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Math Help - Help me please with my revision questions.

  1. #1
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    Help me please with my revision questions.

    Hey all,

    Currently revising for my year 12 maths exam need help with the following 2 problems regarding differentiation. Thanks a bunch.

    1) An open cylindrical plastic container is to be designed so that it has a capacity of 2000 cm^3. Find the dimensions of the container if the manufacturer wants the surface area to be minimised.

    2) A ball is thrown vertically upward with an initial velocity of 19.6m/s. The displacement (in m) of the ball above the ground is s=-4.9t^2+19.6t, where t is the number of seconds elapsed from the moment that the ball is thrown.

    i) Find the initial velocity of the ball as a f(t).
    ii) Find when the velocity is zero
    iii) Find the greatest height reached by the ball.
    iv) What is the acceleration of the ball at any time?


    Thanks a million!
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  2. #2
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    For question number 1, you're gonna need to make a function for the surface area of a cylinder, differentiate it and set it equal to 0, as to find a minimum turning point.
    I gave it a try, although I'm not sure if you're supposed to be differentiating pi, so I don't know how to get any further.
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  3. #3
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    Quote Originally Posted by Ashgasm View Post
    Hey all,

    Currently revising for my year 12 maths exam need help with the following 2 problems regarding differentiation. Thanks a bunch.

    1) An open cylindrical plastic container is to be designed so that it has a capacity of 2000 cm^3. Find the dimensions of the container if the manufacturer wants the surface area to be minimised.

    2) A ball is thrown vertically upward with an initial velocity of 19.6m/s. The displacement (in m) of the ball above the ground is s=-4.9t^2+19.6t, where t is the number of seconds elapsed from the moment that the ball is thrown.

    i) Find the initial velocity of the ball as a f(t). Mr F says: Sub t = 0 into ds/dt.
    ii) Find when the velocity is zero Mr F says: Solve ds/dt = 0.
    iii) Find the greatest height reached by the ball. Mr F says: Sub value of t from ii into s.
    iv) What is the acceleration of the ball at any time? Mr F says: Find the double derivative of s with respect to t.


    Thanks a million!
    1) \pi r^2 h = 2000 \Rightarrow h = \frac{2000}{\pi r^2} .... (1)

    S = 2 \pi r h + \pi r^2.

    Substitute (1) into (2) to get S as a function of r. Solve \frac{dS}{dr} = 0. Test nature of solutions. Get value of r corresponding to minimum of S. Sub this value of r into (1) to get h.
    Last edited by mr fantastic; March 20th 2008 at 10:45 PM. Reason: Replaced the word maximum with the word minimum
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  4. #4
    Junior Member teuthid's Avatar
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    Cylinder

    Let's start with the formulas pertinent to a cylinder:
    V=\pi r^2 h
    S=2\pi r^2+2\pi rh
    where r is the radius, and h is the height.

    In order to minimize S, we need to write it as a function of a single variable. So we need to write h in terms of r. Fortunately, having a specific volume allows us to do just that:

    \pi r^2 h=2000 \\

    \Rightarrow h=\frac{2000}{\pi r^2}

    now we can substitute this new expression for h back into our original formula for surface area:

    S=2\pi r^2+2\pi r\left(\frac{2000}{\pi r^2}\right)

    which we can in turn simplify to:

    S=2\pi r^2+\frac{4000}{r}

    Now we are ready to minimize this surface area function. Minimum values will occur at critical points--values of r such that \frac{dS}{dr}=0. So differentiating we discover:

    \frac{dS}{dr} =4\pi r-4000r^{-2}

    which can be written as:

    \frac{dS}{dr}=\frac{4\pi r^3-4000}{r^2}

    The surface area function will attain an extreme value (minimum or maximum) anywhere that its derivative is zero. solving we find:

    \frac{4\pi r^3-4000}{r^2}=0

    \Rightarrow 4\pi r^3-4000=0

    \Rightarrow 4\pi r^3 = 4000

    \Rightarrow r^3 = \frac{4000}{4\pi}

    \Rightarrow r=\sqrt[3]{\frac{4000}{4\pi}}=\frac{10}{\sqrt[3]{\pi}}

    \Rightarrow r\approx 6.828

    So we see that S is at its minimum (you can use the second derivative or a graph to convince yourself that we're not looking at a maximum here) when r=\frac{10}{\sqrt[3]{\pi}}. To find the accompanying value for h, we can use our value for r:

    h=\frac{2000}{\pi r^2}

    \Rightarrow h=\frac{2000}{\pi \left(\frac{10}{\sqrt[3]{\pi}}\right)^2}

    \Rightarrow h=\frac{20}{\sqrt[3]{\pi}}

    \Rightarrow h\approx 13.656

    Collecting our results we find that the dimensions that would minimize the surface area of this cylinder are:

    Radius: \frac{10}{\sqrt[3]{\pi}} \approx 6.828 cm.

    Height: \frac{20}{\sqrt[3]{\pi}} \approx 13.656 cm.
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  5. #5
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    Thank you all so much!

    I now have another stupid question.

    If I run hot water into my bath, and then add cold water to cool it down, the temperature of the mixture is a function of the amount of cold water added. In once case, adding x litres of cold water reduced the temperature to T*C, where T=20(x+200)/(x+80)

    a) what is the instantaneous rate of change of the water temperature per litre of cold water added? --> I know you have to differentiate this. I couldn't get it to work using the quotient rule. So, I googled it, and used this automatic differentiator thingy, and got this:



    So yeah, how did they arrive at that result??

    Thanks.
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  6. #6
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    Quote Originally Posted by Ashgasm View Post
    Thank you all so much!

    I now have another stupid question.

    If I run hot water into my bath, and then add cold water to cool it down, the temperature of the mixture is a function of the amount of cold water added. In once case, adding x litres of cold water reduced the temperature to T*C, where T=20(x+200)/(x+80)

    a) what is the instantaneous rate of change of the water temperature per litre of cold water added? --> I know you have to differentiate this. I couldn't get it to work using the quotient rule. So, I googled it, and used this automatic differentiator thingy, and got this:



    So yeah, how did they arrive at that result??

    Thanks.
    1. I would have thought getting the derivative was a routine application of the quotient rule. It would assist in helping you if you showed your attempt at getting the answer.

    2. The answer your "automatic differentiator thingy" gave you is an alternative form of \frac{-2400}{(x+80)^2}. As to why it chose to give the answer in the form that it did, computer algebra systems often give unexpected output.

    3. New question, new thread!
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