Calculus: The Language of Change

Exercise Set 1.3
Linearity   in Local Coordinates

by Terry Sargent

Exercise   1

Local  Coordinates   for  a  Line
(a)   Suppose  a  line   goes   through   the  point  (5, -4) and  has  slope  -3.   Define  local coordinates   ( dx,  dy )    with  origin  at  the  point  ( 5 ,  - 4 ) ,  giving  formulas  for  dx  and  dy  in  terms  of  x  and  y.
      What   is the  (  dx , dy )  equation  for  this  line?   
      What   is  the  (x,  y)  equation  of  the  line?
      What   is  the  (x,  y)  slope - intercept  for  ( y  =  m x  +  b)  of  this  line?

(b)  Suppose  a  line  goes  through  the   fixed  point ((Subscript[x, 0 ]) , \(InputForm\`\(\ y\_\(0
	\)\)\))  and  has  slope  m.
       What   are  the  local  ( dx ,  dy )  coordinates  through   ( (Subscript[x, 0  ]), \(InputForm\`\(\ y\_\(0
	\)\)\) ) ?
       What   is  the  ( dx ,  dy )  equation  of the  line ?
       What   is  the  ( x ,  y )  equation  of  the  line?
       What   is  the  (x ,  y)  slope - intercept  form  ( y  = m  x  +  b )  of  this  line    ( b  = ? ) ?

Solution  to  (a)

Let us  look  over   at  definition  1 . 1   on  page  8  again to  get our  first  hint  of  how  to  go  about  solving  this  first  part  of  ( a ) .   The  formula   \(InputForm
      \`\(
	Δy\ \)\/\(Δx
	\) = \ m\)  is  mathematical   short  hand  for saying  that  the  ratio   of  the  change  in  y  over  a  corresponding  change  in  x will always  equal  the  same constant  m (for a linear equation). The  quantities  Δx and   Δy of  Definition  1 . 1  represent any  differences , not  necessarily  those  based  at  ( 5 ,  -4 ) .  So  substitute!   Let's  get rid of the  general   by replacing   it with the  specifics of our problem.   We  are  now  ready  to  find  our  local  coordinates  ( new  x ,  new  y )  =  ( dy ,  dy )  through the  point ( 5 ,  - 4 )  .  Since  we  realize  that  dx  represents  a  change  in  x from  the  point  5  and  dy  represent  a  change  in  y  from  - 4 .   So the  equation   \(InputForm
      \`\(
	Δy\)\/\(Δx
	\)\  = \ m\)   can  be put  into  this  equivalent   form   \(InputForm\`\(
	dy\)\/dx\  = \ m\) .   With  a  little   more  algebra , our  equation  can be rewritten  once  again  as   dy  = m  dx .  We now  have  the  final  form  of  our  answer  once  we    substitute  m  = -3   into  the   equation .   So   dy  =  - 3 dx.

Next  we  make  use  of  one  of  the  intermediate  forms  from  our  previous question
 \(InputForm\`\(
	dy\)\/\(dx
	\)\  = \ m\)   to  answer  our  next  question .   Definition  1 . 2  states that  the  local  coordinates through  the  fixed ( x , y )  point  ( (Subscript[x, 0  ]),  (Subscript[y, 0  ]))  are   given  by
 
    dx  =  x -  (Subscript[x, 0 ])                                               dy =  y -  (Subscript[y, 0])
    
  So  if  we    replace   dy  and  dx  by  their  definition  in  terms  of  x  and  y
we   get   (( 
	y - (Subscript[y, 0]))/( x - (Subscript[x, 0 ]))) = m .   Once  again ,  let's  plug  in  our  values  to  this  equation   
 ((
	 y -(-4)  )/( x-5))  =  -3   to   answer  our  question.

Again,   let's   make  use  of  our  very  last  result   (( 
	y -(-4)  )/( x-5
	)) = -3  and   then   apply  a little algebra   to  rewrite   it  into  a  more  recognizable  form

 y  - ( - 4 )   =  - 3   ( x - 5 )                   ->                           y   +  4  = - 3 x   +  15   
 
  y  =  - 3  x    + 11   is  now  in  the  familiar  slope - intercept   form  ( y  =    m  x  +  b ).

Solution   to  (b)

Let's  go  ahead  and  make   use   of   those   specific examples   that   we  worked  out   for  part  (a).

local  coordinates  are    \(InputForm\`\(
dy\)\/\(dx
	\)\  = \ m\)

Our  ( dx ,  dy )  equation  is   dy  =  m  dx.

The   (x ,  y )  equation  of  the  line  is:
\(InputForm\`\(
dy\)\/\(dx
	\)\  = \ m\)        =        ((y - (Subscript[y, 0]))/( x - (Subscript[x, 0 ]))) = m     ->   
  y -\(\ y\_\(0
	\ \)\)=  m (  x  -  (Subscript[x, 0  ]))

The  slope  intercept  form   y  =  m  x  +  b   is:

 y  -  \(y\_\(0
	\ \)\)=  m  (   x -  (Subscript[x, 0  ]))   or     y  =  m  x  + \(\ \ y\_\(0
	\)\)  -  m  (Subscript[x, 0 ])
This  small section   of   the   equation \(InputForm\`y\_\(\ \ 0
	\)\) -  m  (Subscript[x, 0 ])  gives  us  a  constant ,  so  just  go  ahead  and  call  it   b.

Exercise   2

Point  -  Slope   Form   to   Slope  -  Intercept   Form
   (a)    Algebraically   simplify   the   expression   (( 
	y - 3)/( x + 4))   =   -2    showing   that   it  is  equivalent  to  the  expression  y  =  -2  x  -5  (when  x  ≠  -4)
i)   This  line  has  slope  -2.   Use  both  formulas  to  show this.
ii)  This  line  contains  the  point  (3, -4).   Use  both formulas  to  show  this.
iii) This  line  crosses  the  y-axis  at   y  =  -5.   Use  both  formulas  to  show  this.

(b)   Simply  the   following  expressions  and   give  slope and  y - intercept:
     
i)  (( 
	y - 2  )/( x - 3)) = -4         ii)  (( 
	y - 2  )/( x + 3)) = 4      iii)   (( 
	y + 2  )/( x - 3)) = -4 
iv)  (( 
	y   )/( x - 3)) = 1          v)   ((
	y - 3  )/( x )) = -2     vi)     (( 
	y   )/( x )) = 3

(c)  Simplify  the  expression   ((
	 y - (Subscript[y, 0]))/( x - (Subscript[x, 0 ]) ))= m,  showing  that   it  is  equivalent  to  y  =  m  x  +  b
 ( when  x  ≠   (Subscript[x, 0  ])) ,   where  b  =  \(InputForm\`y\_\(0
	\)\)  -  m  (Subscript[x, 0.])    What   is  the  slope  of  this  line?   
  Use  both  formulas  to show  that   the  line   contains   the   point ((Subscript[x, 0 , ])(Subscript[y, 0])).

Solution   to  part (a)

We   want  to  show  that  the   equation (( 
	y - 3)/( x + 4
	)) = -2   is   equivalent

to   y = - 2  x  - 5  ( when  x ≠ -4) .   First  let's  start by  multiplying  our  first  expression  by     x + 4 and  we  will end   up  with  y - 3 =  - 2  (x +4).  This  is  closer  to  desired   form that  we  want   but  we are  not  quite  there  yet!

We  now  have  y - 3 =  - 2 (x +4)  so  let's  multiply  through  on  the  right  side of  the   equation,  to yeild  y-3 = - 2 x -8,  now even closer  to  our  goal. But if we then jump  the  -3  from  the  right  and  in  the  process  change  its  sign before  it  lands,  we  get 
y= -2 x -8 + 3   or    y = -2 x -5.  In the  first form  of  our  equation  we  were  warned  not  to  let   x = -4,  for if  this were  to  happen  the  equation  would  become    ∞ = -2,  which  is not  true.

Now,  we  use  the  second  form  of  our  equation   y = -2 x -5  and  plug  some  x  values into  it.  When we  plug in x = { 1, -1}  we get  y= -7  and  y = -3.  Now let's plug both  the  x and  corresponding  y values  into  the  first  form ((
	y - 3)/(  x + 4
	
	)) = -2     This gives us ((
	 -7- 3)/( 1 + 4
	)) = -2    which   is  true.  Now make  use  of the
x ,y  pair  (-1, -3)  and plug it into   ((
	 -3 - 3)/( -1 + 4)) = -2  This  is  also true.  We  have  just  utilized  both  equations  to  show  that  each  has a  slope of -2.

Let's  now evaluate the point  (3,-4)  using  the first form  of  our  equation.   ((
	 y - 3)/( x + 4)) = -2    gives 
 ((	 
	-4- 3)/( 3 + 4)) = -2    or   ((
	-7)/( 7)) = -2    or   -1 = -2   which is nonsense.   
 
So the   point  (3, -4)   supplied  by  the  textbook    is wrong, it's   not  on  our   line!   Let's  find  one  that   is.    We  can  do  this  by taking   y = -2 x -5.  By putting   in   x = 0   into the   equation   we get y =  -5.    So  our  line  will cross  the   y-axis  at  the  point (0, -5) .   Setting  y = 0,  gives us   x =  -5/2  and our   line   will  cross  the  x-axis  at  the  point  (-5/2,  0) .  Plugging  either  of  these  points  into  the  first  form  of  our equation gives us the correct  value  for  the  slope, -2.

Using the form y = -2 x -5, the point (-4,3) gives us 3 = 8 - 5, so the textbook probably intended the point to be (-4,3) rather than (3,-4).

Take  the  first  form  of  our  equation  ((
	 y - 3)/( x + 4
	)) = -2 and  then  substitute  x =0  and   y = -5   into it.   Which   give us (( 
	-5 - 3)/(  4
	)) = -2   or (( 
	-8)/(  4
	)) = -2  and  in  it's  final  form   -2 = -2.    And if we plug x = 0 into   the   other   form of our   equation   y = -2 x -5  we   get   y = -5.    This shows that  the  line from our equation   crosses the    y-axis  at    the  point  (0,  -5).

Solution   to part (b)

In  part  (b),   we have  to algebraically  rewrite  each  one of our  problems  into  the  form  y = m x + b.  In this form m is the slope and b is the y-intercept.

i)  Can  be  rewritten  as  y  =  -4 x  +  14.  Here  the  slope is  -4  and  setting  x  =  0  gives  a  y-intercept  of  14 (this  means  that  our  equation  crosses  the  y-axis  at  the point  (0,14).

ii)  Can  be  rewritten  as  y  = 4 x + 14.  Here  the  slope  is  4  and  by  setting  x = 0  gives an y-intercept  of 14.

iii)  Can  be  rewritten  as  y = -4 x + 10.  Here  the  slope  is -4  and  by  setting  x = 0  we  get  an y-intercept  of 10.

iv)   Can  be  rewritten  as  y = x - 3.  Here  the  slope  is  1  and  by  setting  x = 0,  we  get  an y-intercept  of -3.

v)  Can  be  rewritten  as  y  =  -2  x  +  3.  Here  the  slope  is  -2 and  by  setting   x  =  0,  we  get  an  y-intercept   of  3.

vi)  Can  be  rewritten  as  y  =  3 x.  Here  the  slope  is  3  and  by  setting x  =  0,  we  get  an  y - intercept  of  0.   (in  this  case  our  line  will  go  through  the  origin)

Solution   to  part (c)

We  have  been asked  to  show  that  these  two  equations  are   equivalent :  
              ((
	 y - (Subscript[y, 0]))/( x - (Subscript[x, 0 ]) )) =  m     and      y   =  m    x   +   b

Multiplying the equation on the left  by   x - (Subscript[x, 0, ]) gives us:
 
y - \(y\_\(0
	\)\)\(InputForm\`\(
\( = \ \ m\)\)\) 
(x - (Subscript[x, 0) ])

Distributing  the  m:
 
y - \(y\_\(0
	\)\)\(InputForm\`\(
\( = \ \ m\)\)\) 
x - (Subscript[mx, 0 ])

And finally:

 y  =   m x - (Subscript[mx, 0 ])\(InputForm\`\(\ \(+\ y\_\(0
	\)\)\)\)

By setting:

   b   =  \(y\_\(0
	\)\) - (Subscript[mx, 0 ])  
We have:
    
     y   =  m    x   +   b     as desired.  As expected, the slope of this line is m.
     
     We can now use both formulas to show that the line contains the point ((Subscript[x, 0 , ])(Subscript[y, 0])).  
     
Using:     
y - \(y\_\(0
	\)\)\(InputForm\`\(
\( = \ \ m\)\)\) 
(x - (Subscript[x, 0 ]))
we can set x = (Subscript[x, 0]) and solve for y (we should get y = (Subscript[y, 0])):
	
	y - (Subscript[y, 0])  =  m ( (Subscript[x, 0]) - (Subscript[x, 0])) = 0

so	y  =  (Subscript[y, 0])   as desired.

Using:
	    y  =   m x - (Subscript[mx, 0 ])\(InputForm\`\(\ \(+\ y\_\(0
	\)\)\)\)
and as above we can set x = (Subscript[x, 0]) and solve for y:	    

	y = m (Subscript[x, 0])  - m (Subscript[x, 0])  + (Subscript[y, 0])  =  (Subscript[y, 0 ])

So     y = (Subscript[y, 0])  as desired.

Exercise   3

Two   points   Determine   a  Line

     (a)   Suppose  a  line  goes through  the  points  (3,-2)  and (-4, 5).  Find  its  slope.
i)    Use  the  slope  and  the  point  (3,2)  to  find  its  (x, y)  equation.
ii)   Use  the  slope  and  the  point  (-4, 5)  to  find  its  (x, y)  equation.
iii)  Compare  the  two  equations  by  putting  both  in  slope-intercept   form.

     (b)   Find  the  slope-intercept   equation  of  the  line  that  passes  through:   
i)   (1,2)  and  (3,4)     ii)   (2,-1)  and (-3,4)     iii)  (-2,3)  and  (5,7)

Solution   to (a)

First,  we  need  to   compute  the  slope  from  the  given  points  (x, y)  =  (3, -2)   and                 (x,  y)  =  (-4, 5)
\(InputForm
      \`\(\(
	Δy\ \)\/\(Δx
	\) = \ 
        constant\ \ \ \ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ \ \ \)\)\(TextForm
      \`\(\(
\ \ \ \ \ \ \ \ \ 	5 - \((\(-2\))\)\ \)\/\(\(-4\)\  - 3	\) = 
        \ \ \(-1\)\ \)\)

Then  let's  use  the  point  (3,  -2)  and  our  slope  m  =  -1   to find  the  (x, y)  equation :

y -(-2) = -1 ( x - 3)      or       y =  - x  + 1

Now  use  the  point (-4,  5)  and  slope   m =  -1   to  again  find the  ( x,  y)  equation:

y -(5) = -1 ( x - (-4))      or     y =  - x  + 1

We  started   working  with  the  point - slope  form  from  the  start,   once    we  found  our  slope  m.

 y  -\(\ \ y\_\(0\ 
	\ \)\)=   m  (  x  -  (Subscript[x, 0   ]))   then  we   plugged in  both  points    (3, -2 ) and  then  (-4, 5)   separately  into  the  point - slope  form  then  by  doing  a  little  algebra  we were able to get  everything  into  slope intercept  form   y  =  m  x  +  b  or  in  this  case   either    point  gave  us    y  =   - x   +  1

Solution   to (b)

i)   Either point   (1,  2)  or  (3, 4)   will  give  us  the  slope - intercept  equation  of   y  =  x  +  1.

ii)    Either point (2, -1)  or  ( -3,  4)  will  give  us the  slope -intercept  equation  of   
 y =  -x  +  1.
  
iii)   Either point  (-2, 3)  or  (5,  -7)   will  give  us the  slope - intercept  equation  of   
y  =  -10/7 x  +  1/7.

Exercise   4

Graphics   to  Symbolics
Find   an   equation   for  each  of  the  graphs   shown   below.

Solution   to  (a)

By selecting any two  points  that  are  on  the  line,  we  can  determine the slope  for  the  equation.  Then  use  either  of  these  two  points  and  write the  equation  in  the  slope - intercept  form.   Let's take  as  our  (x, y)  points   (-1, 2)  and  (0, 0) .  These  two  points  give  us  a slope of -2 and an  equation   dy = -2 dx.

Solution   to (b)

The  points   (-1,  -1.5)  and  (0, 0)  give  us  the  equation  of the  line:   dy  = \(TextForm\`\(
	3  dx\)\/2\).

Solution  to (c)

The  points    (1.5,  -1)   and   ( 0, 0)   give   us   the  equation   of  the   line    dy  = \(TextForm\`\(
	\(-2\) dx\)\/\(\ 3\)\).

Solution  to (d)

The  points    (2,  0.5)   and   ( 0,  0)   give   us   this  equation   of  the   line    dy  = \(TextForm\`\(
	dx\)\/\(\ 4\)\).

Exercise   5

Numerics  to  Symbolics

    (a)  Find  an  equation  satisfied  by  all  the  points  on  Table  1.1
    (b)  (Optional)  Solve  this  with  the  computer  using  the program   Linear Intro.

Solution   to the  first part

We   want  to  make   use of  the   x  and   y  values  found  in  Table  1.1 :  Linear  data  located  on  page   9  of  our  textbook.  The   data   contained   in  Table  1.1  is  just  a  listing  of  (x,y)  pairs.   It's  easy  to verify  by using  the  Definition  1.1  on  page  8  that   this   data  is  linear.   (For those  of  us  with  really short  term  memories  this  means - that   the ratios  of  all  changes  equal  the  same  constant).   We  can rewrite  our  table  in  a  slightly  different  form  as a  set  of  (x,y)  pairs:
 
 {{1,3},{2,1},{5,-5.},{6.3,-7.6}}

We  can then  select any  two  points  in  the   set. We  know  that   two  points determine   a  slope by taking  the   y   change  over  the   x  change.  No  matter  which  points  we choose  we  get  a  slope  of  -2.  Any one of the  points  together   with  this   slope  then   gives the   point-slope  equation   for  a  linear  function.   So   without  too  much  difficulty  we  come up with  this  equation  from  the   first  two  points  for  our   table:

                               y = 5-2 x

But  how  can  we  check  to  see  if  is  really  the correct   equation?   Let's  plug in  all  the   x  values  from  our  set  of  (x,y)  pairs  into  this  equation.   What   should  we  get? Of  course  we get each  of the   corresponding   y   values {3,1,-5,-7.6}.  The  whole  set  of  data  truly is  linear   if  the  function  determined  by  the  first two  points  matches   at   all  the  points.

Solution  to  the  second part

\(Clear[xList, yList, pairs]; 

      \(xList = {1, 2, 5, 6.3}; \)
\(yList = {3, 1, \(-5\), \(-7.6\)}; \)

          pairs = Transpose[{xList, yList}]; 
Clear[x, y, m, x0, y0]; 

      \(y[x_] := y0 + m\ \((x - x0)\); \)

        \(x0 = xList[[1]]; \)

        \(y0 = yList[[1]]; \)

        \(m = \(yList[[2]] - 
              yList[[1]]\)\/\(xList
                [[2]] - 
              xList[[1]]\); \)

        Print["The first two points say  y = ", Simplify[y[x]]]; 

      Print["Function y[xList]=", y[xList]]; 

      Print["Original yList = ", yList]; 
\)

Exercise   6 - computer exercise.

Exercise   7

Newton's Law of Cooling says that  the  rate of cooling of an object is proportional  to the difference  between the  temperature  of the object and the temperature  of its surroundings.
Let F denote  the  Farenheit   temperature  of the  object and t denote the  time in minutes begining with t=0 at our first observation.  (See Chapters  4 and 8.)

(a)   Suppose an object is  (75^∘) when we first observe it and  it cools to  (60^∘) ten minutes  later.  
           What   is its (average)  rate  of cooling over this time period.

(b)   Sketch the line segment passing through the point  
            (t, F) = (0, 75) and (10, 60).

(c)   What  is the  slope of this  segment?

(d)   What is the equation  of the line through  these two points?

Newton's Law of Cooling does not give the temperature  as a function of the time, but rather  say, 
R = k(F - (Subscript[F, 0]) ), where R is the  rate  of cooling,    (Subscript[F, 0])  is the temperature   of the surroundings, and k is a constant.  The line through the points above is not a good long-term   prediction of temperature because its slope is constant   and that   slope is the   rate   referrred   to in Newton's Law.  Calculus will give us a simple exponential  formula.

Solution   to (a)

We can attack this problem much like we have previous ones.  Here, the rate of cooling can be expressed as the change in temperature divided by the change in time, or:

\(TextForm\`\(

    \(Temp\_final - Temp\_initial\)\/\(Time\_final - Time\_initial\)\)\)

which, with the appropriate values becomes:

((60 - 75)/(10 - 0)) = 1.5 degrees per second

Solution  to (b)

Solution   to (c)

Our  slope  is  -3/2  which  we  obtained  from  the  two points 
 (t, F)  =  (0,  75)  and  (10,  60)  and  by  using  the  General  (x, y)  point - slope  formula

((y1  -  (Subscript[y, 0]))/(x1  -  (Subscript[x, 0]))) =  m

\(InputForm
      \`\(y1\  - \ y\_0\)\/\(x1 - \ x\_0\) = \ 
        \(\(60\  - \ 75\)\/\(10\  - \ 0\) = \ 
          \(\(-15\)\/10 = \ \(-3\)\/2\)\)\)=  -1.5

Solution  to (d)

Our equation for this line is

y = -(3/2) x + 75

To obtain this, we use the slope and the formula

((y  -  (Subscript[y, 0]))/(x -  (Subscript[x, 0]))) =  m

((y  - 75)/(x  - 0)) =  ((-3)/2)

Clear[x,y];
y = -3x/2 + 75;
line = Plot[y,{x,-1,11},AxesLabel->{"x","y"},PlotRange->{50,90},PlotStyle->{RGBColor[1,0,0]}];
Show[line,Graphics[{AbsolutePointSize[5],{Point[{0,75}],Point[{10,60}]}}]];

Exercise   8

The  downward  velocity  v  of  an  object  released  in  vacuum  near  the  surface  of  the  earth  
satisfies  a  linear  equation,  v  =  gt  + (Subscript[v, 0]) ,  where  g  is  Galileo's  universal  gravitational   constant.

g  = \(TextForm\`\(
	9.8\ m\)\/sec\^2\)   or     g  = \(TextForm\`\(
	32\ ft\)\/sec\^2\) .

(See Chapter 10.)

(a)   Suppose  we  first  observe  a  falling  object  traveling  13 \(TextForm\`\(
	\ ft\)\/sec\^2\)  downward.   How  fast  will  it  be  falling  2  seconds  later?

(b)   Suppose  we  first  observe  an  object  thrown  13 \(TextForm\`\(	\ 
	ft\)\/sec\^2\)  upward.   How  fast  will  it  be  falling  2  seconds  later?

Solution   to  (a)

First,  don't  get  excited.   We are  already  pros  at  finding  the  kind  of   answer  that  this  problem  asks of us.
Take  the  equation   v   =   gt   +\(TextForm\`\(\ \ \ v\^\(
0\)\)\)   and  plug  in   values  for  g,  t  and \(TextForm\`\(\ v\^\(
0\)\)\).    So  the  equation  now  becomes  v  =  32 (0)  + 13   or  v  =  +13  at  time  t  =  0.   Do this again,  but  change  the  time  to  t=2  to  get   v = 32 (2)  +13   or   v  =  77  at  time  t = 2.  Remember,  a positive velocity represents an object which is moving downward

Solution  to  (b)

This problem can be addressed in much the same way as the previous one.  Now, however, we must assign a different starting velocity.  Since we said that  a positive velocity is an object traveling downward, it must hold that the opposite is also true. To use the equation for an object which is initially traveling upward, simply change the sign of the initial object's velocity.

At t = 0 this becomes

v= 32(0)-13 = -13

For t = 2

v= 32(2)-13 = 51(ft/sec)