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C4 CONNECTED RATES OF CHANGE- 5 QUESTIONS watch

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    Found these 5 to be slightly tricky, would appreciate any help.

    Thanks
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  1. File Type: doc Differential Equations.doc (78.0 KB, 504 views)
  2. File Type: doc Differential Equations 2.doc (46.5 KB, 302 views)
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    (Original post by melbourne)
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    Found these 5 to be slightly tricky, would appreciate any help.

    Thanks
    Question 10.

    Time in seconds = t
    Surface area of cube = A
    Volume of cube = V

    2t = A³
    hence .. dt/dA = ½A³, This is the rate of change in area due to time.
    A³ = V
    V = (2t)^1/3
    re-arrange to get > (½V)^1/3
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    Question 1:

    C = f(x) = y
    P = (x,y) on curve C

    dy/dx = xy / z, because the rate of change is proportional to the product of x and y, there will be a division of a constant, let the constant be z.

    When P = (4, 2) dy/dx = ½

    So dy/dx = (4 x 2) / z = ½

    rearrange, 8/z = ½

    z = 16, substituite this in previous formula: dy/dx = xy / z,

    dy/dx = xy / 16
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    (Original post by Vijay1)
    Question 1:

    C = f(x) = y
    P = (x,y) on curve C

    dy/dx = xy / z, because the rate of change is proportional to the product of x and y, there will be a division of a constant, let the constant be z.

    When P = (4, 2) dy/dx = ½

    So dy/dx = (4 x 2) / z = ½

    rearrange, 8/z = ½

    z = 16, substituite this in previous formula: dy/dx = xy / z,

    dy/dx = xy / 16
    check your rep comment
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    Question 3:

    Volume into container = 30cm³ per second
    Volume out of container = 2/15Vcm³ per second
    Time in seconds = t

    dV/dt = 30t, is rate of liquid transferred into container at time in seconds
    dV/dt = 2/15V, is rate of liquid leaving container, at time t seconds.

    Putting these together, to get volume in container at a given time:

    dV/dt = 30t - 2/15V

    (dV/ dt = - 2/15V + 30t) x -15

    -15(dV/dt) = 2V - 450t
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    (Original post by melbourne)
    check your rep comment
    Oohh thank you very much :rolleyes:

    I think the last question, I've made a mistake somewhere, its not meant to be a t in the equation! oh well :rolleyes:
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    (Original post by Vijay1)
    Question 3:

    Volume into container = 30cm³ per second
    Volume out of container = 2/15Vcm³ per second
    Time in seconds = t

    dV/dt = 30t, is rate of liquid transferred into container at time in seconds
    dV/dt = 2/15V, is rate of liquid leaving container, at time t seconds.

    Putting these together, to get volume in container at a given time:

    dV/dt = 30t - 2/15V

    (dV/ dt = - 2/15V + 30t) x -15

    -15(dV/dt) = 2V - 450t

    Question 3 (attempt 2 :rolleyes: )

    Volume into container = 30cm³ per second
    Volume out of container = 2/15Vcm³ per second
    Time in seconds = t

    dV/dt = 30cm, is rate of liquid transferred into container at time in seconds
    dV/dt = 2/15V, is rate of liquid leaving container, at time t seconds.

    Putting these together, to get volume in container at a given time:

    dV/dt = 30 - 2/15V

    (dV/ dt = - 2/15V + 30t) x -15

    -15(dV/dt) = 2V - 450
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    QUESTION ELEVEN:

    Model the conical flask as a cone.

    Volume of cone = V = 1/3Ah

    Where A = Area of base = pi x r2

    A slice through the center of the cone is a triangle with angle 30 degrees at the tip.

    tan30 = r/h

    1/√3 = r/h

    r = h/√3

    But A = pi x r2 = pi x (h/√3)2 = pi(h2/3)

    Now V = 1/3Ah = 1/3(pi(h2/3))h = 1/3(pi(h3/3))

    => V = 1/9pih3

    dV/dt = -6

    dV/dh = 1/3pih2

    dh/dt = dh/dV x dV/dt

    dh/dt = 1/(dV/dh) x dV/dt

    dh/dt = -6/(1/3pih2) = -18/(pih2)

    Let -18/pi = k = constant

    So: dh/dt = k/h2

    So rate of change of h is inversely proportional to h2
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    Q 9)

    we are given that dV/dt = -k√V , where k is a +ve constant

    Vol of cylinder is V = πr²h
    dV/dh = πr²

    dh/dt = dh/dV*dV/dt
    dh/dt = (1/πr²)(-k√V)

    and √V = r√(πh)

    giving

    dh/dt = -kr√(πh)/(πr²)
    dh/dt = (-k/r√π)√h
    dh/dt = -K√h
    ==========
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    (Original post by Vijay1)
    Question 10.

    Time in seconds = t
    Surface area of cube = A
    Volume of cube = V

    2t = A³
    hence .. dt/dA = ½A³, This is the rate of change in area due to time.
    A³ = V
    V = (2t)^1/3
    re-arrange to get > (½V)^1/3
    Sorry, VJ, but I don't really follow that.

    Time in seconds = t
    Surface area of cube = A = 6x²
    Volume of cube = V = x³
    where x = length of a side of the cube.

    dA/dx = 12x

    dV/dA = dV/dx*dx/dA
    dV/dA = (3x²)/(12x)
    dV/dA = x/4
    =========

    we are given that dA/dt = 2

    dV/dt = dV/dA*dA/dt
    dV/dt = (x/4)*2 = x/2

    from V = x³
    x = V1/3
    so,
    dV/dt = (1/2)V1/3
    =============
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    (Original post by Fermat)
    Sorry, VJ, but I don't really follow that.

    Time in seconds = t
    Surface area of cube = A = 6x²
    Volume of cube = V = x³
    where x = length of a side of the cube.

    dA/dx = 12x

    dV/dA = dV/dx*dx/dA
    dV/dA = (3x²)/(12x)
    dV/dA = x/4
    =========

    we are given that dA/dt = 2

    dV/dt = dV/dA*dA/dt
    dV/dt = (x/4)*2 = x/3

    from V = x³
    x = V1/3
    so,
    dV/dt = (1/2)V1/3
    =============
    Oh right I see, I wasn't thinking logically there. :rolleyes:

    You've got the correct answer, but I think you've made a mistake her:

    dV/dt = (x/4)*2 = x/3

    should be:

    dV/dt = (x/4)*2 = x/2
 
 
 
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