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    Leaking oil is forming a circular patch on the surface of the sea. The area of the patch is increasing at a rate of 250 square metres per hour. Find the rate at which the radius of the patch is increasing at the instant when the area of the patch is 1900 square metres. Give your answer correct to 2 significant figures.

    How do i approach this question?
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    (Original post by elvss567)
    Leaking oil is forming a circular patch on the surface of the sea. The area of the patch is increasing at a rate of 250 square metres per hour. Find the rate at which the radius of the patch is increasing at the instant when the area of the patch is 1900 square metres. Give your answer correct to 2 significant figures.

    How do i approach this question?
    Translate the words into maths:

    The area of the patch is increasing at a rate of 250 square metres per hour

    So the rate of change of area is +250

    So \frac{dA}{dt} = 250

    Often the units give you a clue. Here square metres per hour suggests \frac{dA}{dt}

    Are you able to carry on from here?
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    (Original post by Notnek)
    Translate the words into maths:

    The area of the patch is increasing at a rate of 250 square metres per hour

    So the rate of change of area is +250

    So \frac{dA}{dt} = 250

    Often the units give you a clue. Here square metres per hour suggests \frac{dA}{dt}

    Are you able to carry on from here?
    so i have to find dA/dt when area=1900.
    dA/dt = dA/dv * dV/dt
    = dA/dr * dr/dV * dV/dt
    is this right so far?
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    (Original post by elvss567)
    so i have to find dA/dt when area=1900.
    dA/dt = dA/dv * dV/dt
    = dA/dr * dr/dV * dV/dt
    is this right so far?
    No you need to find the rate of increase of the radius so you need to find \frac{dr}{dt}. I'm not sure what your V is for?
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    (Original post by Notnek)
    No you need to find the rate of increase of the radius so you need to find \frac{dr}{dt}. I'm not sure what your V is for?
    oh okay. i probably made a mistake. how do i calculate dr/dt? do i use the formula for area of the circle(pir^2) for it or something?
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    (Original post by elvss567)
    oh okay. i probably made a mistake. how do i calculate dr/dt? do i use the formula for area of the circle(pir^2) for it or something?
    Yes that formula gives you A in terms of r and you can use this to find \frac{dA}{dr}. Then you can use the chain rule in a similar way to your previous attempt.
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    (Original post by Notnek)
    Yes that formula gives you A in terms of r and you can use this to find \frac{dA}{dr}. Then you can use the chain rule in a similar way to your previous attempt.
    thankyou
 
 
 
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