2.1 Tables and Trends

Example 2.1

Velocity of a skydiver is given by v(t) = 176(1-(0.834^t)).  Here t is measured in seconds and v has units feet per second.

v[t_]:= 176*(1-(0.834^t))

Part 1

v(2) represents the velocity of the skydiver 2 seconds into free fall, and is 53.58 ft/sec.

v[2]

Part 2

We now want to examine how the velocity of the skydiver over time by creating both a table of the velocities, and a table of the rate of change in these velocities. 

Note that Example 1.4 from Section 1.2 has much of the code required to do this part of this example!

ParData = Table[{i,v[i]},{i,0,60,5}]

TableForm[ParData,TableHeadings->{None,{"t (sec)","v (ft/sec)"}}]

ROCParData = Table[NumberForm[(ParData[[i+1]][[2]]-ParData[[i]][[2]])/(ParData[[i+1]][[1]]-ParData[[i]][[1]]),{4,3}],{i,1,Length[ParData]-1}]

FancyROCParData = Prepend[{ROCParData},{"0 to 5","5 to 10","10 to 15", "15 to 20", "20 to 25", "25 to 30","30 to 35","35 to 40","40 to 45","45 to 50","50 to 55", "55 to 60"}];

Grid[FancyROCParData, Frame-> All];

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

Grid[Transpose[SupFancyROCParData] ,Frame-> All]

As we can see from this last table, the average rate of change in velocity between successive 5 second intervals decreases rapidly, by the time we are to 60 seconds, there is almost no change in velocity.

Part 3

From the data in the first table of Part 2, the terminal velocity appears to be 176 ft/sec.

Part 4

To determine when the skydiver reaches 99% of terminal velocity, we first need to find 99% of 176 ft/sec

0.99*176

So now we need to determine when this happens.  From looking at the first table from Part 2, it appears that this occurs somewhere between t = 25 and t = 30 seconds. So now we create a table for each second between t = 25 and t = 30 seconds.

ParDataTerm = Table[{i,v[i]},{i,25,30,1}]

From the above list, we know we hit 99% of terminal velocity between t = 25 and t = 26 seconds.  So we can break down that interval into 1/10 seconds:

ParDataTerm = Table[{i,v[i]},{i,25,26,0.1}]

No we have narrowed it down into the time interval t = 25.3 to t = 25.4 seconds.  Now, since we have the function stored in Mathematica as v(t), we can use the solve command on it!

NSolve[v[t] == 0.99*176,t, Reals]

Example 2.2

The info we need to know:
A business has 20 canoes to rent out.
The business offers discounts for group rentals.
One canoe rents for $35, two for $34 each, three for $33 each etc...

Part 1

To determine how much money is taken in for renting four canoes, we first need the price, which would be $35-3 = $32 per canoe.  Since they are renting 4 of them, the result is 4*$32 = $128

32*4

Part 2

The formula the price P as a function of the number of canoes n rented would be P(n) = 35-(n-1) = 36 -n.

Part 3

The formula for revenue R as a function of the number of canoes n rented would be R(n) = n*P(n) = n*(36-n)

Part 4

R[n_] := n*(36-n)

CanoeData = Table[{n,R[n]},{n,1,22}];

TableForm[CanoeData,TableHeadings->{None,{"n canoes","Revenue"}}]

It appears the maximum revenue occurs when 18 canoes are rented.

Exercise 7

We have a complicated formula with 3 variables here:

S is the number of sentences
W is the number of words
U is the number of unfamiliar words

The reading level of an individual is given by the following formula, where the grade level is rounded down, and is meant for grades 1 through 3:

Clear[S]

G[S_, W_,U_]:= 0.121 * (W/S) +0.082U+0.659

We will be considering passages with 200 words, 10 of which are unfamiliar, thus: G(S, 200, 10)

Part a.

ReadingGrade  = Table[{S,G[S,200,10]},{S,1,20}];

TableForm[ReadingGrade,TableHeadings->{None,{"# Sentences","Grade"}}]

As the number of sentences increases, the reading level grade decreases.

Part b.

There appears to be no upper bound, as the grade level keeps decreasing the greater the number of sentences becomes.  That being said, we are restricted to 200 words, so the number of sentences cannot be greater than 200.

Part c.

From the above table, we cannot use less than 10 sentences, since this would result in a grade greater than 3.

Exercise 24

There is a discussion about a similar problem in the section, take a gander at it!
The formula for the probability p of rolling exactly 4 sixes given N dice is given by:

p[N_]:= ((N*(N-1)*(N-2)*(N-3))/24)*((((1/6)))^4)*((((5/6)))^(N-4))

Part a.

We are simply asked to compute p(10) rounded to three decimal places.

N[p[10]]

NumberForm[N[p[10]],{1,3}]

The probability is 5%, so out of 1000 rolls, we would expect to roll exactly 4 sixes 50 times.

.05*1000

Part b.

To find the number of dice which will maximize the probability, we can write out a table, or we can plot some data as well!

DiceData = Table[{n,N[p[n]]},{n,1,1000}];

ListPlot[DiceData, PlotRange->{{0,1000},{0,1}}]

DiceDataList = Table[N[p[n]],{n,1,1000}];

Max[DiceDataList]

Position[DiceDataList,Max[DiceDataList]]

N[p[23]]

N[p[24]]

So it appears that using either 23 or 24 dice will give you the best probability of rolling exactly 4 sixes.