BONUS Week 2 Homework

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Georgia Institute Of Technology *

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Mathematics

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Feb 20, 2024

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BONUS Week 2 Homework Due  Jan 26 at 8:59pm   Points  8   Questions  8   Available  Jan 19 at 5am - Jan 29 at 8:59pm   Time Limit  None Instructions Please answer all the questions below. This quiz was locked Jan 29 at 8:59pm. Attempt History Attempt Time Score LATEST Attempt 1 2 minutes 8 out of 8 Score for this quiz: 8 out of 8 Submitted Jan 25 at 2:05pm This attempt took 2 minutes.
Question 1 1 / 1 pts (Lesson 2.1: Derivatives.) BONUS: If �(�)=ℓ�(2�−3), find the derivative �′ (�). a. 2� b. 12ℓ�(2�−3) Correct! c. 2/(2�−3) This follows by the chain rule, �′(�)=[ℓ�(2�−3)]′=(2�−3)′2�−3=22�−2 d. �/2 (c). This follows by the chain rule, �′(�)=[ℓ�(2�−3)]′=(2�−3)′2�−3=22�−2 Question 2 1 / 1 pts (Lesson 2.1: Derivatives.) BONUS: If �(�)=cos(1/�), find the derivative �′ (�). a. cos(1/�2) b. sin(1/�2) c. −1�2sin(1/�) Correct! d. 1�2sin(1/�) By the chain rule, [cos(1/�)]′=−sin(1/�)[1/�]′=1�2sin(1/�) (d). By the chain rule, [cos(1/�)]′=−sin(1/�)[1/�]′=1�2sin(1/�) Question 3 1 / 1 pts
(Lesson 2.2: Finding Zeroes.) BONUS: Suppose that �(�)=�4�−4�2�+4. Use any method you want to find a zero of �(�), i.e., � such that �(�)=0. a. �=0 b. �=1 c. �=ℓ�(2)=0.693 Correct! d. �=12ℓ�(2)=0.347 This doesn't take too much work. Namely, set 0=�(�)=�4�−4�2�+4=(�2�−2)2. This is the same as �2�=2, or �=12ℓ�(2)=0.347 (d). This doesn't take too much work. Namely, set 0=�(�)=�4�−4�2�+4=(�2�−2)2. This is the same as �2�=2, or �=12ℓ�(2)=0.347 Question 4 1 / 1 pts (Lesson 2.3: Integration.) BONUS: Find ∫01(2�+1)2��. a. 1/2 b. 7/2 c. 7/3 Correct! d. 13/3 We have ∫01(2�+1)2��=(2�+1)36|01=276−16=13/3 (d). We have ∫01(2�+1)2��=(2�+1)36|01=276−16=13/3 Question 5
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1 / 1 pts (Lesson 2.3: Integration.) BONUS: Find ∫12�2���. a. 1 b. �2−� Correct! c. 23.6 We have ∫12�2�=12�2�|12=23.60 d. 46.2 (c). We have ∫12�2�=12�2�|12=23.60 Question 6 1 / 1 pts (Lesson 2.3: Integration.) BONUS: Find lim�→0sin(�)−��. a. 1 Correct! b. 0 If we let �(�)=sin(�)−� and �(�)=�, then lim�→0�(�)=0 and lim�→0�(�)=0, so that lim�→0�(�)�(�) seems to get us into a 0/0 issue. Thus, we'll need to employ L'Hôspital's Rule: lim�→0�(�)�(�)=lim�→0�′(�)�′(�)=lim�→0cos(�)−11=01=0. c. d. undetermined
(b). If we let �(�)=sin(�)−� and �(�)=�, then lim�→0�(�)=0 and lim�→0�(�)=0, so that lim�→0�(�)�(�) seems to get us into a 0/0 issue. Thus, we'll need to employ L'Hôspital's Rule: lim�→0�(�)�(�)=lim�→0�′(�)�′(�)=lim�→0cos(�)−11=01=0. Question 7 1 / 1 pts (Lesson 2.4: Numerical Integration.) BONUS: Find the approximate value of the integral ∫02(�−1)2�� using the lesson's form of the Riemann sum with �(�)=(�−1)2,�=0,�=2, and �=4. a. -2 b. 1/3 Correct! c. 3/4 We have ∫02(�−1)2��≈�−��∑�=1��(�+(�−�)��)=24∑�=14(2�4−1)2=3/4 Well, this is sort of close to the true integral of 2/3. Of course, we could've done even better if � had been bigger or if we had used the midpoint of each interval instead of the right endpoint. d. 3 (c). We have ∫02(�−1)2��≈�−��∑�=1��(�+ (�−�)��)=24∑�=14(2�4−1)2=3/4 Well, this is sort of close to the true integral of 2/3. Of course, we could've done even better if � had been bigger or if we had used the midpoint of each interval instead of the right endpoint. Question 8 1 / 1 pts (Lesson 2.6: Simulating Random Variables.) BONUS: Suppose � and � are independent Uniform(0,1) random variables. (You can simulate these using the RAND() function in Excel, for instance.) Consider the nasty-looking random variable
�=−2ℓ�(�)cos(2��), where the cosine calculation is carried out in radians (not degrees). Go ahead and calculate �. . . don't be afraid. Now, repeat this task 1000 times (easy to do in Excel) and make a histogram of the 1000 �'s. What distribution does this look like? Correct! a. Normal This is the Box-Muller method to generate normal random variables. We'll learn much more about this later on. b. Unif(0,1) c. Exponential d. Weibull (a). This is the Box-Muller method to generate normal random variables. We'll learn much more about this later on. Here's some example Matlab code that works well. . . %Matlab code clear all;close all;clc; z_vec=zeros(1000,1); for i = 1:1000 u = rand; v = rand; z_vec(i)= sqrt(-2*log(u))*cos(2*pi*v); end nbins = 30; bins = linspace(-5,5,nbins); histogram(z_vec,bins) Of course, you can do this easily in R, Excel, Python etc. Quiz Score: 8 out of 8
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