95

aveRate(10, 0.1)

-1.995

VECTOR(aveRate(10, 10^(-z)), z, 0, 4)

[-1.95, -1.995, -1.9995, -1.999950001, -1.999994936]

VECTOR(aveRate(20, 10^(-z)), z, 0, 4)

[-0.95, -0.995, -0.9995, -0.9999500024, -0.9999947368]

VECTOR(aveRate(0, 10^(-z)), z, 0, 4)

VECTOR(aveRate(30, 10^(-z)), z, 0, 4)

[0.05, 0.005, 0.0005, 0, 0]

take a look at the major quiz and see where we're heading

Precalculus 1 Quiz 040115:

Sketch a smooth curve through your depth vs. clock time data points.

Estimate the coordinates of three points on your curve but not coinciding with actual data points.

From your curve estimate the clock time at which depth is 53, as well as the depth at clock time t = 10.

Suppose the points are (5, 70), (11,50) and (14, 44).

Plug the coordinates of each point into the form y = a t^2 + b t + c.  You'll get three equations.

The equations are

  25 a +   5 b + c = 70

121 a + 11 b + c = 50

196 a + 14 b + c = 44.

For the first two equations subtract the second from the first

          25 a +   5 b + c = 70

- [ 121 a + 11 b + c = 50 ]

_____________________

    -96 a - 6 b = 20

Subtracting the second equation from the third we get

     75 a + 3 b = -6

This gives us the system

    -96 a - 6 b = 20

     75 a + 3 b = -6

Doubling the second equation gives us the system

    -96 a - 6 b = 20

    150 a + 6 b = -12

which we add to get

     54 a = 8

We easily solve this to get

      a = 8 / 54 = 4 / 27

We plug this into either of the two equations in a and b.  Picking the first of these equations

    -96 a - 6 b = 20

we obtain

    -96 (4 / 27) - 6 b = 20.

Multiplying both sides by 27 to clear the fraction and doing the rest of the obvious stuff we get, if my arithmetic is correct,

    b = - 154 / 27.

Plugging our values of a and b into the first of the original equations we get

    25 (4 / 27) + 5 (-154 / 27) + c = 70

which we solve by familiar but irritating details to get

   c = 2560 / 27.

Our values of a, b and c give us the model

y = 4 / 27 t^2 - 154 / 27 t + 2560 / 27.

We can approximate this equation by

y = 0.148·t^2 - 5.70·t + 94.8

The Excel best-fit model for our actual data is

y = 0.131 t^22 - 6.74 t + 95.4

which should be better than our hand-graphed result.

Class estimates of the clock time at which depth is 53, as well as the depth at clock time t = 10, give us respective averages of t = 7.4 and y = 46.

How do these estimates compare with the predictions of our function

y(t) = 0.148·t^2 - 5.70·t + 94.8?

y stands for depth, so if we plug in y(t) = 53 we obtain the equation

53 = 0.148·t^2 - 5.70·t + 94.8.

If we can solve this equation for t then we have a prediction based on the model.

We can solve this equation by putting it into the form a t^2 + b t + c = 0 and using the quadratic formula.

We end up with two solutions:

t = 9.86  or

t = 28.7.

The t = 28.7 doesn't correspond to anything that happens in the real world, but the t = 9.86 does.  Our graph isn't overly accurate, and our t = 7.4 estimate doesn't compare badly with the t = 9.86 from the equation.

To get the model's prediction of the depth when t = 10 we simply plug t = 10 into

y(t) = 0.148·t^2 - 5.70·t + 94.8

to get

y(10) = 0.148·10^2 - 5.70·10 + 94.8 = 52.6.

Again this compares reasonably well with our sloppy-graph estimate of y = 46.

smooth curve thru all data points ...

Ask specific questions, get specific answers, email first and you need to do the qa; also ask questions at the beginning of class, not stick around after and ask, general phil being that I need to address your questions in a way that benefits everyone and not just you.  You ask me a question, you write out a transcript and share with everyone, anonymously or otherwise, your choice.

another precalc question:  why is the quad model not good for the freq vs. length data; make sure they graph data for all the first three situations.

note how grainy our depth data was but how smooth the model was and how well it approximated when we consider that any of the data points could have been half a unit to the right or left

take a look at the major quiz and see where we're heading

gotta move into trap graphs and interp etc.

challenge problem:  if we know at each instant how quickly temp is changing how can be figure out the temp change over a given time interval? 

maybe first bring up the concept of rate in terms of how fast temperature changes; hate to get into numerical examples but

algebra review topics need to be brought up on a regular and scheduled basic

we're not going to bother with lin eqns but gonna need quad form and applications, techniques for solving power fn models, right from the beginning and we should state in review examples of the kinds of things we can do to solve equations: adding, mult, taking powers of both sides etc.

if y = at^2 + bt + c represents depth vs. clock time (and incidentally note that I need to find the curve fit before the next class) then to find the clock time at which a certain depth occurs

plug depth in for t (maybe say x instead) and evaluate

plug in depth for y and rearrange the equation so the quad form applies then use the quadratic formula