To solve this problem, let's go through each part step-by-step, using the given values and applying differentials.

Given Data

• Edge of the cube, $x = 19$ inches
• Possible error in measurement, $dx = 0.03$ inches

Part (a): Volume of the Cube

1. Volume Formula:
   The volume $V$ of a cube with edge length $x$ is given by: $V = x^3$

2. Differential of Volume (dV):
   Using differentials to approximate the error in volume: $dV = 3x^2 \cdot dx$ Substituting $x = 19$ and $dx = 0.03$: $dV = 3 \times (19)^2 \times 0.03$ Calculating this: $dV = 3 \times 361 \times 0.03 = 32.49 \text{ cubic inches}$

3. Relative Error in Volume:
   The relative error in volume is given by: $\frac{dV}{V} = \frac{3x^2 \cdot dx}{x^3} = \frac{3 \cdot dx}{x}$ Substituting $x = 19$ and $dx = 0.03$: $\frac{dV}{V} = \frac{3 \times 0.03}{19} \approx 0.005$

Part (b): Surface Area of the Cube

1. Surface Area Formula:
   The surface area $S$ of a cube with edge length $x$ is: $S = 6x^2$

2. Differential of Surface Area (dS):
   Using differentials to approximate the error in surface area: $dS = 12x \cdot dx$ Substituting $x = 19$ and $dx = 0.03$: $dS = 12 \times 19 \times 0.03 = 6.84 \text{ square inches}$

3. Relative Error in Surface Area:
   The relative error in surface area is: $\frac{dS}{S} = \frac{12x \cdot dx}{6x^2} = \frac{2 \cdot dx}{x}$ Substituting $x = 19$ and $dx = 0.03$: $\frac{dS}{S} = \frac{2 \times 0.03}{19} \approx 0.003$

Final Answers

• (a) Volume of the Cube

  • Possible error: $\pm 32.490$ cubic inches
  • Relative error: $\pm 0.005$

• (b) Surface Area of the Cube

  • Possible error: $\pm 6.840$ square inches
  • Relative error: $\pm 0.003$

Would you like more details on any part of the solution or further clarification?

Here are five related questions that expand on this topic:

1. How would the errors change if the cube’s edge length measurement had a different possible error, say 0.05 inches?
2. What would the possible error and relative error be if the shape were a sphere with the same radius?
3. How does the relative error in volume compare to the relative error in surface area for larger cubes?
4. What is the significance of relative error in practical applications involving volume and surface area calculations?
5. How would you apply differentials to find the possible error in the diagonal of the cube?

Tip: When using differentials for error approximation, remember that the relative error formula often simplifies by canceling terms, making it easier to compute the ratio without recalculating the entire measurement.