Calculus II Review Exam with Questions and Answers, Exams of Computer Science

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MATH 125G

Calculus II Review

Exam

Q & A

1. A ladder 10 meters long rests on horizontal ground and leans against a vertical wall. The foot of the ladder is pulled away from the wall at the rate of 2 m/s. How fast is the top of the ladder sliding down the wall when the foot of the ladder is 6 m from the wall? (Use trigonometric functions to model the situation and find the answer using differentiation techniques) A) - 1.6 m/s B) - 2.4 m/s C) - 3.2 m/s D) - 4 m/s Answer: C) - 3.2 m/s Rationale: Let x be the distance of the foot of the ladder from the wall and y be the height of the top of the ladder on the wall. Then, by Pythagoras' theorem, x^2 + y^2 = 10^2. Differentiating both sides with respect to time t, we get 2x dx/dt + 2y dy/dt = 0. We are given that dx/dt = 2 m/s and we want to find dy/dt when x = 6 m. Substituting these values into the equation, we get 2(6)(2) + 2y dy/dt = 0, which implies that y dy/dt = - 12. To find y, we use x^2 + y^2 = 10^2 again and get y = sqrt(10^2 - 6^2) = 8 m. Therefore, dy/dt = - 12/8 = - 3/2 m/s when x = 6 m.
2. Evaluate the integral ∫(sin(x))^3 dx from 0 to π/4 using integration by parts. A) (1/3)(sqrt(2) - 1) B) (1/3)(sqrt(2) + 1) C) (1/3)(sqrt(2) - 2) D) (1/3)(sqrt(2) + 2) Answer: A) (1/3)(sqrt(2) - 1) Rationale: To use integration by parts, we need to choose u and dv such that du and v are easy to find. A good choice is u = (sin(x))^3 and dv = dx, since du = 3(sin(x))^2 cos(x) dx and v = x. Then, by the formula ∫udv = uv

• ∫vdu, we get ∫(sin(x))^3 dx = x(sin(x))^3 - ∫x(3(sin(x))^2 cos(x)) dx. To

D) It diverges by the ratio test Answer: A) It converges to 1/ Rationale: To test the convergence of the series, we can use the comparison test with a simpler series that has the same behavior as n goes to infinity. A good choice is ∑(n=1)^∞ 1/n^2, which is a p-series with p = 2 > 1, so it converges. To compare the two series, we need to find the limit of the ratio of their terms, which is lim(n->∞) ((n^2 + 1)/(n^4 + 2n

• 3))/(1/n^2) = lim_(n->∞) (n^4 + n^2)/(n^4 + 2n + 3). This limit is also an indeterminate form of type ∞/∞, so we can use L'Hopital's rule and differentiate both numerator and denominator four times with respect to n. This gives lim_(n->∞) (24)/(24) = 1. Since this limit is positive and finite, the comparison test tells us that both series either converge or diverge together. Since we know that ∑(n=1)^∞ 1/n^2 converges, we can conclude that ∑(n=1)^∞ (n^2 + 1)/(n^4 + 2n + 3) also converges. To find its sum, we can use the formula for the sum of a p-series, which is ∑_(n=1)^∞ 1/n^p = π^(p)/6 for p > 1. Since our series has p = 2, its sum is π^(2)/6. However, we need to multiply this by the limit of the ratio of the terms that we found earlier, which is 1. Therefore, the sum of the series is π^(2)/6 * 1 = π^(2)/6.

1. Which of the following trigonometric functions is equal to the square root of (1 - cos^2x)? A) sinx B) cosx C) tanx D) secx Answer: A) sinx Rationale: Using the Pythagorean identity sin^2x + cos^2x = 1, we can rewrite the given expression as sinx.
2. The integral of e^(3x) dx is:

A) (1/3)e^(3x) + C B) e^(3x) + C C) (3/2)e^(3x) + C D) (1/3)e^(3x^2) + C Answer: A) (1/3)e^(3x) + C Rationale: The integral of e^(3x) with respect to x results in (1/3)e^(3x) + C, where C is the constant of integration.

3. By using integration by parts, solve the integral of xcos(x) dx. A) (xsin(x) + cos(x)) + C B) - xcos(x) + sin(x) + C C) (xcos(x) - sin(x)) + C D) (xsin(x) - cos(x)) + C Answer: B) - xcos(x) + sin(x) + C Rationale: By applying integration by parts, the integral of xcos(x) dx can be solved as - xcos(x) + sin(x) + C.
4. Which of the following expressions represents the indeterminate form 0/0? A) lim x->0 sinx/x B) lim x->1 (x^2 - 1)/(x - 1) C) lim x->∞ x^2 + 1/x D) lim x->π/2 tanx Answer: A) lim x->0 sinx/x Rationale: The expression sinx/x is equivalent to 1 when x approaches 0, leading to the indeterminate form 0/0.
5. Determine the sum of the infinite series Σ(n=1 to ∞) (2/3)^n. A) 3 B) 2 C) 1/ D) 1

c) 1 d) 1/√ Answer: a) √3/ Rationale: The exact value of sin(π/3) is √3/2, which is the y-coordinate of the point on the unit circle corresponding to the angle π/3. Question 2: If tan(θ) = 3/4, what is the value of cos(θ)? a) 3/ b) 4/ c) 5/ d) 5/ Answer: b) 4/ Rationale: Given tan(θ) = 3/4, we can use the Pythagorean identity to find cos(θ) = 4/5. Applications of Integration: Question 3: A tank is in the shape of an inverted circular cone with height 10 meters and base radius 5 meters. If the tank is being filled with water at a rate of 3 cubic meters per minute, how fast is the water level rising when the water is 6 meters deep? a) 0.2 m/min b) 0.3 m/min c) 0.4 m/min d) 0.5 m/min Answer: c) 0.4 m/min Rationale: This problem can be solved using the concept of related rates and the formula for the volume of a cone. The rate at which the water level is rising can be found by differentiating the volume formula with respect to time. Techniques of Integration:

Question 4: Evaluate the integral ∫(2x + 3)/(x^2 + 3x + 2) dx. a) ln|x + 2| + 3tan⁻¹(x + 1) + C b) 2ln|x + 2| + 3tan⁻¹(x + 1) + C c) 2ln|x + 1| + 3tan⁻¹(x + 2) + C d) ln|x + 1| + 3tan⁻¹(x + 2) + C Answer: a) ln|x + 2| + 3tan⁻¹(x + 1) + C Rationale: This integral can be solved by using partial fraction decomposition and the trigonometric substitution method. Indeterminate Forms: Question 5: Find the limit as x approaches 0 of (sinx - x)/(x^3). a) 0 b) 1 c) - 1 d) ∞ Answer: a) 0 Rationale: By applying L'Hôpital's rule, the limit of (sinx - x)/(x^3) as x approaches 0 can be evaluated to be 0. Infinite Series: Question 6: What is the sum of the infinite series ∑(2^n)/(3^(n-1)) from n=1 to ∞? a) 1/ b) 3/ c) 3 d) 2/ Answer: c) 3 Rationale: This series is a geometric series with first term 2/3 and common ratio 2/3, thus the sum can be found using the formula for the sum of an infinite geometric series.