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101Ryan101
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If the length of one a side of a cube is increasing at a rate of 4m/s how fast is the surface area of the cube increasing when the length is 16m.
 3 years ago
 3 years ago
101Ryan101 Group Title
If the length of one a side of a cube is increasing at a rate of 4m/s how fast is the surface area of the cube increasing when the length is 16m.
 3 years ago
 3 years ago

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pk51 Group TitleBest ResponseYou've already chosen the best response.0
This is related rates, right?
 3 years ago

pk51 Group TitleBest ResponseYou've already chosen the best response.0
I get 768 also. Hint. Relate the information, take a derivative, plug in what you know.
 3 years ago

101Ryan101 Group TitleBest ResponseYou've already chosen the best response.0
Oh thanks.. opps.
 3 years ago

Alchemista Group TitleBest ResponseYou've already chosen the best response.1
Its surface area not volume: \[A = 6\cdot l^2\]\[\frac{dl}{dt} = 4\]\[\frac{dA}{dl} = 12 \cdot l \cdot \frac{dl}{dt}\]\[12 \cdot 16 \cdot 4 = 768\]
 3 years ago

101Ryan101 Group TitleBest ResponseYou've already chosen the best response.0
so if I want to know the rate of increase of volume it's increasing much faster than the surface area so ... (I'm a bit confused as to taking the derivative of V here...) I know dL/dt = 4m/s so for volume...when the length is 16m V = L^3 dV/dt =(3L^2)dL/dt = (3)(16)^2)(4) = 3072m/s I assume that's correct. But I'm not used to seeing an extra dL/dt. I guess the explanation for that is that it's a dependent variable.
 3 years ago
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