Exercise 3.3.1.
The United States uses the English system of (length) measurements. The rest of the world uses the metric system. So, people who travel abroad and companies that trade with foreign partners often need to convert English measurements to metric ones and vice versa.
Here is a table that shows the six major units of length measurements of the English system:
| English | metric | ||
|---|---|---|---|
1 inch | = | 2.54 | cm |
1 foot | = | 12 | in. |
1 yard | = | 3 | ft. |
| 1 rod | = | 5(1/2) | yd. |
| 1 furlong | = | 40 | rd. |
| 1 mile | = | 8 | fl. |
Develop the functions inches->cm, feet->inches, yards->feet, rods->yards, furlongs->rods, and miles->furlongs.
Then develop the functions feet->cm, yards->cm, rods->inches, and miles->feet.
Hint: Reuse functions as much as possible. Use variable definitions to specify constants.
Solution
;; inches->cm : number -> number
;; convert inches into cm
(define (inches->cm inches)
(* inches INCH->CM))
;; feet->inches : number -> number
;; convert feet into inches
(define (feet->inches feet)
(* feet FOOT->INCH))
;; yards->feet : number -> number
;; convert yards into feet
(define (yards->feet yards)
(* yards YARD->FOOT))
;; rods->yards : number -> number
;; convert rods into yards
(define (rods->yards rods)
(* rods ROD->YARD))
;; furlongs->rods : number -> number
;; convert furlongs into rods
(define (furlongs->rods furlongs)
(* furlongs FURLONG->ROD))
;; miles->furlongs : number -> number
;; convert miles into furlongs
(define (miles->furlongs miles)
(* miles MILE->FURLONG))
(define MILE->FURLONG 8)
(define FURLONG->ROD 40)
(define ROD->YARD 5.5)
(define YARD->FOOT 3)
(define FOOT->INCH 12)
(define INCH->CM 2.54)
;; feet->cm : number -> number
;; convert feet into cm
(define (feet->cm feet)
(inches->cm
(feet->inches feet)))
;; yards->cm : number -> number
;; convert yards into cm
(define (yards->cm yards)
(feet->cm
(yards->feet yards)))
;; rods->inches : number -> number
;; convert rods into inches
(define (rods->inches rods)
(feet->inches
(yards->feet
(rods->yards rods))))
;; miles->feet : number -> number
;; convert miles into feet
(define (miles->feet miles)
(yards->feet
(rods->yards
(furlongs->rods
(miles->furlongs miles)))))
Exercise 3.3.2.
Develop the program volume-cylinder. It consumes the radius of a cylinder's base disk and its height; it computes the volume of the cylinder.
Solution
;; volume-cylinder : number number -> number
;; calculate volume of cylinder
;; example: (volume-cylinder 4 4) -> 201.06176
(define (volume-cylinder radius height)
(*
(area-circle radius)
height))
;; area-circle : number -> number
;; calculate area of circle
;; example: (area-circle 4) -> 50.26544
(define (area-circle radius)
(* PI (sqr radius)))
(define PI 3.14159)
Exercise 3.3.3.
Develop area-cylinder. The program consumes the radius of the cylinder's base disk and its height. Its result is the surface area of the cylinder.
Solution
;; area-cylinder : number number -> number
;; calculate surface area of cylinder
;; example: (area-cylinder 4 4) -> 201.06176
(define (area-cylinder radius height)
(+
(* 2
(area-circle radius))
(* height
(circumference-circle radius))))
;; circumference-circle : number -> number
;; calculate circumference of circle
;; example: (circumference-circle 4) -> 25.13272
(define (circumference-circle radius)
(* 2
(* PI radius)))
;; area-circle : number -> number
;; calculate area of circle
;; example: (area-circle 4) -> 50.26544
(define (area-circle radius)
(* PI (sqr radius)))
(define PI 3.14159)
Exercise 3.3.4.
Develop the function area-pipe. It computes the surface area of a pipe, which is an open cylinder. The program consumes three values: the pipe's inner radius, its length, and the thickness of its wall.
Develop two versions: a program that consists of a single definition and a program that consists of several function definitions. Which one evokes more confidence?
Solution
;; area-pipe number number number -> number
;; calculate surface area of open pipe
;; example: (area-pipe-multi 3 5 0.1) -> 195.4697298
(define (area-pipe-multi inner-radius length thickness-wall)
(+
(area-pipe-wall inner-radius length)
(area-pipe-wall
(+ inner-radius thickness-wall)
length)
(*
2
(-
(area-circle (+ inner-radius thickness-wall))
(area-circle inner-radius)))))
;; area-pipe-wall : number number -> number
;; calculate area of pipe wall
;; examples:
;; (area-pipe-wall 3 5) -> 94.2477
;; (area-pipe-wall 3.1 5) -> 97.38929
(define (area-pipe-wall radius length)
(* (circumference-circle radius)
length))
;; area-circle : number -> number
;; calculate area of circle
;; examples:
;; (area-circle 3) -> 28.27431
;; (area-circle 3.1) -> 30.1906799
(define (area-circle radius)
(* PI (sqr radius)))
;; circumference-circle : number -> number
;; calculate circumference of circle
;; examples:
;; (circumference 3) -> 18.84954
;; (circumference 3.1) -> 19.477858
(define (circumference-circle radius)
(* 2 PI radius))
(define PI 3.14159)
;; area-pipe : number number number -> number
;; calculate surface area of open pipe
(define (area-pipe radius length thickness-wall)
(+
(* 2 3.14159 radius length)
(* 2 3.14159
(+ radius thickness-wall)
length)
(* 2
(-
(* 3.14159 (sqr
(+ radius thickness-wall)))
(* 3.14159 (sqr radius))))))
Obviously, the solution with the multiple definitions evokes more confidence, because you can develop a higher order function before getting into the details of the lower order functions.
Exercise 3.3.5.
Develop the program height, which computes the height that a rocket reaches in a given amount of time. If the rocket accelerates at a constant rate g, it reaches a speed of g * t in t time units and a height of 1/2 * v * t where v is the speed at t.
Solution
;; height : number -> number
;; calculate height of rocket at certain time
;; example: (height 3) -> 45
(define (height time)
(* 0.5 (speed time) time))
;; speed : number -> number
;; calculate speed of rocket at certain time
;; example: (speed 3) -> 30
(define (speed time)
(* G time))
(define G 10)
Exercise 3.3.6.
Recall the program Fahrenheit->Celsius from exercise 2.2.1. The program consumes a temperature measured in Fahrenheit and produces the Celsius equivalent.
Develop the program Celsius->Fahrenheit, which consumes a temperature measured in Celsius and produces the Fahrenheit equivalent.
Now consider the function
;; I : number -> number
;; to convert a Fahrenheit temperature to Celsius and back
(define (I f)
(Celsius->Fahrenheit (Fahrenheit->Celsius f)))
Evaluate (I 32) by hand and using DrScheme's stepper. What does this suggest about the composition of the two functions?
Solution
;; Fahrenheit->Celsius number -> number
;; examples:
;; (Fahrenheit->Celsius 32) -> 0
;; (Fahrenheit->Celsius 212) -> 100
(define (Fahrenheit->Celsius F)
(* (- F 32) (/ 5 9)))
;; Celsius->Fahrenheit number -> number
;; examples:
;; (Celsius->Fahrenheit 0) -> 32
;; (Celsius->Fahrenheit 100) -> 212
(define (Celsius->Fahrenheit C)
(+ (* C (/ 9 5)) 32))
;; I : number -> number
;; to convert a Fahrenheit temperature to Celsius and back
(define (I f)
(Celsius->Fahrenheit (Fahrenheit->Celsius f)))
(I 32) is Celsius->Fahrenheit applied to (Fahrenheit-Celsius 32)
(Fahrenheit-Celsius 32)
evaluates to
(/ (* (- 32 32) 5) 9)
(/ (* 0 5) 9)
(/ 0 9)
0
(Celsius-Fahrenheit 0)
evaluates to
(+ (/ (* 0 9) 5) 32)
(+ (/ 0 5) 32)
(+ 0 32)
32
DrScheme's Stepper:
(I 32)
(Celsius->Fahrenheit
(Fahrenheit->Celsius 32))
(Celsius->Fahrenheit
(Fahrenheit->Celsius 32))
(Celsius->Fahrenheit
(* (- 32 32) (/ 5 9)))
(Celsius->Fahrenheit
(* (- 32 32) (/ 5 9)))
(Celsius->Fahrenheit
(* 0 (/ 5 9)))
(Celsius->Fahrenheit
(* (/ 5 9)))
(Celsius->Fahrenheit (* 0 5/9))
(Celsius->Fahrenheit (* 0 5/9))
(Celsius->Fahrenheit 0)
(Celsius->Fahrenheit 0)
(+ (* 0 (/ 9 5)) 32)
(+ (* 0 (/ 9 5)) 32)
(+ (* 0 9/5) 32)
(+ (* 0 9/5) 32)
(+ 0 32)
(+ 0 32)
32
The constants 32 and 5/9 could be incorporated in both as variables.
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