School of Mathematics and Statistics
Carleton University
Math. 69.107
SOLUTIONS TO ASSIGNMENT 2
Due: November 20, 1998,
or November 19, 1998 (Dr. Pandey only)

1. [10] Krypton 87 is a radionuclide with a half-life of 1.27 hours. Determine how long it would take for a 5 gram sample to decay to 0.6 grams.

   Solution: Use the Decay Law,

   

   where T is the half-life of the material. In this case N(0) = 5 and T = 1.27. Thus,

   

   and since we want N(t) = 0.6 we need to solve for t, that is,

   

   and so

   

2. [10] Use graph paper to sketch the graph of the function f defined by

   

   Be sure to include all asymptotes and find all critical points, maxima and minima and points of inflection, if any.

   Solution: The first derivative is given by

   

   The critical points are given by the set of numbers

   

   where the two terms involving square roots are the zeros of the quadratic (obtained by the quadratic formula) in the numerator of the expression and the other numbers arise since is not defined at those points.

   The SDT of is given by

   

   Note the ``minus sign'' in the expression for . We have just found the SDT for so we need to change all the signs in the last column to find the correct sign of . As a result we find that f is increasing when -0.87 < x < 1.53 and decreasing when or .

   Next, the second derivative is given by

   

   To test the nature of the critical points at we just calculate

   

   and so gives a local maximum. Similarly we can find that

   

   and so there is a local minimum here.

   Furthermore, there are three vertical asymptotes at and x=1 and one horizontal asymptote, namely y=0, at both since

   

   Finally there are no points of inflection since whenever or 1 < x < 2 and whenever -2 < x < 1 or .

   For example, it suffices to show that the numerator in . To see this note that if , then and this factors as . Its critical points are . The Second Derivative Test shows that x=0 is a local maximum, x=-1/2 is a local minimum and x=3/2 is a local minimum. The values of p(x) at these points are p(0) = 20, p(-1/2) = 153/8, and p(3/2)= 25/8. This means that 25/8 must be the global minimum value of p(x) and so p(x) > 0 for every x. Since the numerator can never be equal to zero it follows that too.

   The graph then looks like:

   

3. Evaluate the following integrals.

   [5] a)

   

   [5] b)

   

   Solution: a) Let , . Then

   

   b) Note that

   

   Let so . When x=1, u = 1 and when x=4, u=2. So the first integral becomes

   

   The second integral is easy and

   

   Combining these two answers we get

   

   
   

4. [10] Evaluate

   

   Use Leibniz's Rule. We know that

   

   Simplifying and passing to the limit (using extended real number arithmetic) we find that the second term tends to 0 and so

   

Total: /40

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