Calculus I, Fall 2015

Quiz 4

Nov 26, 2015

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1. Find lim_{x→0}(\frac{1}{x} − \frac{1}{sin x}).

Ans:

lim_{x→0}(\frac{1}{x} − \frac{1}{sin x}) = lim_{x→0} \frac{sin x−x}{x sin x}(8 pts)
= lim_{x→0} \frac{cos x−1}{sin x+x cos x}(8 pts) = lim_{x→0} \frac{− sin x}{2 cos x−x sin x}(4 pts) = 0.

2. Find the point on y = \sqrt{x}, x ≥ 0 that is closest to (2, 0). Explain why the answer you have is actually a global minimum. Ans: Let D^{2} = (x − 2)^{2} + (y − 0)^{2} = (x − 2)^{2} + x on y = \sqrt{x} and x ≥ 0.(5 pts) Then \frac{dD^{2}}{dx} = 2x − 3, hence x = \frac{3}{2} is a critical point.(5 pts)

Moreover \frac{dD^{2}}{dx} < 0 if x < \frac{3}{2} and \frac{dD^{2}}{dx} > 0 if x > \frac{3}{2}.(5 pts)

Hence D^{2} has absolute minimum at x = \frac{3}{2}, and (\frac{3}{2}, \sqrt{\frac{3}{2}}) is the point on y = \sqrt{x} which is closest to (2,0).(5 pts)

3. Write down Newton’s method that can be used to find \sqrt[3]{2}. Need not give the numerical value. Ans: Let f(x) = x^{3} − 2, then the root of f(x) = 0 gives x = \sqrt[3]{2}. Newtons’ method: Start with a reasonable x_{0}, (for example x_{0} = 1), then iterate x_{n+1} = x_{n} − \frac{x_{n}^{3}−2}{3x_{n}^{2}}.(10 pts)

4. Express ∫_{1}^{2} \frac{1}{1 + x^{2}} dx as a limit of Riemann sum (with uniformly spaced partition

and c_{k} of your choice). Then find the limit of the definite integral using fundamental Theorem of Calculus. Ans:

(1) lim_{n→∞} ∑_{k=1}^{n} \frac{1}{1+c_{k}^{2}} \frac{1}{n} (8 pts) for some c_{k} ∈ [\frac{k−1}{n}, \frac{k}{n}] (2 pts).

(2) Since (tan^{−1} x)′ = \frac{1}{1+x^{2}} (8 pts), by FTC we can know that the definite integral is tan^{−1}(2) − tan^{−1}(1) (2 pts).

5. Evaluate \frac{d}{dx} ∫_{x^{2}}^{0} \sqrt{1 + t^{4}} dt.

Ans: = −\frac{d}{dx} ∫_{0}^{x^{2}} \sqrt{1 + t^{4}} dt (4 pts) = −\sqrt{1 + x^{8}} · \frac{d}{dx}(x^{2}) (12 pts) = −\sqrt{1 + x^{8}} · (2x) (4 pts).