Key Concepts and Formulas

• Surface Area of a Cube: $A = 6a^2$, where $a$ is the side length.
• Volume of a Cube: $V = a^3$, where $a$ is the side length.
• Related Rates and Chain Rule: If $y = f(x)$ and $x = g(t)$, then $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$.

Step-by-Step Solution

Step 1: Define Variables and State Given Information

We define the following variables:

• $a$: side length of the cube (in cm)
• $A$: surface area of the cube (in cm$^2$)
• $V$: volume of the cube (in cm$^3$)
• $t$: time (in seconds)

We are given that $\frac{dA}{dt} = 3.6$ cm$^2$/sec and we want to find $\frac{dV}{dt}$ when $a = 10$ cm.

Step 2: Write Down the Formulas for Surface Area and Volume

The surface area of a cube is given by $A = 6a^2$, and the volume of a cube is given by $V = a^3$.

Step 3: Differentiate the Surface Area Formula with Respect to Time

We differentiate $A = 6a^2$ with respect to $t$ using the chain rule: $\frac{dA}{dt} = \frac{d}{dt}(6a^2) = 6 \cdot 2a \cdot \frac{da}{dt} = 12a \frac{da}{dt}$ This equation relates the rate of change of the surface area to the rate of change of the side length.

Step 4: Differentiate the Volume Formula with Respect to Time

We differentiate $V = a^3$ with respect to $t$ using the chain rule: $\frac{dV}{dt} = \frac{d}{dt}(a^3) = 3a^2 \frac{da}{dt}$ This equation relates the rate of change of the volume to the rate of change of the side length.

Step 5: Solve for $\frac{da}{dt}$ using the given $\frac{dA}{dt}$

We have $\frac{dA}{dt} = 12a \frac{da}{dt}$ and $\frac{dA}{dt} = 3.6$. Substituting the given value, we get: $3.6 = 12a \frac{da}{dt}$ Solving for $\frac{da}{dt}$, we have: $\frac{da}{dt} = \frac{3.6}{12a} = \frac{0.3}{a}$

Step 6: Substitute $\frac{da}{dt}$ into the equation for $\frac{dV}{dt}$

We have $\frac{dV}{dt} = 3a^2 \frac{da}{dt}$. Substituting $\frac{da}{dt} = \frac{0.3}{a}$, we get: $\frac{dV}{dt} = 3a^2 \left(\frac{0.3}{a}\right) = 0.9a$

Step 7: Evaluate $\frac{dV}{dt}$ when $a = 10$ cm

We substitute $a = 10$ into the equation for $\frac{dV}{dt}$: $\frac{dV}{dt} = 0.9(10) = 9$ Therefore, the rate of change of the volume when $a = 10$ cm is 9 cm$^3$/sec.

Common Mistakes & Tips

• Units: Always include the units in your calculations and final answer. This helps in verifying the correctness of your solution.
• Chain Rule: Remember to apply the chain rule when differentiating with respect to time.
• Substitution Timing: Only substitute the value of $a$ after differentiating. Substituting before differentiating will lead to an incorrect result.

Summary

We were given the rate of change of the surface area of a cube and asked to find the rate of change of its volume at a specific instant. We used the formulas for the surface area and volume of a cube, differentiated them with respect to time, and then solved for the unknown rate. The rate of change of the volume when the side length is 10 cm is 9 cm$^3$/sec.

The final answer is \boxed{9}, which corresponds to option (A).