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Math Help - A volume/SA problem

  1. #1
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    A volume/SA problem

    The volume of a cube is decreasing 10cm^3/min. How fast is the Surface Area decreasing when the length of one edge is 30cm?
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  2. #2
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    Quote Originally Posted by nirva
    The volume of a cube is decreasing 10cm^3/min. How fast is the Surface Area decreasing when the length of one edge is 30cm?
    If s is the side of the cube then V=s^3 thus taking the derivative to respect to time from both time we have \frac{dV}{ds}=3s^2\frac{ds}{dx}, conditions of the problem says the edge is 30 thus, s=30 and the rate of change is 10, thus, \frac{dV}{ds}=10. Thus, placing in these values,
    10=3(10)^2\frac{ds}{dt}, thus,
    \frac{ds}{dt}=\frac{1}{30}. But this tells you the instantenous rate of change of the side, the problem asks for the instantenous rate of change of the surface area. Since A=6s^2 for the cube then taking the derivate to respect to time we have,
    \frac{dA}{dt}=12s\frac{ds}{dt} know we know that s=30 for those are the conditions of this problem and in the other problem you found that \frac{ds}{dt}=\frac{1}{30} thus, substituting these values,
    \frac{dA}{dt}=12(30)\frac{1}{30}=12.
    Just placing in the proper unit values to get,
    \frac{dA}{dt}=12\frac{\mbox{cm}^2}{\mbox{min}}
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  3. #3
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    Quote Originally Posted by nirva
    The volume of a cube is decreasing 10cm^3/min. How fast is the Surface Area decreasing when the length of one edge is 30cm?
    Here is one way.

    Cube of side s cm.
    Volume, V = s^3 -----------------------(1)
    Surface area, A = 6(s^2) = 6s^2 -------(2)

    From (1),
    s = V^(1/3)
    Substitute that into (2),
    A = 6(V^(1/3))^2
    A = 6V^(2/3)
    Differentiate both sides with respect to time t,
    dA/dt = 6(2/3)*V^(-1/3)*(dV/dt)
    Plug into that the given dV/dt = -10 cu.cm/min,
    dA/dt = 4*V^(-1/3) *(-10)
    dA/dt = (-40)/[cubrt(V)] ---------------(3)
    So when s=30cm,
    V = (30)^3 cu.cm. Plug that into (3),
    dA/dt = (-40)/[cubrt(30^3)]
    dA/dt = -40/[30] = -4/3 sq.cm/min

    Therefore, at that time, the surface area is decreasing by 4/3 sq.cm/min. ------------answer.

    --------------------------
    Another way.

    V = s^3
    dV/dt = 3s^2 *(ds/dt) --------(i)

    A = 6s^2
    dA/dt = 12s*(ds/dt) ----------(ii)

    When s=30cm, and that dV/dt is always -10cu.cm/min, then, plug those into (i),
    -10 = 3(30^2)*(ds/dt)
    -10 = 2700(ds/dt)
    ds/dt = -10/2700 = -1/270 cm/min.
    Substitute that, and s=30cm, into (ii),
    dA/dt = 12(30)(-1/270)
    dA/dt = -4(10)(1/30)
    dA/dt = -40/30 = -4/3 sq.cm/min. ---------same as above.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker
    If s is the side of the cube then V=s^3 thus taking the derivative to respect to time from both time we have \frac{dV}{ds}=3s^2\frac{ds}{dx}, conditions of the problem says the edge is 30 thus, s=30 and the rate of change is 10, thus, \frac{dV}{ds}=10. Thus, placing in these values,
    10=3(10)^2\frac{ds}{dt}, thus,
    That should be 30^2 not 10^2
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  5. #5
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    Quote Originally Posted by nirva
    The volume of a cube is decreasing 10cm^3/min. How fast is the Surface Area decreasing when the length of one edge is 30cm?
    Hello,

    here is a more practical way: There are 4 squares involved, when the surface area decreases. I've attached a drawing to demonstrate, what I would like to explain.
    So the part of the surface, which is "lost" per minute, is a rectangle with 4*30 cm length and \frac{10\ cm^3}{900\ cm^2} height. That means, it's an area of \frac{1}{90} cm \cdot 120 cm=\frac{4}{3} cm^2 per min.

    Greetings

    EB
    Attached Thumbnails Attached Thumbnails A volume/SA problem-oberfl_cub.gif  
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  6. #6
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    Quote Originally Posted by CaptainBlack
    That should be 30^2 not 10^2
    Sorry, it hate making mistakes like that. Now the guy is going to take the test an get everything wrong because of me.
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