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    hey, please help me to understand/ answer if you can

    " A baloon is filled with air at a constant rate of 80cm^3.
    Assuming the baloon is spherical as it is filled, find the rate at which the radius is increasing at the instant when the radius is 6cm"



    My (probably useless) workings so far:


    dv/dt = 80 .

    v= 4/3 pi r^3

    dv/dr=4pi r^2

    merciverymuchos x
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    I think you need to find dr/dt ?
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    You can use \frac{dr}{dv}=\dfrac{1}{\left(\f  rac{dv}{dr}\right)} and \frac{dr}{dt}=\frac{dr}{dv} \times \frac{dv}{dt} by the chain rule.
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    (Original post by ttoby)
    You can use \frac{dr}{dv}=\dfrac{1}{\left(\f  rac{dv}{dr}\right)} and \frac{dr}{dt}=\frac{dr}{dv} \times \frac{dv}{dt} by the chain rule.
    thanks. dr/dt = 20 pi r^2

    so do i then plug in 6 in to got dr/dt = 720pi as the answer?
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    (Original post by jim666666)
    thanks. dr/dt = 20 pi r^2

    so do i then plug in 6 in to got dr/dt = 720pi as the answer?
    Not quite, you should get dr/dt = 20/ (pi r^2). Then you plug in r=6.
 
 
 
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