Calculus: The Language of Change

Problem CD-3.1
Symbolic Magnification for an Unknown y=f[x]

by  Sharon Portz

Problem CD-3.1

Explain the following statements about figure CD-3.7

1.The vertical distance from the dx-axis up to the curve is
			f[x + Δx] - f[x] 
but the magnified view of this vertical segment measures
			\(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\)

2. The vertical distance from the dx-axis up to the line dy = m dx above the point dx = Δx is 
			mΔx
but the magnified view of this vertical segment measures
			m

3. The magnified gap ϵ measures
			ϵ = \(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\) - m
but the actual change in the function as x moves to x + Δx is
			f[x + Δx] - f[x] = m Δx + ϵ Δx

Solution  to 1

The key to explaining this statement is to remember the fact that when you look at figure CD-3.7 you can describe each point on the graph using two coordinates.  We have the global coordinates (x, y) and the local coordinates (dx,dy).  The explanation of the vertical distance between the dx coordinate and the curve is an expression of global coordinates.

Choose a point on the function (for reference) that is to the right of the dx,dy origin.  This point, in local coordinates, is (dx, dy).  However, in global coordinates it is (x+ Δx, f(x + Δx)).  Why?  Because the dx,dy origin lies at some point (x,y).  To give the location of our reference point, we move Δx units to the right of the dx,dy origin.  That gives us a global x-value of x+Δx.  To get the global y-value, we plug x+Δx into the function f[x].  Thus we have (x+ Δx, f(x + Δx)).

The vertical distance we are trying to find deals only with the y-values. The y-value of the dx axis is f[x].  Confused?  Just keep in mind that the dx,dy origin lies at the point (x,y) and the y-value at this origin is f[x].  So the y-value (or height) of the dx axis is f[x].  From above we know that the y-value in global coordinates is f(x + Δx).  Therefore, if we take 
			f(x + Δx) - f(x)  = f[x + Δx] - f[x]
we find the vertical distance between the curve and the dx axis.

So why is the maginfied view \(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\)?  Remember that when we magnified we multiplied the function by (1/Δ)x.  From above we have f[x + xΔ] - f[x], multiply that by (1/Δ)x and we get
\(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\).

Solution   to 2

Keep in mind what was said above about the global and local coordinates.  Here we are concerned with local coordinates.  The line we are concerned about is dy=mdx.  Choose some point on the dx axis (for reference).   Its coordinates are (Δx, 0). Plug Δx into the equation dy=mdx.  This gives you the dy-value above the point (Δx, 0).  Since we are working with local coordinates, dx has height of 0, dy is the vertical distance between the line and the dx axis.  Therefore dy = mΔx and 
mΔx is the difference between the dx axis and the line.

To find the magnified vertical distance we multiply by (1/Δ)x.  Thus we have ((1/Δ)x)(mΔx) = m.

Solution   to 3

Let's look at what \(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\)- m really means.  It is the vertical distance from the dx-axis to the curve in the magnified graph minus the vertical distance from the dx-axis to the line dy = mdx, in the magnified graph.  To put it simply, it is the vertical difference between the curve and the line in the magnified graph.  And we call it ϵ.

The actual change in the function as x moves to x+Δx is f[x+Δx] - f[x].  This makes sense intuitively because it subtracts the old value of y from the new value of y.  (Assuming that 
f[x+Δx] = new y and f[x] = old y.)  Now, why does f[x+Δx] - f[x] = mΔx + ϵΔx?

We know that	ϵ = \(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\)- m    	If we multiply both sides by Δx and then add mΔx to both sides we obtain the desired result:
		f[x + Δx] - f[x] = m Δx + ϵ Δx