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Related rates and a spherical weather balloon

  1. Jul 20, 2011 #1
    (b]1. The problem statement, all variables and given/known data[/b]

    A spherical weather balloon has a radius of 1m when it is 1500m high.
    You observe that the radius increases at a rate of 2cm/min as it continues to rise.
    At what rate is the surface area increasing when the radius is 4m?

    2. Relevant equations

    I thought volume of a sphere might be useful to help solve this question.
    V = 4/3pi(r^3)



    3. The attempt at a solution
    I was not sure if the first sentence of the question could help me solve this problem or not, so i decided to try and solve it without using the info from there.

    It looks like from the given information that the following might be the case:

    if i let r be the radius then r = 4
    the rate at which the radius increases with respect to time could be shown as:
    dr/dt = 2

    Now, this is where i get stuck; i am not sure if i am correct in making my next step to differentiate the volume of a sphere function, but this is what i arrived at:
    if V=4/3pi(r^3)
    THEN
    dV/dt=4pi(r^2)
    I am stuck at this point, because i don't know how to take it further, help please?
     
    Last edited: Jul 20, 2011
  2. jcsd
  3. Jul 20, 2011 #2
    You need to know the equation for the surface area of a sphere:

    [tex]SA = 4 \pi r^2[/tex]
     
  4. Jul 20, 2011 #3
    thank you for the formula.

    Are my calculations correct?

    DS/dt=8(pi)(r)dr/dt
    =8(pi)(4)(2)
    =201.062cm^2/min
     
    Last edited: Jul 20, 2011
  5. Jul 22, 2011 #4
    Oh, almost! Try adding units into your calculation to see if you can figure out where you went wrong.
     
  6. Jul 22, 2011 #5
    2cm/min is 5.556X10^-6m/s^2
     
  7. Jul 22, 2011 #6
    ooops...4m = 400cm
    so,
    the result is 20106.2cm^2/min
     
  8. Jul 22, 2011 #7
    in the future, how do i know which unit to convert?
     
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