4. Use your figure to study the sign of vr(t) in the time interval [-4 : 4]. Does vr(t) have any root in the interval [-4 : 4]? If yes, estimate the roots graphically.
4. Use your figure to study the sign of vr(t) in the time interval [-4 : 4]. Does vr(t) have any root in the interval [-4 : 4]? If yes, estimate the roots graphically.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![4. Use your figure to study the sign of vr(t) in the time interval [-4 : 4]. Does vr(t) have any root in the
interval [-4 : 4]? If yes, estimate the roots graphically.
5. Manually use Bisection iterative technique with 6 iterations to find a root of vr(t) in the intervals
[0.3 : 0.7] and [-1 : -5]. Calculate the percentage of error. Show details of your steps.
6. Manually use Newton-Raphson iterative technique with 6 iterations to find a root of vr(t) in the
intervals [0.3 : 0.7] and [-1 : -5]. Calculate the percentage of error. Show details of your steps.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b6423c1-63d9-43b9-83dc-687d4d09a21d%2Ff6d26f2f-75c2-42be-a0dc-6310c7478c2d%2Fiirvu39_processed.png&w=3840&q=75)
Transcribed Image Text:4. Use your figure to study the sign of vr(t) in the time interval [-4 : 4]. Does vr(t) have any root in the
interval [-4 : 4]? If yes, estimate the roots graphically.
5. Manually use Bisection iterative technique with 6 iterations to find a root of vr(t) in the intervals
[0.3 : 0.7] and [-1 : -5]. Calculate the percentage of error. Show details of your steps.
6. Manually use Newton-Raphson iterative technique with 6 iterations to find a root of vr(t) in the
intervals [0.3 : 0.7] and [-1 : -5]. Calculate the percentage of error. Show details of your steps.
Transcribed Image Text:Analysis 2: The voltage potential, v(t), builds up on the loops, based on the orientation of the magnetic
field during an MR scan is given by:
v(t) = 0.250t4 + 0.166t3 – 0.500²
and the voltage at time t = 0 is 0.
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