Calculus I, Fall 2021

Brief solutions to Quiz 9

Jan 04, 2022

1. Solve for y′ + (tan x)y = cos x, y(0) = 1.

Ans: Multiply the equation by exp(∫ tan x dx) = \frac{1}{cos x}. \frac{1}{cos x}y′ + \frac{sin x}{cos^{2} x}y = 1, (\frac{y}{cos x})′ = 1, \frac{y}{cos x} = x + C, y(0) = 1, therefore C = 1. y = (x + 1) cos x.

2. Evaluate ∫_{1}^{2} \frac{cosh(ln t)}{t}dt.

Ans: = ∫_{1}^{2} cosh(ln t)d(ln t) = sinh(ln t)|_{1}^{2} = sinh(ln 2)(= \frac{3}{4}).

3. Order the functions n, \sqrt{n} log_{2} n, (log_{2} n)^{5} by rate of growth as n → ∞. You can use the notations a ≪ b to say b grows faster than a, and a ≈ b to say a and b grow at the same rate. Ans: Use L’Hôpital’s Rule to find lim_{n→∞} \frac{ln n}{n^{\frac{1}{8}}} = 0 ⟹ lim_{n→∞} \frac{(ln n)^{4}}{n^{\frac{1}{2}}} = 0 ⟹ \sqrt{n} ≪ (log_{2} n)^{4}.

Therefore (log_{2} n)^{5} ≪ \sqrt{n} log_{2} n.

Similarly, from L’Hôpital’s Rule lim_{n→∞} \frac{ln n}{n^{\frac{1}{2}}} = 0, therefore \sqrt{n} log_{2} n ≪ n.

Answer: (log_{2} n)^{5} ≪ \sqrt{n} log_{2} n ≪ n.