Evaluate the line integral, where c is the given curve. C (x/y) ds, c: x = t3, y = t4, 1 ≤ t ≤ 4.

Answers

Answer 1

We have

x = t³   ===>   dx/dt = 3t²

y = t⁴   ===>   dy/dt = 4t³

Then with the given parameteriztion, the line integral along C of x/y is

[tex]\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{t^3}{t^4} \sqrt{(3t^2)^2 + (4t^3)^2} \, dt[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac1t \sqrt{9t^4 + 16t^6} \, dt[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{\sqrt{t^4}}t \sqrt{9 + 16t^2} \, dt[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{t^2}t \sqrt{9 + 16t^2} \, dt[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \int_1^4 t \sqrt{9 + 16t^2} \, dt[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \frac1{32} \int_1^4 32t \sqrt{9 + 16t^2} \, dt[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \frac1{32} \int_1^4 \sqrt{9 + 16t^2} \, d\left(9+16t^2\right)[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \frac1{32} \cdot \frac23 \left(9+16t^2\right)^{\frac32}\bigg|_1^4[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \frac1{48} \left(265^{\frac32} - 25^{\frac32}\right) = \boxed{\frac{265\sqrt{265}-125}{48}}[/tex]

Answer 2

The line integral will be "[tex]\frac{265\sqrt{265} - 125 }{48}[/tex]". To understand the calculation, check below.

Line integral

According to the question,

Curves, x = t³ or,

            [tex]\frac{dx}{dt}[/tex] = 3t²

             y = t⁴ or,

            [tex]\frac{dx}{dt}[/tex] = 4t³  

Now, the line integral along C will be:

→ [tex]\int\limits_C {\frac{x}{y} } \, ds[/tex] = [tex]\int\limits^4_1 {\frac{t^3}{t^4} }[/tex] √(3t²)² + (4t³)² dt

             = [tex]\int\limits^4_1 {\frac{1}{t} }[/tex] √9t⁴ + 16t⁶ dt

             = [tex]\int\limits^4_1 {\frac{t^4}{t} }[/tex] √9 + 16t² dt

             = [tex]\int\limits^4_1 {\frac{t^2}{t} }[/tex] √9 + 16t² dt

             = [tex]\int\limits^4_1[/tex] t√9 + 16t² dt

             = [tex]\frac{1}{32}[/tex] [tex]\int\limits^4_1[/tex] √9 + 16t² d (9 + 16t²)

             = [tex]\frac{1}{32}\times \frac{2}{3}[/tex] [tex](9+16t^2)^{\frac{3}{2} } |_1^4[/tex]

             = [tex]\frac{1}{48}[/tex] ([tex]265^{\frac{3}{2}} - 25^{\frac{3}{2}[/tex])

             = [tex]\frac{265 \sqrt{265 } -125 }{48}[/tex]

Thus the above approach is correct.

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Explanation:

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The standard deviation is of 7, hence \sigma = 7σ=7 .

The minimum value is the 80th percentile, which means that it is X_mXm when Z has a p-value of 0.8.

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Z = 1.43Z=1.43 has a p-value of 0.9236.

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Using the normal distribution, it is found that the probability that the volunteer selected will receive a certificate of merit given that the  number of hours the volunteer worked is less than 90 is closest to:

Letter B

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}\\Z=\sigma X - \mu[/tex]

It measures how many standard deviations the measure is from the mean. After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X. In this problem:

The mean is of 80, hence [tex]\mu = 80[/tex]. The standard deviation is of 7, hence [tex]\sigma = 7[/tex]. The minimum value is the 80th percentile, which means that it is [tex]X_m[/tex] when Z has a p-value of 0.8.

The probability that the volunteer selected will receive a certificate of merit given that the number of hours the volunteer worked is less than 90, which is the p-value of Z when X = 90 subtracted by the p-value of Z when X = X_mX=Xm , hence the p-value of Z when X = 90 subtracted by 0.8.

[tex]Z = \frac{X - \mu}{\sigma}\\Z=\sigma X - \mu\\Z = \frac{90 - 80}{7}\\Z=790 - 80\\Z = 1.43Z=1.43[/tex]

Has a p-value of 0.9236.

[tex]0.9236 - 0.8 = 0.1236[/tex]

Hence closest to 0.123.

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A global resources company uses data-intensive, cloud-based simulation software, but users in remote locations find that the responsiveness is poor due to the amount of data being transferred to and from the cloud. In response, the company decides to deploy multiple instances of the application in locations closer to the end users.

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