amonoconnor
  • amonoconnor
Is this the correct answer? "The formula for the surface area or a sphere is: SA = 4pi*r^2. If a snowball melts so that its SA decreases at a rate of 1cm^2/min, find the rate at which the radius decreases when the diameter is 1 meter. Any and all help is greatly appreciated!
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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misty1212
  • misty1212
HI!! ( again )
misty1212
  • misty1212
\[S=4\pi r^2\\ S'=8\pi rr'\]
amonoconnor
  • amonoconnor
\[\frac{dr}{dt} = \frac{1}{400\pi} cm/\min\]

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More answers

misty1212
  • misty1212
your are told \(S'=1\) put \[1=8\pi rr'\] solve for \(r'\)
misty1212
  • misty1212
if the diameter is \(1\) then the radius is \(\frac{1}{2}\)
misty1212
  • misty1212
so no, i don't think that answer is correct
misty1212
  • misty1212
oooh i see the units have changed doe
misty1212
  • misty1212
very tricky
misty1212
  • misty1212
ok now your answer looks a lot better
misty1212
  • misty1212
\[1=8\times 50\pi r'\] so yeah
amonoconnor
  • amonoconnor
Here's what I did: \[SA = 4\pi*r^2\] \[\frac{dSA}{dt} = 4\pi*2r*\frac{dr}{dt}\] \[(1cm^2/\min) = 4\pi(2*50cm)*\frac{dr}{}\] \[1 = 400\pi*\frac{dr}{dt}\] \[\frac{dr}{dt} = \frac{ 1 }{400\pi}\]
amonoconnor
  • amonoconnor
Is that wrong? :/
misty1212
  • misty1212
yeah right lot easier to write \(r'\) instead of \(\frac{dr}{dt}\) but yes you are correct i did not notice they changed the units from cm to mm very tricky (well not really, if i knew how to read carefully) you are right as far as i can see
amonoconnor
  • amonoconnor
Thanks again misty;)
misty1212
  • misty1212
\[\color\magenta\heartsuit\]

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