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3 The extension,

y

, of a material with an applied force,

F

, is given by $y=e...

Feb 2, 2024

3 The extension,

y

, of a material with an applied force,

F

, is given by

y=e^{F \times 1 \times 10^{-3}}

. a) Calculate the work done if the force increases from

100 \mathrm{~N}

500 \mathrm{~N}

using: i) An analytical integration technique ii) A num erical integration technique [Note: the work done is given by the area under the curve] b) Compare the two answers c) Using a computer spreadsheet increase the number of values used for your num erical method d) Analyse any affect the size of num erical step has on the result.

Generated Graph

Solution by Steps

step 1

To find the work done analytically, we integrate the function

y = e^{F \times 1 \times 10^{-3}}

with respect to

F

from 100 to 500

step 2

The indefinite integral of

e^{F \times 1 \times 10^{-3}}

with respect to

F

1000 e^{F \times 1 \times 10^{-3}} + C

, where

C

is the constant of integration

step 3

Evaluating the definite integral from 100 to 500, we get

1000 (e^{0.5} - e^{0.1})

step 4

Plugging in the values, the work done is

1000 (e^{0.5} - e^{0.1}) \approx 543.55

joules

Answer

Analytical technique: Approximately 543.55 joules

Key Concept

Analytical Integration Technique

Explanation

The work done is calculated by integrating the force-extension function over the given range of force. The analytical method provides an exact value for the work done.

step 1

To find the work done numerically, we use a numerical integration technique such as the left Riemann sum approximation

step 2

The left Riemann sum approximation is given by $ \frac{400 (e^{2/5} - 1) e^{1/10}}{(e^{2/(5 n)) - 1) n} $, where $ n $ is the number of subintervals

step 3

Assuming subintervals of equal length, the approximation for

n

subintervals is

1000 (e^{2/5} - 1) e^{1/10} - \frac{200 ((e^{2/5} - 1) e^{1/10})}{n}

plus higher order terms

step 4

For a large number of subintervals, the numerical result will approach the analytical result of approximately 543.55 joules

Answer

Numerical technique: Approaches 543.55 joules as

n

increases

Key Concept

Numerical Integration Technique

Explanation

The numerical method approximates the work done by summing the areas of rectangles under the curve. The accuracy improves as the number of subintervals increases.

step 1

To compare the two answers, we look at the results obtained from the analytical and numerical methods

step 2

The analytical method gave us an exact value of approximately 543.55 joules

step 3

The numerical method will approach this value as the number of subintervals increases

step 4

The comparison shows that the numerical method can provide a good approximation to the analytical result, especially with a higher number of subintervals

Answer

The numerical result approaches the analytical result of approximately 543.55 joules as the number of subintervals increases.

Key Concept

Comparison of Integration Techniques

Explanation

The comparison illustrates that while the analytical method gives an exact value, the numerical method can approximate this value closely with a sufficient number of subintervals.

step 1

To analyze the effect of the numerical step size on the result, we can use a computer spreadsheet to increase the number of values used in the numerical method

step 2

As the number of subintervals (and thus the number of values) increases, the step size decreases, leading to a more accurate approximation of the integral

step 3

A smaller step size means that the rectangles used in the Riemann sum are narrower and fit the curve more closely, reducing the error in the approximation

step 4

Therefore, increasing the number of values used in the numerical method will generally result in a more accurate estimate of the work done

Answer

Increasing the number of values (decreasing step size) in the numerical method improves the accuracy of the result.

Key Concept

Effect of Numerical Step Size

Explanation

The size of the numerical step affects the accuracy of the numerical integration; smaller steps yield more accurate results by better approximating the area under the curve.