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Math Help - Growing rate

  1. #1
    RAz
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    Question Growing rate

    A drop of water is a perfect sphere and by condensation, the droplet picks up moisture at a rate proportional to its surface area. Show that the drops radius increases at a constant rate.

    What sort of calculus would I use to tackle this problem? Thanks
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  2. #2
    MHF Contributor

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    Quote Originally Posted by RAz View Post
    A drop of water is a perfect sphere and by condensation, the droplet picks up moisture at a rate proportional to its surface area. Show that the drops radius increases at a constant rate.

    What sort of calculus would I use to tackle this problem? Thanks
    Differential Calculus! More specifically, "related rates". Use the formula for volume of a sphere- V= \frac{4}{3}\pi r^3- and the formula for area of the surface- A= 4\pi r^2. Differentiate the volume formula, with respect to t, using the chain rule, and set it equal to a constant times the area. Solve for dr/dt.
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  3. #3
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    Rates of change

    Hello RAz
    Quote Originally Posted by RAz View Post
    A drop of water is a perfect sphere and by condensation, the droplet picks up moisture at a rate proportional to its surface area. Show that the drops radius increases at a constant rate.

    What sort of calculus would I use to tackle this problem? Thanks
    We can re-write the phrase
    the droplet picks up moisture at a rate proportional to its surface area
    as "the rate of change of its volume is proportional to its surface area"; or, with the usual notation:

    \frac{dV}{dt}= kS, for some constant k.

    So, using V = \tfrac43\pi r^3 and S = 4\pi r^2, we get:

    \frac{d}{dt}\Big(\tfrac43\pi r^3\Big)=4k\pi r^2

    \Rightarrow \frac{d}{dr}\Big(\tfrac43\pi r^3\Big)\times\frac{dr}{dt}=4k\pi r^2

    \Rightarrow 4\pi r^2 \frac{dr}{dt}=4k\pi r^2

    \Rightarrow \frac{dr}{dt}=k

    \Rightarrow r is increasing at a constant rate.

    Grandad
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