Calculus: The Language of Change

Exercise Set CD 3.1
Algebra of Microscopes  - CD

by Sharon Portz

Exercise   1

1.  Let ϵ = 0.33 and Δx = 0.01.  Draw a line with a scale of 1 unit = 1 cm on the left half of your paper.  Put dots at x=0, Δx=0.01, and ϵΔx = 0.0033.  (What  is the difficulty when you draw in centimeter   units?)  On the right half of your paper, draw a line starting at 0 magnified by 100.  How far in centimeters  on the magnified picture  is the dot for Δx and ϵΔx?

Solution

The problem with drawing such a line in cm units is that they are very small and it is hard to measure such tiny measurements like 0.01cm and 0.033cm.  The Δx dot on the magnified picture should be 1.0 cm from the x=0 dot.  The ϵΔx dot should be 3.3 cm from the x = 0 dot.  And, the distance between the Δx dot and ϵΔx dot is 3.3 cm - 1.0 cm = 2.3 cm.

Exercise   2

Your first magnification

1. Sketch the graph of y=(x^3) for -1.5 ≤ x ≤ 1.5 using equal x and y scales with 1 cm = 1 unit.

2. We want to focus our microscope over the  x-value x=2/3.  What is the corresponding y-value?

3. Draw the local (dx,dy)-axes with its origin at the (x,y)-point, (x,y)=(2/3, 8/27) on the same graph.  The (dx,dy) origin lies at the center of your microscope.  We will think of x as fixed and vary dx and dy.

4. The straight line in local coordinates dy=(4/3)dx is tangent to y=(x^3) at x=2/3.  Sketch this as a dotted line on the same graph.

5. Magnify a portion of your graph by 10 and draw the microscopic view with the same scale you used for your original (x,y)-plot.  This way (1/10) will appear unit size, that is, measure 1 cm on your graph paper.

Your result from Exercise CD-3.3.2 should look something like the figure CD-3.6.  Compare your work to the computer animation in the program SecantGapZ.

Solution   to  2  part 1

The graph should look like this :

Solution   to  2  part 2

The value at  x = (2/3) of y = (x^3) is (8/27).  This can be seen by tracing the graph or calculating it directly using the function.

Solution   to  2  part 3

The graph should look like this:

Notice here that the origin is at the point (x,y) = ((2/3),(8/27))

Solution   to  2  part 4

The graph below combines the original origin and axes with the new origin and axes and the line tangent at the point (x,y) = ((2/3),(8/27)).

Solution   to  2  part 5

Your graph should look like the one on the right .

Exercise   3

Measuring the Specific Gap

Carefully draw the curve y=(x^3) between the values x=2/3 and  x+Δx = 2/3 + 1/10 on a scale where 1/10 of a unit is 1 cm, that  is, magnified by 10 if 1 cm is the original unit.  The curve y=(x^3) lies above your tangent line dy=4/3 dx, where the local dx-dy axes lie at the center of the "microscope" magnifying the figure.

1. What are the (x,y)-corrdinates  of the point where (dx,dy)=(0,0)?

2. The x-coordinate of the point dx= 1/10=0.1 is 0.7666..., because dx=0 corresponds to x=2/3=0.6666...  What is the y-coordinate on y=(x^3) when x=0.7666...? What is the change in y from the dx axis to the point on the curve above x=2/3 + 1/10?

3. What is the change in dy from dy=0 to the point on dy=(4/3)dx above dx=Δx=1/10?

4. What we see in the microscope as ϵ on our centimeter scale is the magnified difference between the answer to Part 2 and the answer to Part 3, so its actual size is given by applying the formula from Exercise 1 to this difference.  How much is it?

Solution   to  3  part 1

The x,y-coordinates of the point (dx, dy) = (0,0) are ((2/3), (8/27)).  They can be found by plugging
x = (2/3) into the function y =(x^3).

Solution   to  3  part 2

The y-coordinate is 0.4506.  We get this by plugging the x-coordinate, 0.7666= 2/3 + 1/10 into the function y = (x^3). The change in y can be calculated by subtracting the new y-value of 0.4506 from the original y-value of (8/27) so   0.4506 -  (8/27)  = 0.1543.

Solution   to  3  part 3

For this part of the problem we use the equation dy = (4/3)dx.  To find the change in dy we must first find out what dy is when dx is (1/10).  So dy = ((4/3))((1/10)) = (2/15) = 0.1333.  To get the change in dy we subtract the old dy from the new dy.  This is 0.1333 - 0 = 0.1333.

Solution   to  3  part 4

We subtract the answer to part 3 from the answer in part 2.   0.1543 - 0.1333 = 0.0210.   On a magnified graph we would obtain this by multiplying ϵ times Δx.

Exercise   4

A Gap in Function Notation

Let f[x]=(x^3) in the following computations, and verify that the function procedures produce the answers to the measurement of ϵ as in the examples and exercises above, especially Exercise CD-3.3.3.  Basic units are measured in centimeter, whereas magnified units appear 10 times larger.

1. When x=2/3 and the (dx,dy)-coordinates are centered on y=f[x] above this value of x, then the (x,y)-value of (dx,dy)=(0,0) is (x,f[x])=(2/3,f[2/3])=(2/3, ??).

2. On the sketch of the portion of y=f[x] between x=2/3 and x+Δx = 2/3 + 1/10 with a scale of 1/10 unit = 1 cm, the length of the vertical segment from the dx-axis to the (x,y)-point
 (x + Δx, f[x + xΔ]) is 10 times the unit value of
 			f[x + xΔ] - f[x] = ((((2/3) + (1/10)))^3)- ((((2/3)))^3) = ??
 or simply,
 			\(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\)

3. If we sketch the line dy=mdx, with m=4/3, the length of the verical segment (in original units) between the dx-axis and the point above dx = Δx = 1/10 is dy = mΔx.  If we magnify by 1/Δx so that  Δx appears unit size or specifically, 1/10 unit equals 1cm, then this distance measures m = 4/3.

4. The magnified gap ϵ measures ϵ = \(TraditionalForm
      \`\(f[x\  + \ xΔ]\  - \ f[x]\)\/Δx\) - m

5. The actual unmagnified size of the gap between the curve and its tangent approximation above x + Δx  is
   f[x + xΔ] - f[x] - m Δx = ϵ * Δx

Solution   to  4  part 1

This is just like the problem above, only the notation is different.
(x,f[x]) = (2/3, f[2/3]) = (2/3, 8/27)

Solution   to  4  part 2

[((((2/3) + (1/10)))^3)- ((((2/3)))^3)]/0.1 =  1.543

Solution   to  4  part 3

As indicated, m = 4/3.

Solution   to  4  part 4

ϵ = \(TraditionalForm
      \`\(f[x\  + \ xΔ]\  - \ f[x]\)\/Δx\) - m = (0.1543/0.1)- 4/3 = 0.21

Solution   to  4  part 5

ϵ * Δx = (0.21)(0.1) = 0.021