Calculus: The Language of Change

Exercise Set 7.1
Differentiation with Parameters

by John Robeson

Exercise 1) Approximate Roots

(a)

Compute the derivative(dx/db) (considering a and c as parameters).  When is this defined?  In the cases when it is not defined, what is going on in the original root-finding problem?  Consider some special cases to help such as (a,b,c) = (1,-3,2), (a,b,c) = (1,-2,1), (a,b,c) = (0,-2,1).

Solution to (a)

The quadratic equation is
		(ax^2) + bx + c
and the largest root is
		x =((-b + (Sqrt[(b^2) - 4 ac]))/(2 a)) = -(b/(2  a)) + (1/(2  a)) (Sqrt[(b^2) - 4 ac]) 


so		(dx/db) = -(1/(2  a)) + (b/(2  a (Sqrt[ (b^2) - 4 ac]))) 

This is defined when:
		i)  a ≠ 0
and		ii)  (b^2) - 4 ac ≥  0

If		i)  a = 0, then we have (0/0) in the original 			root-finding equation.
If		ii)  (b^2) - 4 ac = 0, then the two roots are equal 		and if (b^2) - 4 ac ≺ 0, then there are no real roots.

Consider special case (a,b,c) = (1,-3,2):

		(dx/db) = -(1/(2  a)) + (b/(2  a (Sqrt[ (b^2) - 4 ac])))  = -(1/2) - (3/2) = -2

Consider special case (a,b,c) = (1,-2,1):

		(dx/db) = -(1/(2  a)) + (b/(2  a (Sqrt[ (b^2) - 4 ac])))  =-(1/2) - (2/(2*0))
This is undefined because (Sqrt[(b^2) - 4 ac]) = 0 and we are dividing by zero in the second term.

Consider special case (a,b,c) = (0, -2, 1):

		(dx/db) = -(1/(2  a)) + (b/(2  a (Sqrt[ (b^2) - 4 ac])))  = -(1/0) - (2/(0*2))
This is also undefined because we are again dividing by zero.

(b)

Consider the case (a,b,c) = (1,-3,2) and denote the error in measuring b by db.  Suppose the magnitude of db could be as large as 0.01.  Use the differential approximation to estimate the resulting error in the root.

Solution to (b)

Refer to the text page 107 to review differential approximations.

In our case

	(dx/db) = -(1/(2  a)) + (b/(2  a (Sqrt[ (b^2) - 4 ac])))  = -(1/2) - (3/2) = -2
	
and	dx = -2 db = -2*0.01 = -0.02
so the resulting error in the root could be as large as -0.02.

(c)

The following are for practice differentiating with respect to different letters.

Compute the derivative (dx/da) (considering b and c as parameters).  When is this defined?

Solution to (c)

x =((-b + (Sqrt[(b^2) - 4 ac]))/(2 a)) = -(b/(2  a)) + (1/(2  a)) (Sqrt[(b^2) - 4 ac]) 


(dx/da) = (b/(2 (a^2))) + 
        ((-4 c)/((2 a)* 2 (Sqrt[ (b^2) - 4 ac]))) + 
        (((-1) (Sqrt[(b^2) - 4 ac]) )/(2 (a^2)))

	= -(c/(a (Sqrt[(b^2) - 4 a c]))) + 
        ((b - (Sqrt[(b^2) - 4 a c]))/(2 (a^2)))

(d)

Compute the derivative (dx/dc) (considering a and b as parameters).  When is this defined?

Solution to (d)

x =((-b + (Sqrt[(b^2) - 4 ac]))/(2 a)) = -(b/(2  a)) + (1/(2  a)) (Sqrt[(b^2) - 4 ac]) 


(dx/dc) = 0 + 
        (( 1 (-4 a))/((2 a) 2 (Sqrt[ (b^2) - 4 ac])))


	= -(1/(Sqrt[(b^2) - 4 ac]))

(e)

Check your differentiation with the computer.

Solution to (dx/db):

\(\(

WalkD[\(\(-b\) + \@\(b\^2 - 4\ a\ c\)\)\/\(2\ a\), b]\)\)

Solution to (dx/da):

\(\(

WalkD[\(\(-b\) + \@\(b\^2 - 4\ a\ c\)\)\/\(2\ a\), a]\)\)

Solution to (dx/dc):

\(\(

WalkD[\(\(-b\) + \@\(b\^2 - 4\ a\ c\)\)\/\(2\ a\), c]\)\)

Exercise 2) Planck's Radiation Law

Planck's radiation law can be written

		I = a((ω^3)/((e^(bω/T)) - 1))
		
for constants a and b.  This expresses the intensity I of radiation ω for a body at absolute temperature T.  Suppose T is also fixed.  Express I as a chain or composition of functions (of the variable ω with parameters) and products of fuctions to which the rules of this chapter apply.  For example, you can use the Exponential Rule, ((d ((e^u)))/du) = (e^u).  What is the formula for (dI/d)ω?  Check your work with the DfDx program.

Solution to Exercise 2

If we let u[ω] = bω/T  and  v[ω] = a(ω^3), we have

		I = v[ω]((((e^u)[ω] - 1))^(-1))	
		
		u'[ω] = b/T 
		
		v'[ω] = 3  a(ω^2)
		
and by the Product Rule and Chain Rule

		(dI/d)ω = v[ω](-(((e^u)[ω] u'[ω])/(((((e^u)[ω] - 1))^2)))) + ((v'[ω])/((e^u)[ω] - 1))
		
Substituting u, v, u' and v' we obtain:

		= a(ω^3) (
        -(((e^(bω/T))  b/T 
              )/((((e^(bω/T)) - 1))^2))) + (( 3  a(ω^2))/((e^(bω/T)) - 1))

Checking this with the DfDx program, we obtain an equivalent answer:

\(\(u = \ b\ ω/T; 
v = \ a\ ω\^3; 

    D[v\ \((E\^u - 1)\)\^\(-1\), ω]
\)\)