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
$\frac{\mathrm{ft}^{2}}{\sec }$
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Step 1 ：Given \( \frac{dV}{dt} = 190 \) 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 \( A = 6s^2 \)

Step 4 ：Differentiate volume with respect to time \( t \) to find \( \frac{ds}{dt} \): \( \frac{dV}{dt} = 3s^2 \frac{ds}{dt} \)

Step 5 ：Solve for \( \frac{ds}{dt} \): \( \frac{ds}{dt} = \frac{\frac{dV}{dt}}{3s^2} = \frac{190}{3(4)^2} = \frac{190}{48} = 3.958 \) feet per second

Step 6 ：Differentiate surface area with respect to time \( t \) to find \( \frac{dA}{dt} \): \( \frac{dA}{dt} = 12s \frac{ds}{dt} \)

Step 7 ：Plug in values of \( s \) and \( \frac{ds}{dt} \): \( \frac{dA}{dt} = 12(4)(3.958) = 190.368 \) square feet per second

Step 8 ：\(\boxed{\frac{dA}{dt} = 190.368} \) square feet per second