Answer to Question #286176 in Calculus for Celik

Question

1) The extension, 𝑦, of a material with an applied force, 𝐹, is given by

𝒚 = 𝒆 𝑭×(𝟏×𝟏𝟎−𝟑) .

a) Calculate the work done if the force increases from 100N to 500N using:

i) An analytical integration technique

ii) Simpsons Rule

[Note: the work done is given by the area under the curve]

b) Compare the two answers

c) Increase the number of values used for your numerical method and analyse any affect the size of numerical step has on the result

2) Use numerical integration and integral calculus to analyse the results of a complex engineering problem.

The work done by a mechanism is given by:

𝒚 = 𝒙 𝐜𝐨𝐬 𝒙

a) Use integration by parts to determine the area under the curve between the limits of x = 7 and x =5 and hence the work done.

b) Choose a suitable strip width and use Simpsons rule to determine the area under the curve and hence the work done.

c) Evaluate the answers for a and b. Does one method verify the results of the other?

Answer 1

1)

i)

 $y=e^{F\cdot 10^{-3}}$

$dy=10^{-3} e^FdF$

$W=\int_{100}^{500}Fdy$

$W=10^{-3}\int_{100}^{500}Fe^{F\cdot 10^{-3}}dF$

$\int Fe^{F\cdot 10^{-3}}dF=10^3Fe^{F\cdot 10^{-3}}-10^3\int e^{F\cdot 10^{-3}}dF=$

$=10^3Fe^{F\cdot 10^{-3}}-10^6e^{F\cdot 10^{-3}}=10^3e^{F\cdot 10^{-3}}(F-10^3)$

$W= e^{F\cdot 10^{-3}}(F-10^3)]_{100}^{500}=-500\cdot1.65+900\cdot1.11=169.65$ J

ii)

$f(F)=10^{-3}Fe^{F\cdot 10^{-3}}$

$\Delta F=100$

$n=4$

$W=\frac{\Delta F}{3}(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+f(x_4))=$

$=\frac{100}{3}(f(100)+4f(200)+2f(300)+4f(400)+f(500))=170.29$ J

b)

error:

$\frac{170.29-169.65}{169.65}=0.0038=0.38\%$

c)

$n=8$

$\Delta F=50$

$W=\frac{100}{3}(f(100)+4f(150)+2f(200)+4f(250)+2f(300)+4f(350)+$

$+2f(400)+4f(450)+f(500))=170.29$ J

In our case, increasing the number of rectangles for Simpsons Rule (from n=4 to n=8) practically has not effect. In both cases we have same result (170.29 J).

2)

a)

$\int xcosx dx=xsinx-\intop sinx dx=xsinx+cosx$

work done by a mechanism:

$W=(xsinx+cosx)|^7_5=5.35+4.51=9.86$ J

b)

$n=6$

$\Delta x=1/3$

$f(x)=xcosx$

$W=\frac{\Delta x}{3}(f(5)+4f(16/3)+2f(17/3)+4f(6)+2f(19/3)+4f(20/3)+f(7))=$

$=9.86$ J

c)

We have same results in a) and b). So, one method verify the results of the other.