Homework Assignment - Section 1.1

Solutions (according to Karl)

Problem 2

The revenue function is given by M(t) = 1.19t+13.22
Here t is measured in years starting with 2000 at t=0, and M is measured in billions of dollars.

M[t_]:=1.19*t+13.22

M[t]

Part a.

In practical terms, M(5) is the revenue for McDonalds for the year 2005

Part b.

In functional notation, this is simply M(10)

Part c.

We simply need to compute M(10), but we have already defined M(t) at the start of this problem

M[10]

Thus, the revenue in 2010 for McDonalds is 25.12 billion dollars.

Problem 3

The skid mark problem, here S(L) = 5.05 (Sqrt[L]), here L is the length of the skid (feet) and S is the speed (mph) right when the brakes were applied.  We will call our function ‘Skid’, as defined below:

Skid[L_]:=5.05*Sqrt[L]

Skid[L]

Part a.

In functional notation, we simply want to know the speed when L=60, so this is S(60). We evaluate this in the command next, and it is 39.1 mph.

Skid[60]

Part b.

So S(100) is the speed required so that the braking process takes 100 feet.  The problem does not ask it, but we calculate that it is 50.5 pmh.

Skid[100]

Problem 4

The Harris-Benedict Formula -- a function of three variables:

M(w,h,a) = 66+6.3w+12.7h-6.8a
W(w,h,a) = 655+4.3w+4.7h-4.7a

Here w is weight (lbs), h is height (in), and a is age (years). 

Since I am a male, I will define and use the male formula.  I cleared all the variables first, you might not have to.

Clear[M,w,h,a]

M[w_,h_,a_]:=66+6.3*w+12.7*h-6.8*a

M[w,h,a]

I will now use my biometric data:  I weigh 175 lbs, am 72” tall, and am 43 years old.  Thus, I wish to compute M(175,72,43) which is computed below.  My basal metabolic rate is 1790.5, in whatever those units are measured in.

M[175,72,43]

Problem 7

Tax problem.

T(I) = 0.11I-500

I is taxable income, T is the tax owed, both measured in dollars.  We will define this function a bit differently here, as I is a reserved variable:

Tax[Inc_]:=0.11*Inc-500

Tax[Inc]

Part a.

In functional notation, this would be T(13000), which is computed to be $930 below.

Tax[13000]

Part b.

So here, we need to compute the difference: T(14000)-T(13000), which is $110, so our taxes increase by $110 if our income increases from $13,000 to $14,000.

Tax[14000]-Tax[13000]

Part c.

So here, we need to compute the difference: T(15000)-T(14000), which is $110, so our taxes increase by $110 if our income increases from $14,000 to $15,000, the same as  in part b.

Tax[15000]-Tax[14000]

Problem 9

The flying ball problem

V(t) = 40-32t

Here t is measured in seconds and V is velocity measured in m/s.

Clear[V,t]

V[t_]:=40-32*t

V[t]

Part a.

In functional notation, this is V(1), which we compute to be 8 m/s as below.  Since it is positive, the ball is moving upwards (rising).

V[1]

Part b.

Similarly, we need V(2), we see V(2) = -24 which is negative, so the ball is falling.

V[2]

Part c.

V[1.25]

Solve[V[t]==0.0,t]

At t=1.25, the ball has reached the maximum height above the ground, at which point it has no velocity.  Note we could use the Solve command to find this value as well!

Part d.

So we need to compute V(2)-V(1), V(3)-V(2), and V(4)-V(3)

V[2]-V[1]

V[3]-V[2]

V[4]-V[3]

Since the change in time is the same for each, this tells us that the change in velocity is constant!  We happen to know that gravity is measured as -32 ft/(s^2), which we have just verified.

Problem 11

A population of deer in a preserve... The formula is which models the population is given by

N(t) = (12.36/(0.03+(0.55^t)))

Here t is measured in years since the herd was introduced, and N is the total number of deer.
We will not use N as the function variable, as N is already a function here (N[] approximates numbers in decimal form).

Deer[t_]:=12.36/(0.03+0.55^t)

Deer[t]

Part a.

To find the initial population we need N(0), or for us Deer[0]

Deer[0]

So there were 12 deer at the start.

Part b.

N(10) is the number of deer 10 years after the program started.  We have N(10) = 379.9 or rounded to 380 deer.

Deer[10]

Part c.

Similar to part b, this is just N(15) = Deer[15] = 410.26, which we will round to 410 deer.

Deer[15]

Part d.

We simply compute N(15)-N(10) = Deer[15]-Deer[10] = 30.33.

Deer[15]-Deer[10]

So there was approximately a 30 deer increase over the 5 years from 10 to 15 years after the start of the project.

Problem 12

The cost (in $) of operating a car is a function of two variables, g, d where g is the price of a gallon of gas ($) and d distance to drive (miles)

Cost(d,g) = ((g d)/32)

Clear[Cost,g,d]

Cost[g_,d_]:=g*d/32.0

Cost[g,d]

Part a.

Here we simply want to let g= $0.98 and d = 230 miles, so this is Cost(0.98, 230)

Cost[0.98,230]

So the cost is $7.04

Part b.

Cost[3.53,172]

What we have calculated is that it costs $18.98 to drive 172 miles if gas costs $3.53 per gallon.