Question

The volume of a cube decreases at a rate of 0.5ft cubed per minute. What is the rate of change of the side length when the side lengths are 12 ft? (Looking for how to do it not an answer)

Answer 1

The solution to the problem is as follows:

a cube has sides s 

V = s^3 ---> differentiating 

dV/dt = 3s^2 * ds/dt 

0.5 ft^3 / min = 3 * (12ft)^2 * ds/dt 

(1/2) ft^3 / min = 3 * 144 ft^2 * ds/dt 

(144/6) ft^3 / ( ft^2 min ) = ds/dt 

24 ft / min = ds/dt 

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Answer 2

Answer:

0.0016 feet per minute

Step-by-step explanation:

Volume =$(side)^{3}$

V=   $(x)^{3}$

taking derivative both sides ,we get

$\frac{dV}{dt} = 3x^2\frac{dx}{dt}$

$\frac{dV}{dt} = 0.5$     [ given ]

 $0.5= 3x^2\frac{dx}{dt}$   at x =12

$0.5= 3{(12)}^2\frac{dx}{dt}$

   $\frac{0.5}{3(12)^2}=\frac{dx}{dt}$

           $\frac{dx}{dt}=0.0016 ft per minute$