Calculus Test 2

Instructor: Prof. A. Mingarelli
Feb. 18, 1997

NOTATION: Pi = 3.14159... is our usual Greek number

1. Find the derivative of the following functions:

Solution:

Part 1: Use the Chain Rule and the sum-rule: Then y' = 2x / (1 + x4) + x / (sqrt(1-x2)) + Arcsin(x).

Part 2: In this case use the Fundamental Theorem of Calculus...Note that y '(x) = sqrt(sin(x2)). Thus, y ' (0) = 0.

2. Cobalt 60 is a radioactive isotope tat is used extensively in radiology. It has a half-life of 5.27 years. If an initial sample has a mass of 400 grams, how many years will it take for 95 % of it to decay?

Solution:Let N(t) = Cekt be the amount of material at time t (in years). The C = N(0). Now N(5.27) = N(0)/2 and this gives "k" as k = - ln(2)/5.27. So, N(t) takes the form N(t) = N(0)/2t/5.27 = 400/ 2t/5.27. If we want 95 % of it to decay then there must be 5 % left...but 5 % of 400 grams is 20 grams, right? Thus, we want to find a time t so that N(t) = 20. Solving for "t" gives t = (5.27) ln(20)/ln(2) years ~ 22.78 years.

3. Evaluate the definite integral of the function f(x) = |x-5| between the limits x = 2 and x = 7.

Solution: Note that |x-5| = x-5 if x>=5 and |x-5| = -(x-5) = 5-x if x-5 < =0, by definition. So the range of of integration, [2,7], needs to be broken down into two intervals: one is [2,5] and the other is [5,7]. Thus the original integral becomes a sum of 2 integrals, namely, the integral of (5-x) over the interval [2,5] and the integral of (x-5) over the interval [5,7]. These are easily evaluated to give 13/2.

4. Find the area under the curve y = x - sin(x) between the values x = 0 and x = Pi/2.

Solution: First we check that y >= 0 in the interval [0,Pi/2]. This is easy because its derivative y ' = 1 -cos(x) >= 0 in this interval so that y is increasing and since y(0) = 0, it follows that y is not negative. Next, the area is given by the integral of the function x - sin(x) over the interval [0,Pi/2]. This is not a difficult integral ...it has x2/2 + cos(x) as an antiderivative which we evaluate between the limits x=0 and x=Pi/2. This gives Pi2/8 as the area.

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