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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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

Mathematics
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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
Okay but how do I find dA/dt @welshfella
check my post i found dA/ds in that

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dA/dt = ds/dt * dA/ds = 1.5 * 24
Ohhh so it would be 1.5(24)
yea
But - because it's falling?
oh right my mistake its -36
Okay I understand thank you for your help!
yw

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