4. A certain computer algorithm used to solve very complicated differential equations uses an iterative method. That is, the algorithm solves the problem the first time very approximately, and then uses that first solution to help it solve the problem a second time just a little bit better, and then uses that second solution to help it solve the problem a third time just a little bit better, and so on. Unfortunately, each iteration (each new problem solved by using the previous solution) takes a progressively longer amount of time. In fact,
the amount of time it takes to process the k-th iteration is given by T(k) = 1.2k
seconds.
A. Use a definite integral to approximate the time (in hours) it will take the computer algorithm to run through 60 iterations. (Note that T(k) is the amount of time it takes to process just the k-th iteration.) Explain your reasoning.
B. The maximum error in the computer's solution after k iterations is given by Error = 2k^-2. Approximately how long (in hours) will it take the computer to process enough iterations to reduce the maximum error to below 0.0001?
A: 1/3600 ∫[0,60] T(k) dk
see what you can do with that
A. To approximate the time it will take for the computer algorithm to run through 60 iterations, we can use a definite integral. The time it takes to process each iteration is given by T(k) = 1.2k seconds.
We need to find the total time taken for all 60 iterations, which can be represented by the definite integral ∫(0 to 60) 1.2k dk.
Integrating the function 1.2k with respect to k, we get 0.6k^2 evaluated from 0 to 60.
Plugging in the upper and lower limits, we get: 0.6(60)^2 - 0.6(0)^2 = 0.6(3600) = 2160 seconds.
To convert seconds to hours, we divide by 3600 (since there are 3600 seconds in an hour): 2160/3600 = 0.6 hours.
Therefore, it will take approximately 0.6 hours for the algorithm to run through 60 iterations.
B. The maximum error in the computer's solution after k iterations is given by Error = 2k^-2.
We want to find the number of iterations required to reduce the maximum error to below 0.0001.
Setting up the inequality 2k^-2 < 0.0001, we can solve for k.
Simplifying the inequality, we get 2/k^2 < 0.0001.
Taking the reciprocal of both sides, we obtain k^2 > 2/0.0001 = 20000.
Taking the square root of both sides, we get k > √20000.
Approximately, √20000 is equal to 141.42.
Since k represents the number of iterations, we can round up to the nearest whole number. So, it will take approximately 142 iterations to reduce the maximum error to below 0.0001.
From part A, we know that each iteration takes 1.2k seconds. Substituting k = 142, we get T(142) = 1.2(142) seconds.
To convert seconds to hours, we divide by 3600: 170.4/3600 ≈ 0.0472 hours.
Therefore, it will take approximately 0.0472 hours for the computer to process enough iterations to reduce the maximum error to below 0.0001.
A. To approximate the time it will take the computer algorithm to run through 60 iterations, we can use a definite integral. Let's break down the steps:
First, we need to find the time it takes for each iteration. Given that the time for the k-th iteration is T(k) = 1.2k seconds, we want to find the total time for all the iterations up to k = 60. We can express this sum as:
T_total = T(1) + T(2) + T(3) + ... + T(60)
Now, let's convert seconds to hours by dividing each term by 3600 (since there are 3600 seconds in an hour).
T_total (in hours) = (T(1) / 3600) + (T(2) / 3600) + (T(3) / 3600) + ... + (T(60) / 3600)
The expression inside the sum is in the form T(k) / 3600. We can also rewrite this as (1.2k / 3600).
Now, we can rewrite the total time equation as a definite integral:
T_total (in hours) = ∫[(1.2k / 3600)] from k = 1 to 60.
To calculate this definite integral, we need to evaluate the antiderivative of (1.2k / 3600) and evaluate it at the upper and lower limits:
T_total (in hours) = [(1.2/3600) * (k^2) / 2] from k = 1 to 60.
Now, substitute the upper limit (60) and lower limit (1) into the expression and evaluate:
T_total (in hours) = [(1.2/3600) * (60^2) / 2] - [(1.2/3600) * (1^2) / 2].
Calculate this expression to find the approximate time it will take the computer algorithm to run through 60 iterations in hours.
B. The maximum error in the computer's solution after k iterations is given by Error = 2k^-2. We want to find the number of iterations it takes to reduce the maximum error to below 0.0001.
Given the error formula, we can set it to be less than 0.0001 and solve for k:
2k^-2 < 0.0001.
Rearranging the inequality and multiplying both sides by k^2 gives:
2 < 0.0001 * k^2.
Divide both sides by 0.0001:
2 / 0.0001 < k^2.
Simplify the left side:
20000 < k^2.
Taking the square root of both sides (and considering the positive square root):
sqrt(20000) < k.
Simplify the square root:
141.42 < k.
Therefore, it will take more than 141 iterations to reduce the maximum error to below 0.0001.
To find the time it will take in hours, we can plug this value of k into the time formula T(k) = 1.2k. However, we need to convert k to seconds by multiplying it by 3600, since the time formula is given in seconds:
T = 1.2 * 141 * 3600.
Calculate this expression to find the approximate time it will take the computer to process enough iterations to reduce the maximum error to below 0.0001 in hours.
time for 60 iterations
= ∫ 1.2k dk from 0 to 60
= [.6k^2] from 0 to 60
= .6(60^2) - 0 = 2160 seconds
2k^-2 < .0001 , assuming k is still in seconds
k^-2 < .00005
k^2 < 1/.00005
k^2 < 20,000
k < 141.42 = appr .039 hours
if 140 seconds, error = 2(140^-2) = .00010204 , which is > .0001
if 141 seconds, error = 2(141^-2) = .0000502.. , which is < .0001