Homework Assignment - Section 1.2

Solutions (according to Karl)

Problem 1

The data is given as follows:

MinWageData = {{1950,7.01},{1960,7.59},{1970,9.28},{1980,8.46},{1990,6.66},{2000,6.90},{2010,7.67}};

TableForm[MinWageData,TableHeadings->{None,{"y = Year","m = Minimum Wage ($)"}}]

Part a.

The value m(1990) = $6.66 is the 5th row of the above table.

Position[MinWageData,1990]

MinWageData[[5]]

Part b.

In functional notation, this is simply m(1985)

Part c.

We simply need to compute m(1985) by taking the average of the the data points for 1980 and 1990

(MinWageData[[4]]+MinWageData[[5]])/2.0

Using the data we have, in 1985 the minimum wage was estimated to be $7.56.

Problem 4

In 2000, the mobile handset sales totaled $414.99 million, and in 2005 the total was $778.75 million.  We are asked to compute the average rate of change per year:

(778.75-414.99)/(2005-2000)

The average rate of change was calculated to be $72.75 million dollars in handset sales per year.

Problem 7

The American Food Dollar

FoodData = {{1969,25},{1989,30},{2009,34}};

TableForm[FoodData,TableHeadings->{None,{"d = Year","P = Percent spent away from home"}}]

Part a.

Here, P(1989) = 30 is the middle row of the above table.

Part b.

P(1999) is the percent of the American food dollar that was spent on eating away from home in 1999.  We can find this by taking the average of the last two points, since 1999 is halfway between 1989 and 2009.

N[(FoodData [[3]]+FoodData [[2]])/2]

From the computation above, we see in 1999, the percentage should have been 32%, which makes sense, since it is halfway between 30% and 34%.

Part c.

For the average rate of change, we calculate ((ΔP)/(Δd)) for the second two points in the list.

N[(FoodData [[3]][[2]]-FoodData [[2]][[2]])/(FoodData [[3]][[1]]-FoodData [[2]][[1]])]

From the above computation, the average rate of change in the percentage is 0.2% per year.

Part d.

To compute P(2004), which corresponds to the percent of the American food dollar spent on eating out, we simply use the fact that the average rate of change computed in Part c was 0.2% per year

FoodData [[3]][[2]]+0.2*(2004-FoodData [[3]][[1]])

The result here is 33%, which makes sense since 5*0.2% = 1%, which means 5 years after the 1999, we should have an extra 1%.

Part e.

We can continue this idea using the same formula as in Part d.

FoodData [[3]][[2]]+0.2*(2014-FoodData [[3]][[1]])

Assuming the average rate of change holds constant up until 2014, the answer is 35% of the American food dollar is spent on eating out.

Problem 9

Internet Access Problem

InternetData = {{2000,113},{2003,166},{2009,196}};

TableForm[InternetData,TableHeadings->{None,{"t = Year","I = Millions of Americans"}}]

Part a.

From the table, I(2000) = 113 million Americans with Internet access.

Part b.

We use the last two points in the data set to determine the average rate of change ((ΔI)/(Δt)) from 2003 to 2009

N[(InternetData[[3]][[2]]-InternetData[[2]][[2]])/(InternetData[[3]][[1]]-InternetData[[2]][[1]])]

The average rate of change is 5 million people per year

Part c.

To estimate the number of Americans with internet access in 2007, we start with the value in 2003, and use the average rate of change from Part b. by adding to this value four times the rate of change to get us to the number in 2007 .

InternetData[[2]][[2]]+5*(2007-2003)

So our answer is 186 millions have Internet access in 2007.

Problem 10

Winter weather in Oklahoma!

OKWeathData = {{0,62},{1,59},{2,38},{3,26},{4,22}};

TableForm[OKWeathData,TableHeadings->{None,{"h = Hours since 4 P.M.","T = Temperature (F)"}}]

Part a.

In functional notation, this is T(1.5) since time is measure in hours past 4 P.M.  To compute this value, we take the average of T(1) and T(2), which results in a temperature of 48.5 F.

Part b.

To find the average rate of change between 5 P.M. and 6 P.M. we use the second and third data points:

(OKWeathData[[2]]+OKWeathData[[3]])/2.0

(OKWeathData[[3]][[2]]-OKWeathData[[2]][[2]])/(OKWeathData[[3]][[1]]-OKWeathData[[2]][[1]])

%/60.0

The average rate of change over this time period is -21 F / hour, which is also -0.35 F / minute.

Part c.

To estimate the temperature at 5:12 P.M., we start at the temperature at 5:00 P.M. and to it 12 times the aver rate of change (per minute).  This gives a temperature of 54.8 F.

OKWeathData[[2]][[2]]-0.35*12

Part d.

The temperature got to freezing somewhere between 6:00 and 7:00 P.M.  And note that if we take the average of those two temperatures, we get 32 F.  Thus, we expect the temperature got to freezing at 6:30 P.M.

(OKWeathData[[3]]+OKWeathData[[4]])/2.0

Problem 12

Falling WITH a parachute (because the parachute is important!)

JumperData = {{0,0},{1,16},{2,19.2},{3,19.84},{4,19.97}};

TableForm[JumperData,TableHeadings->{None,{"t = Seconds ino the fall","v = velocity (ft/sec)"}}]

Part a.

We will compute the average rate of change between each successive pair of points, and then slap them into a table:

ROCJumperData =Table[{StringForm[" `` to `` ",JumperData[[ i]][[1]],JumperData[[i+1]][[1]]],(JumperData[[i+1]][[2]]-JumperData[[i]][[2]])/(JumperData[[i+1]][[1]]-JumperData[[i]][[1]])},{i,1,Length[JumperData]-1}];

TableForm[ROCJumperData  ,TableHeadings->{None,{"Time Increment (seconds)","Change in velocity (ft/sec^2)"}}]

So from this table, we see that the change in velocity it clearly dropping off to zero, thank goodness!  Thus, we must be approaching a terminal velocity.

Part b.

By inspecting the original table, it appears that the limiting velocity is 20 ft/sec.

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...

Problem 18

A growing boy!!!

WeightData = {{4,36},{8,54},{12,81},{16,128},{20,156},{24,163}};

TableForm[WeightData,TableHeadings->{None,{"t = Age (years)","W = Weight (lbs)"}}]

Part a.

Similar to problem 12, we compute the average rates of change and put them into a table.

ROCWeightData =Table[{StringForm[" `` to `` ",WeightData[[ i]][[1]],WeightData[[i+1]][[1]]],N[(WeightData[[i+1]][[2]]-WeightData[[i]][[2]])/(WeightData[[i+1]][[1]]-WeightData[[i]][[1]])]},{i,1,Length[WeightData]-1}];

TableForm[ROCWeightData ,TableHeadings->{None,{"Time Increment (years)","Change in weight (lbs/year)"}}]

Part b.

From the table at the end of Part a, we see that the average change in weight continued to increase until the age range of 12 to 16 years, where it was at a max, then it slows down afterwards.

Part c.

To approximate the age at 30 years old, we take the weight at 24 years, and use the average rate of change between 20 and 24 years to get us to 30:

WeightData[[-1]][[2]]+1.75*(30-24)

We have approximated the weight at 30 years old to be 173.5 lbs.

Part d.

So the average rate of change from 4 to 8 years old is 4.5 lbs/year.  So we can start at the weight at 4 years, and subtract off four times this amount:

WeightData[[1]][[2]]-4*4.5

This implies the child was 18 lbs at birth, maybe a world record?  Obviously assuming a constant rate of change from birth to 4 years old is not valid.