SOLUTION: A spherical ballon is being filled with air at a constant rate of 2cm^3 per second. By using the chain rule dr/dt equals to 1/2(pi)(r^2) where t is in seconds. Find The rate at whi

Algebra ->  Distributive-associative-commutative-properties -> SOLUTION: A spherical ballon is being filled with air at a constant rate of 2cm^3 per second. By using the chain rule dr/dt equals to 1/2(pi)(r^2) where t is in seconds. Find The rate at whi      Log On


   

Question 1142721: A spherical ballon is being filled with air at a constant rate of 2cm^3 per second. By using the chain rule dr/dt equals to 1/2(pi)(r^2) where t is in seconds. Find The rate at which the radius is increasing when r=54
Answer by ikleyn(52890) About Me  (Show Source):
You can put this solution on YOUR website!
.

What is written in the post - is INCORRECT. 


The volume of a sphere   V(r) = %284%2F3%29%2Api%2Ar%5E3,  as EVERYBODY knows.


From the formula,  %28dV%29%2F%28dt%29 = %28dV%29%2F%28dr%29.%28%28dr%29%2F%28dt%29%29 = %284%2F3%29%2A3%2Api%2Ar%5E2.%28%28dr%29%2F%28dt%29%29 = 4%2Api%2Ar%5E2.%28%28dr%29%2F%28dt%29%29.


Substitute here  %28dV%29%2F%28dt%29 = 2 cm^3/s,  which is given.  You will get


                 2 = 4%2Api%2Ar%5E2.%28%28dr%29%2F%28dt%29%29,


hence,


                %28%28dr%29%2F%28dt%29%29 = %281%2F2%29%2A%281%2F%28pi%2Ar%5E2%29%29,


totally different from what is written in your post.


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To get the answer to your question, simply substitute r= 54 into my last formula.