Related rates and a spherical weather balloon

  • Thread starter 1irishman
  • Start date
  • #1
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(b]1. Homework Statement [/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?

Homework Equations



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

Answers and Replies

  • #2
51
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You need to know the equation for the surface area of a sphere:

[tex]SA = 4 \pi r^2[/tex]
 
  • #3
243
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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:
  • #4
51
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Oh, almost! Try adding units into your calculation to see if you can figure out where you went wrong.
 
  • #5
243
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2cm/min is 5.556X10^-6m/s^2
 
  • #6
243
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ooops...4m = 400cm
so,
the result is 20106.2cm^2/min
 
  • #7
243
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in the future, how do i know which unit to convert?
 

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