A spherical balloon is inflated with helium at a rate of 140pift^3/min How fast is the balloon’s radius increasing when the radius is 7ft

Question

A spherical balloon is inflated with helium at a rate of 140pift^3/min How fast is the balloon’s radius increasing when the radius is 7ft

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Nick 3 years 2021-08-20T00:26:55+00:00 1 Answers 9 views 0

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  1. sori
    0
    2021-08-20T00:28:02+00:00
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    Answer:

    dr = 0.71 ft/min

    the balloon’s radius is increasing at 0.71 ft/min when the radius is 7ft.

    Explanation:

    Given;

    Rate of inflation dV = 140pift^3/min

    Radius r = 7 ft

    Change in radius = dr

    Volume of a spherical balloon is;

    V = (4/3)πr^3

    The change in volume can be derived by differentiating both sides;

    dV = (4πr^2)dr

    Making dr the subject of formula;

    dr = dV/(4πr^2)

    Substituting the given values;

    dr = 140π/(4π×7^2)

    dr = 0.714285714285 ft/min

    dr = 0.71 ft/min

    the balloon’s radius is increasing at 0.71 ft/min when the radius is 7ft.

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