Related Rates (Two Formulas)
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The volume of a cube is increasing at a constant rate of 190 cubic feet per second. At the instant when the side length of the cube is 4 feet, what is the rate of change of the surface area of the cube? Round your answer to three decimal places (if necessary).
Answer Attempt 1 out of 2
$\frac{{ft}^{2}}{\sec}$
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Answer

$\frac{d A}{d t} = 190.368$ square feet per second

Steps

Step 1 ：Given $\frac{d V}{d t} = 190$ cubic feet per second and side length $s = 4$ feet

Step 2 ：Volume of a cube $V = s^{3}$

Step 3 ：Surface area of a cube $A = 6 s^{2}$

Step 4 ：Differentiate volume with respect to time $t$ to find $\frac{d s}{d t}$ : $\frac{d V}{d t} = 3 s^{2} \frac{d s}{d t}$

Step 5 ：Solve for $\frac{d s}{d t}$ : $\frac{d s}{d t} = \frac{\frac{d V}{d t}}{3 s^{2}} = \frac{190}{3 (4)^{2}} = \frac{190}{48} = 3.958$ feet per second

Step 6 ：Differentiate surface area with respect to time $t$ to find $\frac{d A}{d t}$ : $\frac{d A}{d t} = 12 s \frac{d s}{d t}$

Step 7 ：Plug in values of $s$ and $\frac{d s}{d t}$ : $\frac{d A}{d t} = 12 (4) (3.958) = 190.368$ square feet per second

Step 8 ： $\frac{d A}{d t} = 190.368$ square feet per second

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