Consider a spherical cow that is consuming a great deal of hay. He is fattening himself up at a rate of 200 liters/day, so how quickly is her radius growing (in cm/da) when he can't fit through the circular stall door (that's 2 meters wide)?
@zimmah
So you know to use the volume of a sphere?
Ya, that's what we covered
So you are talking about rates, so differentiate the volume equation.
let me pull it up one sec
4/3*pi*r^3 right?
right, since we are talking about rates you will need to differentiate that
one sec i will
4(pi)(r)^3/3 right?
No do you know how to find derivative of r^3?
oh r^3 ok one sec
3r^2
ok so the 3 on top cancels the 3 on bottom
\[V'=\frac{4}{3} \pi (3 r^2) r' =4 \pi r^2 r'\]
but don't forget the derivative or r is actually r' since r is a function of time
V' and r are given to you in your problem. Plug those in and solve for r'
find the derivative of 4pir^2?
No ...I'm asking you to plug in the numbers that were given to you.
at what rate was the volume changing?
200 liters/day i think
Replace V' with that.
Ah, yes, the infamous spherical cow, found in jokes such as this one: Milk production at a dairy farm was low, so the farmer wrote to the local university, asking for help from academia. A multidisciplinary team of professors was assembled, headed by a theoretical physicist, and two weeks of intensive on-site investigation took place. The scholars then returned to the university, notebooks crammed with data, where the task of writing the report was left to the team leader. Shortly thereafter the physicist returned to the farm, saying to the farmer "I have the solution, but it only works in the case of spherical cows in a vacuum."
and what is the diameter? Divide that by 2 to find the radius
What would the diameter be. imma little lost now
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