Calculus: The Language of Change

Exercise CD-3.4
Symbolic Increments - CD

by Michelle Nguyen

Exercise Set  CD-3.4

1.  Check Your General Formula with y=(x^3)

When  f[x] =(x^3), the  microscopic gap is ϵ = δx × (3x + δx).

a. Use this formula for ϵ to check the  errors  you measured in Exercise CD-3.3.3 and Problem 3.3.

b. We are  interested  in the  size of the  gap ϵ at  x+ 1/100.  How much is this gap when in a microscope of power 100 focused at  the point x=2/3?  How much is it really?  Use your general formula  to show that  ϵ= 0.0201, and ϵ × (δx) = 0. 000201 (See Figure CD-3.9).

2. Your General Formula  with y= (x^4).  Let  f[x] =(x^4) and show that  the microscope equation 
                                          f[ x+ δx] - f[x] = f'[x]× δx + ϵ × (δx)
    becomes                                      
                                         (((x + δx))^4) - (x^4) = 
        4 (x^3)×δx + ϵ×δx
    with ϵ= δx ×(6(x^2) + 4xδx + δ(x^2)).  
    
    
3.  There  is only one possible line that can 'fit' a graph y=f[x] when magnified.  To see this, let  y= f[x]= (1/2)x  near  x=0, but suppose the approximation  we try is dy= dx, or m=1, so the microscope  equation is 
                           f[ x+Δx]-f[x]= Δx+ ϵ × Δx
Show that no matter what magnification, we always have ϵ = -1/2 above the point that appears to be one unit away from the intersection. (Do this by writing f[x] and f'[x] explicitly in the microscope equation and solving for ϵ.)

Solution  to 1

a) When  f[x] =(x^3), the  microscopic gap is ϵ = δx × (3x + δx).
We  know δx=(1/10), x=(2/3)  as given in Homework 3.3(ProbCD-1.3).nb.
So, we substitute these values into the formula above to get ϵ.

ϵ= (1/10)(3((2/3))+(1/10))

ϵ=(21/100)

b) We are given that δx= (1/100), and  x= (2/3)
From  the equation:  ϵ = δx × (3x + δx)  we obtain:
            ϵ =(1/100) ×(3 × (2/3) + (1/100))

            ϵ = (1/100) × (((200  + 1)/100))
            
            ϵ= (201/10000) = 0.0201
            
Then  ϵ × (δx) = 0.0201 x (1/100) = 0.000201

Solution   to 2

For  f[x] =(x^4) we expand:
f[ x+ δx] - f[x]	= (((x + δx))^4) - (x^4)
		=  (x^4) + 4(δx) (x^3)+ 6(((δx))^2)(x^2) +4(((δx))^3)x + (((δx))^4) - (x^4)
		= 4(δx) (x^3)+ 6(((δx))^2)(x^2) +4(((δx))^3)x + (((δx))^4)
		= 4 (x^3)(δx) + δx[6(δx)(x^2) +4(((δx))^2)x + (((δx))^3)]
		= f'[x](δx) + ϵ δx
with f'[x] =4 (x^3) and ϵ= δx ×(6(x^2) + 4xδx + δ(x^2))

Solution   to 3

f[ x+Δx]-f[x]= Δx+ ϵ × Δx
or		(1/2)( x+ Δx) - (1/2)x =   Δx+ ϵ × Δx
so		(1/2)x + (1/2)Δx -(1/2)x - Δx = ϵ × Δx
and		-(1/2)Δx  = ϵ × Δx
therefore	ϵ = -(1/2) 
So no matter what value we select for Δx, ϵ is always -(1/2)