1.2 Functions Given by Tables

Example 1.3

The following is the data from this example:

({{d = year, 1970, 1980, 1990, 2000, 2010}, {W = number in millions, 31.5, 45.5, 56.8, 66.3, 71.9}})

Part 1

W(1970) = 31.5, which is the number of million women employed outside the home in the year 1970

Part 2

W(1975) is the number of million women employed outside the home in the year 1975. We do not have a value which corresponds to this in the table, but we can take the average of the two values equidistant from 1975, which is 1970 and 1980.  It turns out to be 38.5 million women.

ListPlot[{{1970,31.5},{1980,45.5},{1990,56.8},{2000,66.3},{2010,71.9}},AxesOrigin->{1970,0},PlotRange->{{1968,2012},{0,80}},PlotStyle->Blue, PlotMarkers->{"▽",16}, Joined-> True]

(31.5+45.5)/2

(W(1980)+W(1970))/2

Part 3

To compute W(1972) we need the average yearly change between 1970 and 1980
AROC = ((W(1980)-W(1970))/(1980-1970))

AvgRateOfChange = (45.5-31.5)/(1980-1970)

So the average increase in employed women outside the household is an additional 1.4 million women per year.  To get the estimate for 1972, we start with the value at 1970, and add 2 times the average increase we just calculated.  This gives 34.3 million women, as computed below.

Value1972 = 31.5+2*1.4

71.9-66.3

Example 1.4

Data from this example:
({{Time t (seconds), 0, 10, 20, 30, 40, 50, 60}, {Velocity v (ft per second), 0, 147, 171, 175, 175.8, 176, 176}})

Part 1

starts with no velocity, and accelerates quickly and then levels off by 20 seconds, and tends to a constant velocity of 176 ft/sec after 40 seconds

Part 2

We are asked to compute v(15), t=15 is halfway between t=10 and t=20, so we average the two values to get v(15) = 159 ft/sec:

(147+171)/2.0

Part 3

Terminal velocity!!!!!

ParData = {{0,0},{10,147.0},{20,171.0},{30,175.0},{40,175.8},{50,176.0},{60,176.0}};

ListPlot[ParData,AxesOrigin->{0,0},PlotRange->{{-1,65},{0,200}},PlotStyle->Blue, PlotMarkers->{"▽",16}, Joined-> True]

Part 4

First we are going to introduce some fancy code to make tables from existing data!

ROCParData = Table[(ParData[[i+1]][[2]]-ParData[[i]][[2]])/(ParData[[i+1]][[1]]-ParData[[i]][[1]]),{i,1,Length[ParData]-1}]

FancyROCParData = Prepend[{ROCParData},{"0 to 10","10 to 20","20 to 30", "30 to 40", "40 to 50", "50 to 60"}]

Grid[FancyROCParData, Frame-> All]

SupFancyROCParData = MapThread[Prepend,{FancyROCParData,{"Time Interval","Avg ROC in v"}}];

Grid[SupFancyROCParData ,Frame-> All]

Now that we have the table, we see that the average rate of change in velocity (which we call acceleration and measured in ft/sec/sec, or ft/(sec^2)) goes to zero by the time interval 50 to 60 seconds, thus velocity is constant by then!

Part 5

Answering this for a third time, we clearly see that terminal velocity is 176 ft/sec

Exercise 2

The following is the dataset from the problem.

MortData = {{1975,9.05},{1980,13.74},{1985,12.43},{1990,10.13},{1995,7.93},{2000,8.05},{2005,5.87},{2010,4.69},{2015,3.84}};

TableForm[MortData,TableHeadings->{None,{"y=year","r=Mortgage rate"}}]

Part a

r(2003) is the 30-year fixed mortgage-rate in 2003.

Part b

We need to find the average rate of change from 2000 to 2005, then use that to approximate the value at 2003.

{MortData[[6]],MortData[[7]]}

AvgROC = (MortData[[7]][[2]]-MortData[[6]][[2]])/(MortData[[7]][[1]]-MortData[[6]][[1]])

(5.87-8.05)/5

So the average rate of change in the mortgage rate from 2000 to 2005 is -0.436% per year.  So we multiply this by three to get the mortgage rate in 2003:

MortData[[6]][[2]]+3*(-0.436)

So we have a 30-year fixed mortgage-rate in 2003 of 6.74%, which is close to halfway between the rates from 2000 and 2005, so this makes sense.

Exercise 13

Here is the data for this exercise:

C14Data = {{0,5},{5,2.73},{10,1.49},{15,0.81},{20,0.44}};

TableForm[C14Data,TableHeadings->{None,{"t=Thousands of years","C=Grams remaining"}}]

Part a

To determine the average yearly rate of change over the first 5000 years, we use the first two data points, and modify the denominator by multiplying by 1000.

(C14Data[[2]][[2]]-C14Data[[1]][[2]])/((C14Data[[2]][[1]]-C14Data[[1]][[1]]))

(C14Data[[2]][[2]]-C14Data[[1]][[2]])/(1000*(C14Data[[2]][[1]]-C14Data[[1]][[1]]))

So we have that the average yearly rate of change over the first 5000 years is -0.00045400000000000003` grams/year

Part b

We have done all the hard work needed in Part a.  We have the average yearly rate of change between the two data points needed to approximate the amount left after 1236 years -- we start with the initial 5 grams, then multiply the above answer by 1236 years to get the grams left:

C14Data[[1]][[2]]+1236*(-0.000454)

We can expect there to be 4.44 grams left after 1236 years.

Part c

We should expect the limiting value to be 0 grams!  The data shows that about 1/2 the amount is left over each 5000 year interval, thus the amount left can become arbitrarily small given enough time...