Calculus: The Language of Change

Exercise Set 6.8
Review - Inside the Microscope

by Michelle Nguyen

Exercise 1

A Partial  View of the "Bell Shaped" Curve:

The derivative  of the function  f[x] = (e^(-(x^2))) is f'[x]= -2 (xe^(-(x^2))) (as you may verify using rules of differentiation .)  This question asks, So what? (or what does this tell us mathematically?)  You answer it as follows:  Draw microscopic views of the graph 
y= (e^(-(x^2)  ))when the microscope is focused on the graph over the x-points, x= 0, ±0.1, ±1, ±10.  Give the numerical  values of the derivatives and sketch the slopes to scale on equal axes.

Solution

Evaluate the cells below to obtain the value of f'[x] and the graphs.

a) x= 0, then  f'[0]=-2 *0* (e^(-(0^2)))

f'[0]= 0*1

f'[0]=0

\(Clear[f, x, dx, dy]; 
f[x_] := \ E\^\(-x\^2\); 

    dy\  = \ \(f'\)[x]\ dx; 
Print["The graph of ", f[x]]; 

    Print[" with differential dy = ", dy]; 
x\  = \ 0; 

    Print["and its tangent dy = ", dy]; 

    Print["at (x,y)= (", x, ",", f[x], ")"]; 

    Plot[{f[x + dx] - f[x], dy}, {dx, \(-1\), 1}, 
      AxesLabel -> {"dx", "dy"}, 
      PlotStyle -> {{Thickness[0.02], Red}, Blue}]; \)

b) x= 0.1, then  f'[0.1]=-2* 0.1* (e^(-(((0.1))^2)))

\(\(N[\ \(-2\)*\ 0.1*\ E\^\(-\((0.1)\)\^2\)]
\)\)

\(Clear[f, x, dx, dy]; 
f[x_] := \ \ E\^\(-x\^2\); 

    dy\  = \ \(f'\)[x]\ dx; 
Print["The graph of ", f[x]]; 

    Print[" with differential dy = ", dy]; 
x\  = \ 0.1; 

    Print["and its tangent dy = ", dy]; 

    Print["at (x,y)= (", x, ",", f[x], ")"]; 

    Plot[{f[x + dx] - f[x], dy}, {dx, \(-1\), 1}, 
      AxesLabel -> {"dx", "dy"}, 
      PlotStyle -> {{Thickness[0.02], Red}, Blue}]; \)

c) x = -0.1 then,  f'[-0.1]=-2*(- 0.1)* (e^(-(((-0.1))^2)))

N[-2*(- 0.1)* (E^(-(((-0.1))^2)))]

\(\(
Clear[f, x, dx, dy]; 
f[x_] := \ E\^\(-x\^2\); 

    dy\  = \ \(f'\)[x]\ dx; 
Print["The graph of ", f[x]]; 

    Print[" with differential dy = ", dy]; 
x\  = \(-\ 0.1\); 

    Print["and its tangent dy = ", dy]; 

    Print["at (x,y)= (", x, ",", f[x], ")"]; 

    Plot[{f[x + dx] - f[x], dy}, {dx, \(-1\), 1}, 
      AxesLabel -> {"dx", "dy"}, 
      PlotStyle -> {{Thickness[0.02], Red}, Blue}]; 
\)\)

d) x=1, then f'[1]=-2 *1*(e^(-(((1))^2)))

   f'[1] = -2*(e^(-1))
   
   f'[1] =((-2)/e)

\(\(N[\(-2\)\ *1*E\^\(-\((1)\)\^2\)]
\)\)

\(\(
Clear[f, x, dx, dy]; 
f[x_] := \ E\^\(-x\^2\); 

    dy\  = \ \(f'\)[x]\ dx; 
Print["The graph of ", f[x]]; 

    Print[" with differential dy = ", dy]; 
x\  = 1.0; 

    Print["and its tangent dy = ", dy]; 

    Print["at (x,y)= (", x, ",", f[x], ")"]; 

    Plot[{f[x + dx] - f[x], dy}, {dx, \(-1\), 1}, 
      AxesLabel -> {"dx", "dy"}, 
      PlotStyle -> {{Thickness[0.02], Red}, Blue}]; 
\)\)

e) x=-1, then f'[-1]= - \(\(2\ \((\(-1\))\)\ e\^\(-\((\(-1\))\)\^2\)
\)\)
    f'[-1] = 2* \(TraditionalForm\`\(e\(^\^\(-1\)\)\)\)
    
    f'[-1]=(2/e)

N[- 2 (-1) (E^(-(((-1))^2)))]

\(\(
Clear[f, x, dx, dy]; 
f[x_] := \ E\^\(-x\^2\); 

    dy\  = \ \(f'\)[x]\ dx; 
Print["The graph of ", f[x]]; 

    Print[" with differential dy = ", dy]; 
x\  = \(-1.0\); 

    Print["and its tangent dy = ", dy]; 

    Print["at (x,y)= (", x, ",", f[x], ")"]; 

    Plot[{f[x + dx] - f[x], dy}, {dx, \(-1\), 1}, 
      AxesLabel -> {"dx", "dy"}, 
      PlotStyle -> {{Thickness[0.02], Red}, Blue}]; 
\)\)

f) x=10, then  f[10]= -2 *10 *(e^(-(10^2))) = -(20/(e^100))

\(\(
N[\(-2\)\ *10\ *E\^\(-10\^2\)]\)\)

\(\(
Clear[f, x, dx, dy]; 
f[x_] := \ E\^\(-x\^2\); 

    dy\  = \ \(f'\)[x]\ dx; 
Print["The graph of ", f[x]]; 

    Print[" with differential dy = ", dy]; 
x\  = 10.0; 

    Print["and its tangent dy = ", dy]; 

    Print["at (x,y)= (", x, ",", f[x], ")"]; 

    Plot[{f[x + dx] - f[x], dy}, {dx, \(-1\), 1}, 
      AxesLabel -> {"dx", "dy"}, 
      PlotStyle -> {{Thickness[0.02], Red}, Blue}]; 
\)\)

g) x= -10, then  f[-10] = -2 (-10) (e^(-(((-10))^2)))=(20/(e^100))

\(\(N[\(-2\)\ \((\(-10\))\)\ E\^\(-\((\(-10\))\)\^2\)]
\)\)

\(Clear[f, x, dx, dy]; 
f[x_] := \ \ E\^\(-x\^2\); 

    dy\  = \ \(f'\)[x]\ dx; 
Print["The graph of ", f[x]]; 

    Print[" with differential dy = ", dy]; 
x\  = \(-\ 10.0\); 

    Print["and its tangent dy = ", dy]; 

    Print["at (x,y)= (", x, ",", f[x], ")"]; 

    Plot[{f[x + dx] - f[x], dy}, {dx, \(-1\), 1}, 
      AxesLabel -> {"dx", "dy"}, 
      PlotStyle -> {{Thickness[0.02], Red}, Blue}]; \)

Exercise 2

a

Find the  (x,y) -equation  of the tangent  to  y = (x^3) at  x = -2.

Solution  of part a

From page 107 of the text:

To find the (x,y) equation of the tangent line, do the following steps.

a) Compute  f'[x]  by rules :

 		y= (x^3) ⇒ (dy/dx)= 3 (x^2)

b) Substitute  the  fixed  x = 
        a (x = -2, in this case) to\(find\ the\ numerical\ \ value\ of\ the\ \ slope\ \ m = 
          \(\(f'\)[x] : 
		
		\(f'\)[\(-2\)] = 
            \(3*\((\(-2\))\)\^2 = \ 
              \(m
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 

                \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = \ 3*\ 4\)\)\)\)                                               m   = 12

c) Compute the specific y value, y= f[x] at the point of tangency.  This gives you a specific point (x,y) = (a,b) that  lies on both the curve and tangent line:

For x = -2, y= (x^3) = -8

So  (a, b) =  (-2,  -8)

d) Change the local equation of the line dy= mdx to the point-slope (x,y) form of a line ((y - b)/(x - a)) = m and simplify to the slope-intercept  form  y= mx+ i:

To find i:  -8 = 12(-2) +i
		i= 16
So, the  slope-intercept  form  is
		y= 12x+ 16

b

Find the (x, y) - equation of the tangent   to  y =  
          Sin[x] at  x  =  (π/(3 . ))

Solution of part b

Taking  the derivative  of y =  Sin[x](dy/dx)  =  Cos [x]

\(TextForm\`\(

At\ x\  = π\/3\ , \ \ 
    dy\/dx = \ \(Cos[π\/3]\  = \ 1\/2\)\)\)

so  m = (1/2)
Calculate  the  value of y : \(
y = \ Sin\ [\ π\/3]\) = ((Sqrt[3])/2)

Slope - intercept    form  :  y =  mx + i\(Now\ we\ determine\ \ the\ \ value\ of\ \(i : 
		
\@3\/2\) = 
        1/2*π\/3\  + \ i\)((Sqrt[3])/2)  -  (π/6)  =  i
Slope -  Intercept form is :    y =  
                (1/2)*x  +  ((Sqrt[3])/2) - (π/6)

c

Find the (x,y)-equation  of  the  tangent   to y= Log[x] at  x = 1.

Solution of part c

Taking the derivative:

(dy/dx)= (1/x)

For   x=1 ⇒  (dy/dx)= 1

So  m=1

y[1] = Log[1]

y[1] = 0

(a,b)= (1,0)

y= mx + i

0 = 1(1) + i

-1 = i

Slope - Intercept form:    y= x -1

d

Plot these pairs of curves with the computer program Tangents.

Solution of part d

Evaluate the cells below

A)

\(Clear[f, x, dx, dy]; 
f[x_] := \ x\^3\ ; 
dy\  = \ \(f'\)[x]\ dx; 

    Print["The graph of ", f[x]]; 

    Print[" with differential dy = ", dy]; 
x\  = \ \(-2\); 

    Print["and its tangent dy = ", dy]; 

    Print["at (x,y)= (", x, ",", f[x], ")"]; 

    Plot[{f[x + dx] - f[x], dy}, {dx, \(-1\), 1}, AspectRatio -> 1, 
      AxesLabel -> {"dx", "dy"}, 
      PlotStyle -> {{Thickness[0.02], Red}, Blue}]; \)

B)

\(Clear[f, x, dx, dy]; 
f[x_] := \ Sin[x]\ ; 
dy\  = \ \(f'\)[x]\ dx; 

    Print["The graph of ", f[x]]; 

    Print[" with differential dy = ", dy]; 
x\  = \ π\/3; 

    Print["and its tangent dy = ", dy]; 

    Print["at (x,y)= (", x, ",", f[x], ")"]; 

    Plot[{f[x + dx] - f[x], dy}, {dx, \(-1\), 1}, PlotRange -> {\(-2\), 2}, 
      AxesLabel -> {"dx", "dy"}, 
      PlotStyle -> {{Thickness[0.02], Red}, Blue}]; \)

C)

\(Clear[f, x, dx, dy]; 
f[x_] := Log[x]\ ; 
dy\  = \ \(f'\)[x]\ dx; 

    Print["The graph of ", f[x]]; 

    Print[" with differential dy = ", dy]; 
x\  = \ 1; 

    Print["and its tangent dy = ", dy]; 

    Print["at (x,y)= (", x, ",", f[x], ")"]; 

    Plot[{f[x + dx] - f[x], dy}, {dx, \(-1\), 1}, PlotRange -> {\(-2\), 2}, 
      AxesLabel -> {"dx", "dy"}, 
      PlotStyle -> {{Thickness[0.02], Red}, Blue}]; \)

Exercise 3

Differential Approximation  :

Approximate     
      ((1, 000, 000, 000, 000, 002)^(1/3)) - ((1, 000, 000, 000, 000, 000)^(1/3)) using the differential  increment  of the function f[x]= (x^(1/3)),   x= 1, 000,000,000,000,000 and δx=2  (Two is not infinitesimal, but it is small compared to (10^15).  Computers have a very hard  time with this kind of computation because they work in fixed length decimal approximation.)  We only need a simple way to estimate  f[x+δx]- f[x], since we know f[x]= 100,000 when  x=(10^15).

How many decimals of your approximation are accurate  in this case?  Try it with your calculator or the computer;  you will get the wrong answer unless you wok with very very high- precision arithmetic  .  The differential  is very accurate.

Solution

We follow the procedure from page 107 of the text.

For:		f[x]= (x^(1/3))

		f'[x]=(1/(3 (( (x^2))^(1/3))))

When		x = (10^15)
              
		f'[x]=(1/(3 ((((((10^15)))^2))^(1/3))))
              
		m  =  f'[(10^15)]=(1/30000000000)

		df[x] = m dx = (2/30000000000)

So		f[x + dx] - f[x] ≈ d f[x] =  (2/30000000000)

\(\(N[2/30000000000, 20]
\)\)

How close is our answer?

\(\(N[1000000000000002^\((1/3)\), 35] - \ 100000
\)\)

Our approximation appears to be accurate to 25 decimals.