Final answer:
The rate of change of the cube's surface area when the side length is 11 cm and the volume is decreasing by 21 cm^3/s is approximately -7.645 cm^2/s.
Step-by-step explanation:
The student is asking about the rate of change of the surface area of a cube when the volume is decreasing at a constant rate. To find the rate at which the surface area changes, we use the relationship between volume and surface area of a cube and apply calculus, specifically related rates.
Let s be the side length of the cube. The volume V is given by V = s^3, and the surface area SA is given by SA = 6s^2. Given that the volume is decreasing at a rate of 21 cm^3/s, we have dV/dt = -21 cm^3/s. We want to find dSA/dt when s = 11 cm.
Differentiating both sides of V = s^3 with respect to time t, we get 3s^2 * (ds/dt) = dV/dt. Substituting the given rate of change of volume and solving for ds/dt, we find:
3 * 11^2 * (ds/dt) = -21
ds/dt = -21 / (3 * 11^2)
ds/dt = -0.05785123966942149 cm/s
We then differentiate the surface area with respect to s: dSA/ds = 12s, and use ds/dt to find dSA/dt:
dSA/dt = dSA/ds * ds/dt
dSA/dt = 12 * 11 * (-0.05785123966942149)
dSA/dt ≈ -7.6446281 cm^2/s
Therefore, the surface area of the cube is decreasing at approximately -7.645 cm^2/s when the side length is 11 cm.