MATH 180 FALL 1996

ASSIGNMENT #19

3.4, #10. Change in income = ∫_{0}^{12} r(t)dt = ∫_{0}^{12} 40(1.002)^{t}dt = $485.80

3.4, #14. Notice that each square on the graph represents 10/6, or 5/3 miles. At t=1/3 hours, v = 0. The area between the v graph and the t-axis over the interval 0 < t < 1/3 is − ∫_{0}^{1/3} vdt which is about one square or 5/3 miles. At t = 1/3 she is about 5 − 5/3 or about 10/3 miles from the lake. v is positive on the interval 1/3 ≤ t ≤ 1, so she is moving away from the lake on that interval. About 8 squares are between the v graph and the t−axis. At t = 1,

∫_{0}^{1} v dt = ∫_{0}^{1/3} v dt + ∫_{1/3}^{1} v dt ≈ −\frac{5}{3} + 8 · \frac{10}{6} = \frac{35}{3},

and the cyclist is about 5 + \frac{35}{3} = \frac{50}{3} = 16\frac{2}{3} miles from the lake. Since, starting from the moment t = \frac{1}{3}, she moves away from the lake, the cyclist will be farthest away from the lake at t = 1. The maximal distance is 16\frac{2}{3} miles.

3.5, #2.

lim_{t→∞} t^{3}e^{−t} = 0

A viewing window with 0 ≤ x ≤ 20 (or larger x) and 0 ≤ y ≤ 2 suggests these limits. (In fact, t^{n}e^{−1} → 0 as t → ∞ for any n > 0. This expresses in another way the fact that exponential functions dominate power functions.)

3.5, #3.

lim_{x→∞} \frac{sin x}{x} = 0

A viewing window with 0 ≤ x ≤ 20 (or larger x) and −2 ≤ y ≤ 2 suggests these limits. It is easy to give a direct argument why the limit is 0. The numerator sin x is always between −1 and 1 while the denominator x gets arbitrarily large as x → ∞.

3.5, #5. If the exponent were a fixed number, like 3, we could bring the limit inside the expression that is raised to the exponent. However, the exponent n is not fixed, but rather is going to infinity at the same time that the 1+\frac{1}{n} is going to one. So we cannot bring the limit inside the exponent.

Now, if we were to ask the calculator to evaluate (1 + \frac{1}{n})^{n} for n = say 10^{20}, it will try to evaluate 1 + \frac{1}{10^{20}} and it will have no choice but to round off the answer to exactly 1. Then it will take 1^{10^{20}}, which is 1.