anonymous
  • anonymous
If a snowball melts so that its surface area decreases at a rate of 10 cm2/min, find the rate at which the diameter decreases when the diameter is 8 cm.
Mathematics
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SOLVED
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katieb
  • katieb
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campbell_st
  • campbell_st
so this is a related rates question so is the snowball a sphere...
baru
  • baru
answered same ques ^

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anonymous
  • anonymous
Yes it is a sphere campbell_st
campbell_st
  • campbell_st
ok... so using the diameter in stead of the radius in the surface area of a sphere it becomes \[S = \pi \times (\frac{d}{2})^2 ~~or~~ S = \frac{ \pi \times d^2}{4}\] does that make sense...?
anonymous
  • anonymous
Oh! thanks baru...It's not exactly the same but I'll see
anonymous
  • anonymous
Shouldn't it be 4pi(d/2)^2 ? Campbell_st
campbell_st
  • campbell_st
oops forgot the 4 \[S = 4 \pi (\frac{d}{2})^2 ~~~or~~S = \frac{4\pi d^2}{4} \] so the surface area in terms of diameter is \[S = \pi d^2\] sorry about the error
campbell_st
  • campbell_st
so can you find the derivative od S with respect to d...?
anonymous
  • anonymous
That will be 2pi*d right?
campbell_st
  • campbell_st
correct so you now know \[\frac{dS}{dD} = 2\pi d\] you also know \[\frac{dS}{dt} = 10 \] you could use -10... but everything is decreasing... so you need to use \[\frac{dD}{dt} = \frac{dD}{dS} \times \frac{dS}{dt}\] so find the reciprocal of \[\frac{dS}{dD} ~~then~~multiply~~by~~\frac{dS}{dt}\] to get the equation forthe rate of change in the diameter with respect to time... then just substitute d = 8 for the answer... hope it helps
anonymous
  • anonymous
So then the final answer would be 5/8pi ?
campbell_st
  • campbell_st
That's what I got... and it should be cm/min
anonymous
  • anonymous
Thank you so much Campbell!!! :-D
campbell_st
  • campbell_st
glad to help

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