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shann803 Group Title

A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.3cm/min. At what rate is the volume of the snowball decreasing when the diameter is 14 cm? (Note the answer is a positive number)

  • one year ago
  • one year ago

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  1. amistre64 Group Title
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    is there an equation for the volume of a sphere?

    • one year ago
  2. shann803 Group Title
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    no :(

    • one year ago
  3. amistre64 Group Title
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    ... im pretty sure there is. spheres have been studied now for quite some time, and some ancient mathmatikers are bound to have come up with some sort of formula that tells them how to determine the volume of a sphere. If not, i know a way that we can become very very rich :)

    • one year ago
  4. shann803 Group Title
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    haha i meant there's none that came with the problem! Not that none exist :)

    • one year ago
  5. amistre64 Group Title
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    ;) you hunt down a formula for it, and then we can see what can be done to find an answer to this question then

    • one year ago
  6. shann803 Group Title
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    it says online that it is 4/3nr^3 so i'm not really sure what that means but thats what i got :) I hope it helps!

    • one year ago
  7. shann803 Group Title
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    the n is a pi

    • one year ago
  8. amistre64 Group Title
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    thats the one i was thinking of to find out how fast the volume is changing, we need to take its derivative; and we are in luck that we only have 1 variable and a power rule to deal with \[V=\frac43\pi~r^3\]

    • one year ago
  9. shann803 Group Title
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    ok sounds good but i have a class right now :( so I dont know how you wanna do this because I'll be in class untill 12 :(

    • one year ago
  10. amistre64 Group Title
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    as long as you know what the power rule for derivatives is, that would help out alot

    • one year ago
  11. shann803 Group Title
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    ok the radius is half the diameter right?

    • one year ago
  12. amistre64 Group Title
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    correct

    • one year ago
  13. shann803 Group Title
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    i can find the derivative pretty easily but i'm just unsure what the radius would be

    • one year ago
  14. shann803 Group Title
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    ok let me get you the derivative real quick

    • one year ago
  15. shann803 Group Title
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    the derivative is 4pir^2 so that would be 4pi(7^2) and that would give me 615.44 about

    • one year ago
  16. amistre64 Group Title
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    very good, but there is one other part that needs to be addressed; the chain rule pops out an r'

    • one year ago
  17. shann803 Group Title
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    ok... im lost

    • one year ago
  18. amistre64 Group Title
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    since the radius is half the diamter; the rate of change of the radius is half the rate of change of the diamter so multiply that result by .3/2

    • one year ago
  19. shann803 Group Title
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    so that is 923.16 or so?

    • one year ago
  20. amistre64 Group Title
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    \[\frac d{dt}(V=\frac43\pi~r^3)\] \[\frac {dV}{dt}=\frac43\pi~3r^2*\frac{dr}{dt}\]

    • one year ago
  21. amistre64 Group Title
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    4pi(7^2)*.3/2 = about 92.36

    • one year ago
  22. shann803 Group Title
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    lol alright i calculated wrong and thanks a million!

    • one year ago
  23. amistre64 Group Title
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    with a little practice, these should make more sense :) good luck

    • one year ago
  24. shann803 Group Title
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    thanks :) ill be posting 2 more on here because I completely suck at this so maybe if you're on later you can help me again :) You actually help you don't just say the answer I like that :) Thank you! I have class now so hopefully I'll talk to you later!

    • one year ago
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