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Math Help - Finding dr/dt. Chain Rule Trouble

  1. #1
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    Finding dr/dt. Chain Rule Trouble

    Hello ppl. This is the problem:
    A section of blood vessel has a flow-rate F in cubic millimetres per
    minute proportional to the fourth power of its radius, R. The blood vessel can be considered perfectly cylindrical and is expandable. At some instant the
    flow rate is 30 cubic millimetres per minute when the radius is 0.5 millimetres. At that same instant the flow-rate is increasing at rate of 0.24 cubic millimetres per minute squared.
    (i) What is the rate of increase of the radius at that instant?
    (ii) If the rate of increase of the radius is constant how long after the time when the radius is 0.5 mm is the radius 1 mm?
    I'm really confused, was rummaging over it for hours this morning, I've got the F equation, the constant, but simply cant find dr/dt. Plus my friend was saying that F=dV/dt but i said no F=V/t. I'm now confused about the whole question, someone please help asap.
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  2. #2
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    Quote Originally Posted by kangaroo View Post
    I've got the F equation, the constant...
    What did you come up with?

    Thank you!
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  3. #3
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    F=kr^4

    k=1/480 from values (0.5, 30)

    I'm confused with whether F = V/t or dv/dt, i was thinking the former (by looking at the units)

    Other equation: V= (pi)(r^4)h, where h is, i was thinking, the length of the vessel? :s

    tha main part of the question is: dV/dt = dV/dr * dr/dt
    I was trying to even use Volume=Ft.
    all the variable have confused me.
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  4. #4
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    Quote Originally Posted by kangaroo View Post
    Hello ppl. This is the problem:
    I'm really confused, was rummaging over it for hours this morning, I've got the F equation, the constant, but simply cant find dr/dt. Plus my friend was saying that F=dV/dt but i said no F=V/t. I'm now confused about the whole question, someone please help asap.

    "At that same instant the flow-rate is increasing at rate of 0.24 cubic millimetres per minute squared."
    Think about what an increasing flow rate means. It means that the flow rate is increasing with respect to time (ie: dF/dt).

    This is also agreed by the units as flow rate would be in (length)^3 (time)^{-1} whereas this is given in (length)^3 (time)^{-2} which means a dt must be there something

    Using the chain rule as you said:

    \frac{dF}{dt} = \frac{dF}{dr} \times \frac{dr}{dt}

    \frac{dF}{dr} = 4kr^3 (I didn't check your working for k )

    \frac{dr}{dt} = \frac{dF}{dt} \times \frac{dr}{dF} = 0.24 \times \frac{1}{4kr^3} = 5 \times 10^{-4} ms^{-1}


    edit: I get k = 480 instead of 1/480

    F = kr^4

    30 = k \times 0.5^4

    k = \frac{30}{0.5^4} = 480
    Last edited by e^(i*pi); April 24th 2009 at 06:46 AM. Reason: added in answer
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  5. #5
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    ok fixing up it all up, i get this (first time i use latex!):

    \frac{dF}{dt} = \frac{dF}{dr} \times \frac{dr}{dt}

    F=kr^4

    F=480r^4 (from (0.5,30))

    \frac{dF}{dr} = 1920r^3

    \frac{dr}{dt} = \frac{dF}{dt} \times \frac{dr}{dF} = 0.24 \times \frac{1}{1920(0.5)^3} = 0.24 \times \frac{1}{120} = 28.8 mms^{-1}

    hope that looks ok.


    for second part how would i find t when r=1mm? i dont have any equations with t in it. (im guessing r=28.8t ?)
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  6. #6
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    Quote Originally Posted by kangaroo View Post
    ok fixing up it all up, i get this (first time i use latex!):

    \frac{dF}{dt} = \frac{dF}{dr} \times \frac{dr}{dt}

    F=kr^4

    F=480r^4 (from (0.5,30))

    \frac{dF}{dr} = 1920r^3

    \frac{dr}{dt} = \frac{dF}{dt} \times \frac{dr}{dF} = 0.24 \times \frac{1}{1920(0.5)^3} = 0.24 \times \frac{1}{120} = 28.8 mms^{-1}

    hope that looks ok.


    for second part how would i find t when r=1mm? i dont have any equations with t in it. (im guessing r=28.8t ?)
    I don't get where you got 1/120 from? I make 1/(1920x0.5^3) = 1/240.
    Also your unit of time should be the minute

    \frac{dr}{dt} = \frac{dF}{dt} \times \frac{dr}{dF} = 0.24 \times \frac{1}{1920(0.5)^3} = 0.24 \times \frac{1}{240} = 10^{-3} mm(min)^{-1}

    I think to find t you need to integrate dr/dt at r = 1mm but I'm not sure
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  7. #7
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    oops i used r^4 and didnt reciprocal it

    any1 else idea on (ii)?
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  8. #8
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    Hello, kangaroo!

    A section of blood vessel has a flow-rate F in mm≥/min
    . . is proportional to the fourth power of its radius, R.
    The blood vessel can be considered perfectly cylindrical and is expandable.

    At some instant the flow rate is 30 mm≥/min when the radius is 0.5 mm.
    At that same instant, the flow-rate is increasing at rate of 0.24 mm≥/min≤

    (a) What is the rate of increase of the radius at that instant?

    We have: . F \:=\:kR^4

    We are told: . F = 30,\;R = 0.5

    . . Hence: . 30 \:=\:k(0.5^4) \quad\Rightarrow\quad k \,=\,480

    The function is: . F\:=\:480R^4


    Differentiate with respect to time: . \frac{dF}{dt} \:=\:1920R^3\,\frac{dR}{dt}

    . . We have: . 0.24 \:=\:1920(0.5^3)\,\frac{dR}{dt} \quad\Rightarrow\quad \frac{dR}{dt} \:=\:0.001 mm/min.




    (b) If the rate of increase of the radius is constant,
    how long after the time when the radius is 0.5 mm is the radius 1 mm?
    Am I misreading the question?


    The radius increases from 0.5 mm to 1.0 mm ... an increase of 0.5 mm.

    If the radius increases at a constant 0.001 mm/min,

    . . it will take: . \frac{0.5\text{ mm}}{0.001\text{ mm/min}} \:=\:500\text{ minutes.}


    Is it really that simple?

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