Precalculus I Class 05/01

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Problem Number 6

If the daily cost of heating or cooling a house when outside temperature is T is given by the function

cost(Temp) = $(. 46 + . .025 `sqrt( ( 75 - Temp)^2 ) ,

and if the average daily temperature is given as a function of the number t of days since Jan 1 as

Temp(t) = 49 + 20 sin( 2`pi ( t - 80)/365 ),

then what is the average daily cost on days t = 0, 90, 180 and 270?

We plug t = 0, 90, 180, 270 into Temp(t) to get temperatures 29.37387059, 52.42586288, 68.77355180, and 46.42503645.

We then plug these temperatures into cost(Temp) to get costs  1.600653235, 1.024353427, 0.6156612048 and 1.174374088.

What is the composite function cost( Temp(t) ) ?

We substitute 49 + 20 sin( 2`pi ( t - 80)/365 ) for temp in cost(temp) to get

• cost(temp(t)) = (. 46 + . .025 `sqrt( ( 75 - (49 + 20 sin( 2`pi ( t - 80)/365 )) )^2 )

= (. 46 + . .025 `sqrt( ( 26 - 20 sin( 2`pi ( t - 80)/365 )) )^2 ) .

Evaluate your composite function at t = 0, 90, 180 and 270 to check your expression.

If we substitute t = 0, 90, 180, 270 into cost(temp(t)) = (. 46 + . .025 `sqrt( ( 26 - 20 sin( 2`pi ( t - 80)/365 )) )^2 ) we get costs 1.600653235, 1.024353427, 0.6156612048 and 1.174374088, in agreement with our previous results.

Question about discriminant:

The discriminant is b^2 - 4 a c.  This is the thing you take the square root of in the quadratic formula, then apply the +-.

When the discriminant of a quadratic equation is negative there is no solution because you can't take the square root of a negative.

When the discriminant is zero the square root is zero and both the + and the - give you the same result.  This gives you one repeated solution to the equation.  This means the equation is of the form (x - z)^2, where z is the single zero given by the formula.

When the discriminant is positive you get two different solutions, one for the + and one for the -.

Problem Number 7

If a patient starts with no drug in her body takes a 400 mg dose every 6 hours, losing 30 % of the after-dose amount during that time, then how much drug will remain 6 after the initial dose?

• How much will remain just before the third dose, assuming that she takes her second 400 mg dose at the appropriate time?
• How much will remain just before each of the next three doses, assuming that each dose is taking on schedule?
• What sort of function do you think would model the just-before-dose drug concentration?

The amounts are:

• 400 mg after first dose
• 70%(400 mg) = 280 mg before 2d dose
• 400 mg + 280 mg = 680 mg just after 2d dose.
• 70%(680 mg) = 476 mg just before 3d dose.
• 400 mg + 476 mg = 876 mg just after 3d dose.
• 70%(876 mg) = 613.2 mg just before 4th dose.
• etc.

The just-before dose amounts are 0, 280, 476, 613.2 mg.

We will look at sequence of ratios, then if necessary at sequence of ratios of differences to see that the function is exponential.

• For this problem the ratio of the amounts is not constant, but the ratio of differences is constant.  You should verify this.
• We conclude that the function is exponential with a nonzero asymptote.

The asymptote is the before-dose maintenence-level concentration such that the 30% lost is equal to the dose taken.

Let L be the after-dose maintenance level.  Then

• amt lost = dose

so

• 30% * L = 400 mg

We easily solve to get L = 1333 1/3 mg.

This is the after-dose maintenance level.  The before-dose level is therefore 400 mg less:

• before-dose maintenance level (asymptote) = 1333 1/3 mg - 400 mg = 933 1/3 mg.

The before-dose concentrations 280, 476, 613.4 mg approach 933 1/3 mg as an asymptote.

See notes from the preceding class for more detail on how to obtain the exponential function

• Recall that we find the differences between the values 280, 476, 613.4 etc. and 933 1/3, and we find an exponential function y = A b^n which fits these results. 
• The exponential function for the drug concentrations is then C(n) = 933 1/3 - A b^n..

Problem Number 8

If f(x) = x^-3.004 and g(x) = .49 ^ x, find the formula for (f + g) ( x ).  Find the average rate of change of the function (f + g) ( x) between x = 1 and x = 1.2.

(f+g)(x) = f(x) + g(x) = x^-3 + .49 ^ x.

The ave rate of change between x = 0 and x = .2 is the change in the function value divided by the change in x:

• (f+g)(1) = 1^-3 + .49^1 = 1.4900.
• (f+g)(1.2) = 1.0036.
• The change in (f+g) is 1.0036 - 1.4900 = -.4864.
• The change in x is 1.2 - 1 = .2.

Ave rate of change is therefore

• ave roc = 'dy / 'dx = -.4864 / (.2) = -2.432.

Problem Number 2

A certain breed of rabbit requires a month to mature, after which each pair produces 1 pair of newborns at the end of the next month, 4 pairs the month after that, 1 pair the next month, then 4 pairs the next, etc..  Rabbits never die.  If we start with 1 pair of newborn rabbits, how many will we have at the end of each of the first 10 months?  What recurrence relation will give us the populations for subsequent years?

We begin by making a table.  There are now three categories of rabbits: immature, mature producing 1 pair, and mature producing 4 pairs.

An immature pair will produce a mature pair producing 1 pair.

A mature pair producing 1 pair will produce that pair in the next month and in that month will become a mature pair producing 4 pairs.

A mature pair producing 4 pairs will produce those pairs in the next month and in that month will become a mature pair producing 1 pair.

Starting at month 0 with 1 immature pair only:

• In the subsequent month that immature pair will move into the 'mature producing 1' category.
• There were no mature pairs in month 1 so there will be no immature pairs in month 2.

Starting at month 2 with 1 mature pair producing 1:

• The mature pair producing 1 will move to the mature producing 2 category, and will produce their 1 pair at the beginning of month 3.
• Since in month 2 there were no immature rabbits and no mature producing 3, there will be no rabbits in month 3 in the mature producing 1 category.

Starting at month 3 with 1 immature pair and one mature pair producing 4:

• The immature pair will move into the mature producing 1 category for month 4.
• The mature producing 4 pair will produce 4 immature pairs at the beginning of month 4, and will move into the immature producing 1 column for month 4.
• In month 4 we will therefore have 2 pairs in the mature producing 1 category and 4 immature pairs.  Since there were in month 3 no mature producing 1 pairs there are no mature producing 4 pairs in month 4.

Starting at month 4 with 4 immature pairs and 2 mature pairs

• The 4 immature pairs will move to the mature producing 1 column for month 5.
• The 2 mature producing 1 pairs will produce 2 immature pairs and will move into the mature producing 4 column for month 5.

Starting at month 5 with 2 immature pairs, 4 mature pairs producing 1 and 2 mature pairs producing 4

• The 2 immature pairs will move at month 6 into the mature producing 1 column, where they will be joined by the 2 mature producing 4 pairs from month 5 for a total of 4 immature roducing 1 pairs at month 6.
• The 4 mature producing 1 pairs from month 5 will produce 4 immature pairs, and the 2 mature producing 4 pairs will produce their 8 pairs, giving us 4 + 8 = 12 immature pairs in month 6; these 4 mature pairs will move into the mature producing 4 column for month 6.

The process continues.

Hint on setting up the recurrence relation:  Let a(n) = total number of pairs, b(n) = number of mature pairs producing 1 in month n.  Express i(n) and m4(n), the numbers of immature and of mature producing 4, in month n in terms of b(n-1), b(n-2), b(n-3), b(n-4) etc., whichever terms are appropriate.  Then note that a(n) = i(n) + b(n) + m4(n) and get your expression for a(n) in terms of the values of b(n), b(n-1), b(n-2), b(n-3), b(n-4) etc., as appropriate.

With this model we would get a(n) = b(n) + 2 b(n-1) + 4 b(n-2).   You just need to find the rule for b(n).

Suppose the breed takes a month to mature and every mature pair produces 2 pairs of newborns every month.  What is the recurrence relation?

Let a(n) be the number at the end of each month.  The figure below is similar to the figure obtained in the preceding class, where the total number in month n-2 becomes the number of mature in month n-1 which in turn produce the immature pairs in month n-2.

In this case each mature pair produces 2 immature pairs so the recurrence relation is a(n) = a(n-1) + 2 a(n-2).

Problem Number 8

The difference between room temperature and the temperature of the potato will be an exponential function.

This difference starts at 80 deg - 7 deg = 73 deg.  After 4 minutes the temperature difference  has dropped to 48 deg - 7 deg = 41 deg.

So Tdiff is an exponential function with the two data points (0, 73) and (4, 41), representing Tdiff vs. t.  We find the function for Tdiff as follows:

• We first assume the form Tdiff = A b^t, and we plug in the data points to get the two equations

73 = A * b^0

41 = A * b^4.

• We divide the second equation by the first to obtain

41 / 73 = (A b^4) / (A b^0) or

41/73 = b^4.  Taking the 1/4 power of both sides we get

b = (41/73)^(1/4) = .866.

• Substituting b = .866 into the first equation we get

73 = A * .866^0, or since .866^0 = 1

73 = A.

• Our model is therefore

Tdiff = 73 * .866^t.

The temperature is 7 degrees + Tdiff so our temperature function is

• Temp(t) = 7 + 73 * .866^t.

To find when the temperature first reaches 12 degree we let Temp(t) = 12 degrees and solve for t:

• 12 = 7 + 73 * .866^t.  Adding -7 to both sides we have

5 = 73 * .866^t.  Dividing both sides by 73 we get

5/73 = .866^t.  Taking the log of both sides we get

log(5/73) = t log(.866) so that

t = log(5/73) / log(.866) = 18.6.

We conclude that the temperature first reaches 12 degrees at t = 18.6.