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tmagloire1

  • one year ago

An ice block is melting at the rate of 1.5 inches per hour. How fast is the surface area of the ice cube changing at the instant the ice block has a side length of 2 inches? −12 in2/hour −18 in2/hour −36 in2/hour −24 in2/hour

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  1. welshfella
    • one year ago
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    if side lenght = s Area A = 6s^2 ds/dt = 1.5 dA/ds = 12x = 24 at s = 2 Rate of change of surface area at s = 2 = dA/dt is dA/dt = ds/dt *dA/ds

  2. tmagloire1
    • one year ago
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    Okay but how do I find dA/dt @welshfella

  3. welshfella
    • one year ago
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    check my post i found dA/ds in that

  4. welshfella
    • one year ago
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    dA/dt = ds/dt * dA/ds = 1.5 * 24

  5. tmagloire1
    • one year ago
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    Ohhh so it would be 1.5(24)

  6. welshfella
    • one year ago
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    yea

  7. tmagloire1
    • one year ago
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    But - because it's falling?

  8. welshfella
    • one year ago
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    oh right my mistake its -36

  9. tmagloire1
    • one year ago
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    Okay I understand thank you for your help!

  10. welshfella
    • one year ago
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    yw

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