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

Question

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

welshfella · Answer

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

tmagloire1 · Answer

Okay but how do I find dA/dt @welshfella

welshfella · Answer

check my post i found dA/ds in that

welshfella · Answer

dA/dt = ds/dt  * dA/ds = 1.5 * 24

tmagloire1 · Answer

Ohhh so it would be 1.5(24)

welshfella · Answer

yea

tmagloire1 · Answer

But - because it's falling?

welshfella · Answer

oh right my mistake its -36

tmagloire1 · Answer

Okay I understand thank you for your help!

welshfella · Answer

yw