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Math Help - Calculus paper help needed

  1. #1
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    Calculus paper help needed

    Hi over the festive period i have been trying some past papers, just to get ready for the summer, ive got stuck on a few of the questions and wondering if somebody can help me, the link to the paper is bellow i hope that takes you to it.
    https://www.kent.ac.uk/library/onlin...2009/MA301.pdf

    The first page is fine, however going on to section B it gets a bit more tricky.
    Question 7 using the definition of the derivative i curreny have got the following
    a) ((x+h)^-2-(x)^-2))/h
    = ((1/(x+h)^2-1/(x)^2)*1/h
    =((1/x^2+2hx+h^2)-1/(x^2))*1/h
    =((x^2-(x+h)^2)/(x^2(x+h)^2)*1/h

    Basically trying to simplify it down in order to get the needed -2x^-3, somehow i need to cancell the h outside, then i think it should be ok as h tends to infinity.
    Have the same problem with the other question in 7.

    Question 8
    Im stuggling with (i) managed to get (ii) via completing the square then using a change of variable.
    I know on (i) the function will integrate to something arctan as ive put it into a integration calculator, just wondering if im missing something obvious here.

    Also B i wouldnt know where to start.

    Question 9 im just wondering if my answers are correct really, i believe i have definately got b, however with a i have:
    partial df/dx=e^x(siny) partial df/dy=0 second partial df/dx=0 and second partial df/dy=0
    Therefore shwoing laplaces equation = 0
    if that is correct, i very much doubt it though, how to obtain the fourth partial derivative without any calculations.

    As for question 10 im not sure if i should go there just yet as a few questions have already been asked and might get a bit messy.

    Many thanks in advance once more.
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  2. #2
    Senior Member BAdhi's Avatar
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    your link is asking for a user name and password
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  3. #3
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    Hello, breitling!

    I can't see that site, but I can guess the first problem.


    Question 7: using the definition of the derivative.
    i currently have got the following:

    a)\;\dfrac{(x+h)^{-2}- x^{-2}}{h}

    . . . \displaystyle =\;\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}

    . . . \displaystyle =\;\frac{1}{h}\cdot\left(\frac{1}{(x+h)^2}-\frac{1}{x^2}}\right)

    . . . \displaystyle =\;\frac{x^2-(x+h)^2}{h\,x^2(x+h)^2} . Why did you stop?

    You have: . \displaystyle \frac{x^2 - (x^2 + 2xh + h^2)}{h\,x^2(x+h)^2} \;=\;\frac{-2hx - h^2}{h\,x^2(x+h)^2}

    . . . . . . . . . . \displaystyle =\;\frac{-h(2x+h)}{h\,x^2(x+h)^2} \;=\;-\frac{2x+h}{x^2(x+h)^2}


    Then: . \displaystyle \lim_{h\to0}\left[-\frac{2x+h}{x^2(x+h)^2}\right] \;=\; -\frac{2x+0}{x^2(x+0)^2} \;=\;-\frac{2x}{x^4} \;=\; - \frac{2}{x^3}

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, breitling!

    I can't see that site, but I can guess the first problem.


    You have: . \displaystyle \frac{x^2 - (x^2 + 2xh + h^2)}{h\,x^2(x+h)^2} \;=\;\frac{-2hx - h^2}{h\,x^2(x+h)^2}

    . . . . . . . . . . \displaystyle =\;\frac{-h(2x+h)}{h\,x^2(x+h)^2} \;=\;-\frac{2x+h}{x^2(x+h)^2}


    Then: . \displaystyle \lim_{h\to0}\left[-\frac{2x+h}{x^2(x+h)^2}\right] \;=\; -\frac{2x+0}{x^2(x+0)^2} \;=\;-\frac{2x}{x^4} \;=\; - \frac{2}{x^3}
    My laptop charger is currently broke, so on my computer and all im seeing is a lot of x's never a good sign ha. however im getting a new charger on monday so will put the paper on then and discuss some solutions, many thanks so far.
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, breitling!

    I can't see that site, but I can guess the first problem.



    You have: . \displaystyle \frac{x^2 - (x^2 + 2xh + h^2)}{h\,x^2(x+h)^2} \;=\;\frac{-2hx - h^2}{h\,x^2(x+h)^2}

    . . . . . . . . . . \displaystyle =\;\frac{-h(2x+h)}{h\,x^2(x+h)^2} \;=\;-\frac{2x+h}{x^2(x+h)^2}


    Then: . \displaystyle \lim_{h\to0}\left[-\frac{2x+h}{x^2(x+h)^2}\right] \;=\; -\frac{2x+0}{x^2(x+0)^2} \;=\;-\frac{2x}{x^4} \;=\; - \frac{2}{x^3}

    I finally got my laptop charger so im ready to go, i have tried uploading it directly this time so hope this works.
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  6. #6
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    \displaystyle \int{\frac{dx}{4x^2 + 12x + 34}} = \frac{1}{4}\int{\frac{dx}{x^2 + 3x + \frac{17}{2}}}

    \displaystyle = \frac{1}{4}\int{\frac{dx}{x^2 + 3x + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 + \frac{17}{2}}}

    \displaystyle = \frac{1}{4}\int{\frac{dx}{\left(x + \frac{3}{2}\right)^2 + \frac{25}{4}}}.


    Now make the substitution \displaystyle x + \frac{3}{2} = \frac{5}{2}\tan{\theta} so that \displaystyle dx = \frac{5}{2}\sec^2{\theta}\,d\theta and the integral becomes

    \displaystyle = \frac{1}{4}\int{\frac{\frac{5}{2}\sec^2{\theta}\,d  \theta}{\left(\frac{5}{2}\tan{\theta}\right)^2 + \frac{25}{4}}}

    \displaystyle = \frac{5}{8}\int{\frac{\sec^2{\theta}\,d\theta}{\fr  ac{25}{4}\tan^2{\theta} + \frac{25}{4}}}

    \displaystyle = \frac{5}{8}\int{\frac{\sec^2{\theta}\,d\theta}{\fr  ac{25}{4}(\tan^2{\theta} + 1)}}

    \displaystyle = \frac{5}{8}\cdot \frac{4}{25}\int{\frac{\sec^2{\theta}}{\sec^2{\the  ta}}\,d\theta}

    \displaystyle = \frac{1}{10}\int{1\,d\theta}

    \displaystyle = \frac{1}{10}\theta + C

    \displaystyle = \frac{1}{10}\arctan{\left(\frac{2}{5}x + \frac{3}{5}\right)} + C
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  7. #7
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    Quote Originally Posted by Prove It View Post
    \displaystyle \int{\frac{dx}{4x^2 + 12x + 34}} = \frac{1}{4}\int{\frac{dx}{x^2 + 3x + \frac{17}{2}}}

    \displaystyle = \frac{1}{4}\int{\frac{dx}{x^2 + 3x + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 + \frac{17}{2}}}

    \displaystyle = \frac{1}{4}\int{\frac{dx}{\left(x + \frac{3}{2}\right)^2 + \frac{25}{4}}}.


    Now make the substitution \displaystyle x + \frac{3}{2} = \frac{5}{2}\tan{\theta} so that \displaystyle dx = \frac{5}{2}\sec^2{\theta}\,d\theta and the integral becomes

    \displaystyle = \frac{1}{4}\int{\frac{\frac{5}{2}\sec^2{\theta}\,d  \theta}{\left(\frac{5}{2}\tan{\theta}\right)^2 + \frac{25}{4}}}

    \displaystyle = \frac{5}{8}\int{\frac{\sec^2{\theta}\,d\theta}{\fr  ac{25}{4}\tan^2{\theta} + \frac{25}{4}}}

    \displaystyle = \frac{5}{8}\int{\frac{\sec^2{\theta}\,d\theta}{\fr  ac{25}{4}(\tan^2{\theta} + 1)}}

    \displaystyle = \frac{5}{8}\cdot \frac{4}{25}\int{\frac{\sec^2{\theta}}{\sec^2{\the  ta}}\,d\theta}

    \displaystyle = \frac{1}{10}\int{1\,d\theta}

    \displaystyle = \frac{1}{10}\theta + C

    \displaystyle = \frac{1}{10}\arctan{\left(\frac{2}{5}x + \frac{3}{5}\right)} + C
    Yes i managed to get that via completing the square, then doing a change of veriable, so that i could integrate the function to obtain arctan etc.
    I was stuck on the one before it 8i, i got the differentiation by first principles for 7i after looking at sorobans relpy many thanks for that, but still stuck on 7ii,and 7b.... 8i im finding tough and also 8b. Like i said i think i may of got parts of 9, but not sure really if its correct, 10 also brings some difficulty the last bit especially!
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  8. #8
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    Well for 8i) \displaystyle \int_{\frac{3}{2}}^{\sqrt{3}}{\frac{dx}{\sqrt{3-x^2}}}

    make the substitution \displaystyle x = \sqrt{3}\sin{\theta} so that \displaystyle dx = \sqrt{3}\cos{\theta}\,d\theta.

    Note that would mean if \displaystyle x = \frac{3}{2} then \displaystyle \theta = \frac{\pi}{3} and if \displaystyle x = \sqrt{3} then \displaystyle \theta = \frac{\pi}{2}.

    After making all the necessary substitutions the integral becomes

    \displaystyle \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\sqrt{3  }\cos{\theta}\,d\theta}{\sqrt{3 - (\sqrt{3}\sin{\theta})^2}}}

    \displaystyle = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\sqrt{3  }\cos{\theta}\,d\theta}{\sqrt{3 - 3\sin^2{\theta}}}}

    \displaystyle = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\sqrt{3  }\cos{\theta}\,d\theta}{\sqrt{3(1 - \sin^2{\theta})}}}

    \displaystyle = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\sqrt{3  }\cos{\theta}\,d\theta}{\sqrt{3}\sqrt{1 - \sin^2{\theta}}}}

    \displaystyle = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\cos{\t  heta}\,d\theta}{\sqrt{\cos^2{\theta}}}}

    \displaystyle = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\cos{\t  heta}\,d\theta}{\cos{\theta}}}

    \displaystyle = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{1\,d\theta}

    \displaystyle = \left[\theta\right]_{\frac{\pi}{3}}^{\frac{\pi}{2}}

    \displaystyle = \frac{\pi}{2} - \frac{\pi}{3}

    \displaystyle = \frac{\pi}{6}.
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  9. #9
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    Quote Originally Posted by Prove It View Post
    Well for 8i) \displaystyle \int_{\frac{3}{2}}^{\sqrt{3}}{\frac{dx}{\sqrt{3-x^2}}}

    make the substitution \displaystyle x = \sqrt{3}\sin{\theta} so that \displaystyle dx = \sqrt{3}\cos{\theta}\,d\theta.

    Note that would mean if \displaystyle x = \frac{3}{2} then \displaystyle \theta = \frac{\pi}{3} and if \displaystyle x = \sqrt{3} then \displaystyle \theta = \frac{\pi}{2}.

    After making all the necessary substitutions the integral becomes

    \displaystyle \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\sqrt{3  }\cos{\theta}\,d\theta}{\sqrt{3 - (\sqrt{3}\sin{\theta})^2}}}

    \displaystyle = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\sqrt{3  }\cos{\theta}\,d\theta}{\sqrt{3 - 3\sin^2{\theta}}}}

    \displaystyle = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\sqrt{3  }\cos{\theta}\,d\theta}{\sqrt{3(1 - \sin^2{\theta})}}}

    \displaystyle = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\sqrt{3  }\cos{\theta}\,d\theta}{\sqrt{3}\sqrt{1 - \sin^2{\theta}}}}

    \displaystyle = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\cos{\t  heta}\,d\theta}{\sqrt{\cos^2{\theta}}}}

    \displaystyle = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\cos{\t  heta}\,d\theta}{\cos{\theta}}}

    \displaystyle = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{1\,d\theta}

    \displaystyle = \left[\theta\right]_{\frac{\pi}{3}}^{\frac{\pi}{2}}

    \displaystyle = \frac{\pi}{2} - \frac{\pi}{3}

    \displaystyle = \frac{\pi}{6}.
    Many thanks for that i think i have it now, ive seen a similar question appear on most of the papers so this great many thanks.
    Wondering if theres any help out there for 9 and 10

    9a
    Im slightly stuck on whether to keep the e^x and partial differentiate the:
    ycosy + xsiny
    the goal is to obtain the laplaces equation
    If holding the e^x when doing partial df/dx then the e^X would stay the same and multiply the partial df/dx of ycosy+xsiny
    therefore e^x(siny)
    I very much doubt this is correct, when doing the same for partial df/dy the e^x would go to 0, therefore the differential would be 0 and so on...

    9b was fine, found it a bit more straight forward than the previous, and got the correct answer.

    10a
    My partial df/dx i obtained:
    1/(x-y)-ysin(xy)+y^2e^x

    My partial df/dy i obtained:
    -1/(x-y)-xsin(xy)+2ye^x

    Then forming df/dt by using the chain rule, calculating dt/dx and dy/dt noticing dt/dx is -y and dy/dt is x?
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  10. #10
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    10 c)
    \displaystyle \int_1^2{\int_1^{y^2}{\frac{y}{x + y}\,dx}\,dy} = \int_1^2{\left[y\ln{|x+y|}\right]_1^{y^2}\,dy}

    \displaystyle = \int_1^2{y\ln{|y^2+y|} - y\ln{|1 + y|}\,dy}

    \displaystyle = \int_1^2{y\ln{\left|\frac{y^2 + y}{1 + y}\right|}\,dy}

    \displaystyle = \int_1^2{y\ln{|y|}\,dy}

    \displaystyle = \left[\frac{1}{2}y^2\ln{|y|} - \int{\frac{1}{2}y^2\,\frac{1}{y}\,dy}\right]_1^2

    \displaystyle = \left[\frac{1}{2}y^2\ln{|y|} - \int{\frac{1}{2}y\,dy}\right]_1^2

    \displaystyle = \left[\frac{1}{2}y^2\ln{|y|} - \frac{1}{4}y^2\right]_1^2

    \displaystyle = \frac{1}{2}\cdot 2^2\cdot \ln{|2|} - \frac{1}{4}\cdot 2^2 - \left(\frac{1}{2} \cdot 1^2 \cdot \ln{|1|} - \frac{1}{4}\cdot 1^2\right)

    \displaystyle = 2\ln{2} - \frac{3}{4}.
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  11. #11
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    Quote Originally Posted by breitling View Post
    8. (b) Let \alpha be a real number. Without evaluating the integral, show that

    \displaystyle \int_{0}^{\infty}\frac{e^{-\alpha{x}}-\left(1+\alpha{x}\right)^{-1}}{x}\;{dx} = \int_{0}^{\infty}\frac{e^{-{x}}-\left(1+{x}\right)^{-1}}{x}\;{dx}

    and so is independent of \alpha. Explain why the value of this integral is negative.
    Put t = \alpha{x}, then dx = \frac{1}{\alpha}\;{dt}, thus:

    \begin{aligned} \displaystyle & \int_{0}^{\infty}\frac{e^{-\alpha{x}}-\left(1+\alpha{x}\right)^{-1}}{x}\;{dx} = \int_{0}^{\infty}\frac{e^{-\alpha{x}}-\left(1+\alpha{x}\right)^{-1}}{\alpha{x}}\;{dt} \\& =\int_{0}^{\infty}\frac{e^{-t}-\left(1+t\right)^{-1}}{t}\;{dt} = \int_{0}^{\infty}\frac{e^{-x}-\left(1+x\right)^{-1}}{x}\;{dx}.\end{aligned}

    ... where the last step is because t is just a dummy variable...
    I'll leave for you to show that the value of the integral is negative.
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