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Differential Calculus - Word Problems

  1. Apr 14, 2010 #1
    1. The problem statement, all variables and given/known data

    A spherical balloon is being inflated at the rate of 10 cu in/sec. Find the rate of change of the area when the balloon has a radius of 6 in.

    2. Relevant equations

    [tex] V = \frac {4}{3} \pi r^{3} [/tex] and [tex] A = 4 \pi r^{2} [/tex]

    3. The attempt at a solution

    [tex] \frac {dV}{dt} = \frac{4}{3} \pi 3r^{2}\frac{dr}{dt} [/tex]

    [tex] \frac {dA}{dt} = 4 \pi 2r\frac{dr}{dt} [/tex]

    the value of dV/dt is given in the question so

    [tex] \frac {dV}{dt} = 10 in^{3}/sec [/tex]

    If we substitute the value into the volume equation we can find dr/dt like so

    [tex] 10 = \frac {4}{3} \pi 3r^{2}\frac{dr}{dt} [/tex]

    [tex] \frac {10}{\frac {4}{3} \pi 3r^{2}} = \frac {dr}{dt} [/tex]

    then set r = 6 we get

    [tex] \frac {dr}{dt} = \frac {10}{452.39} = .0221 [/tex]

    Then move on to solve this equation for dA/dt

    [tex] \frac {dA}{dt} = 4 \pi 2r\frac{dr}{dt} [/tex]

    substituting dr/dt value from other equation and setting r = 6 again

    [tex] \frac {dA}{dt} = 10/3 ~= 3.33 in^{2}/sec [/tex]

    Word, problem solved
     
    Last edited: Apr 14, 2010
  2. jcsd
  3. Apr 14, 2010 #2

    Dick

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    Second line. dV/dt=(4/3)*pi*(3*r^2*dr/dt). The r is squared.
     
  4. Apr 14, 2010 #3
    oh yeah it totally is!
     
  5. Apr 14, 2010 #4
    Yeah that was all, thanks man, totally solved. if you could take a look at the other one, I think that I have solved the cone problem as much as I could.. either way though, thanks
     
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