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    A hailstone falls from rest through a cloud under gravity. Initially it is spherical with radius a. As it falls it accumulates mass at a rate (pi)pr^2, where p is its contant density, but remains spherical in shape.

    What is the radius of this hailstone at time t?

    How do you do this? Cheers
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    dm/dt = (pi)pr^2
    dm/dr dr/dt = (pi)pr^2

    I don't know where to go from here. I am not sure whether this method is correct by the way
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    Ok, you are given at the start that:

    dm/dt = p(pi)(r^2)

    Now, since the mass of the hailstone is it's density multiplied by it's volume we can form this:

    m = p(4/3)(pi)(r^3)

    Then we can setup the differentiation to get:

    dm/dr = p(4)(pi)(r^2)

    Now we can use the chain rule:

    dr/dt = (dm/dt)(dr/dm)

    Which leads to:

    dr/dt = [(p)(pi)(r^2)] / [4(p)(pi)(r^2)] = 1/4

    Now can you see how to use calculus to form an equation where r =.....?
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    I don't know if this is just me being stupid here, but the units of the expression for the rate don't include 'per second' so I don't see how it can be a rate.
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    (Original post by laeti)
    I don't know if this is just me being stupid here, but the units of the expression for the rate don't include 'per second' so I don't see how it can be a rate.
    Nor is the mass given in kilograms, the density in kilograms per cubic metre, etc... it's all implied (principally so you don't have to sit there fiddling with units when there are more important things in life).
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    There is a different part to this question...

    find v(t) when

    dv/dt + (3v)/(t+4a) = g
    The answer I got is long and doesn't seem right. I get....

    v = [(g/4)(t^4) + (4a)(t^3)(g) + (24a^2)(t^2)(g) + (64)(a^3)(gt)] / (t+4a)^3
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    Note that you can integrate

    \displaystyle \int (t + 4a)^3 dt

    without expanding out, since (t + 4a) is linear.

    Anyway, you're answer is a solution; note that it can also be written

    v = g\left(\frac{t + 4a}{4} - \frac{64a^4}{(t+4a)^3}\right)

    The only catch is there are infinitely many more solutions, because of the constant that arises in the integration. One other (a perhas a simpler one) is

    \displaystyle v = \frac{g(t+4a)}{4}.
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    (Original post by ukgea)
    Note that you can integrate

    \displaystyle \int (t + 4a)^3 dt

    without expanding out, since (t + 4a) is linear.

    Anyway, you're answer is a solution; note that it can also be written

    v = g\left(\frac{t + 4a}{4} - \frac{64}{(t+4a)^3}\right)

    The only catch is there are infinitely many more solutions, because of the constant that arises in the integration. One other (a perhas a simpler one) is

    \displaystyle v = \frac{g(t+4a)}{4}.
    The initial conditions seem to be that when t = 0, v = 0 if you read the question above. so constant of integration = -64ga^4. Am I correct?
    Cheers
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    (Original post by holst123)
    The initial conditions seem to be that when t = 0, v = 0 if you read the question above. so constant of integration = -64ga^4. Am I correct?
    Cheers
    Depends on how you integrate it; different ways of integrating will have different constants. What's your final answer?

    (But yes, if you're integrating it the way I think you are, it is correct.)
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    (Original post by ukgea)
    Depends on how you integrate it; different ways of integrating will have different constants. What's your final answer?

    (But yes, if you're integrating it the way I think you are, it is correct.)
    (v)(t + 4a)^3 = int(g(t+4a)^3)
    = (g(t+4a)^4)/4 + C

    v = (g(t+4a))/4 + C/(t+4a)^3

    t = 0, v=0

    0 = g(4a)/4 + C/(64a^3)

    So C = -64ga^4

    So v = g(t+4a)/4 - 64ga^4/(t+4a)^3
 
 
 
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