Answered

The surface area of a cube is given by $S=6 x^2$ where $x$ is the length of its edges measured in mm. The surface area is decreasing at a constant rate of 2 $\text{mm}^2/\text{second}$. At the instant when $x=8$, at what rate (in $\text{mm}/\text{second}$) is $x$ changing?

Verified Answer :

  • Use the surface area formula \(S = 6x^2\) to relate the change in surface area to the change in edge length.
  • Differentiate with respect to time using the chain rule to obtain \(\frac{dS}{dt} = 12x \frac{dx}{dt}\).
  • Substitute the given rate \(\frac{dS}{dt} = -2 \, \text{mm}^2/\text{second}\) and the edge length \(x = 8 \, \text{mm}\) into the derivative equation.
  • Solve the resulting equation to find that the rate of change is \(\boxed{-\frac{1}{48} \, \text{mm}/\text{second}}\).

Explanation

  1. Problem Analysis

The surface area \(S\) of a cube with edge length \(x\) is given by the formula \(S = 6x^2\). We are given that the surface area \(S\) is decreasing at a constant rate of \(2 \, \text{mm}^2/\text{second}\), which means the rate of change of \(S\) with respect to time \(t\) is \(\frac{dS}{dt} = -2 \, \text{mm}^2/\text{second}\). We want to find the rate at which the edge length \(x\) is changing, \(\frac{dx}{dt}\), at the specific instant when the edge length \(x = 8 \, \text{mm}\).

  1. Differentiating the Surface Area Formula

To find the rate of change of the edge length, we relate the surface area \(S\) to the edge length \(x\) through the given equation \(S = 6x^2\). Since both \(S\) and \(x\) are functions of time \(t\), we differentiate both sides of this equation with respect to \(t\) using the chain rule:
\[\frac{dS}{dt} = \frac{d}{dt}(6x^2) = 12x \cdot \frac{dx}{dt}\]

  1. Substituting Values and Solving for the Rate of Change

With the derivative established as \(\frac{dS}{dt} = 12x \cdot \frac{dx}{dt}\), we can isolate \(\frac{dx}{dt}\) to solve for the rate of change of the edge length:
\[\frac{dx}{dt} = \frac{\frac{dS}{dt}}{12x}\]
Now, we substitute the known values into this equation: \(\frac{dS}{dt} = -2 \, \text{mm}^2/\text{second}\) and \(x = 8 \, \text{mm}\).
\[\frac{dx}{dt} = \frac{-2}{12 \cdot 8} = \frac{-2}{96}\]

  1. Final Result and Interpretation

Simplifying the fraction \(\frac{-2}{96}\), we get \(\frac{-1}{48}\). Therefore, the rate at which the edge length \(x\) is changing at the instant \(x = 8 \, \text{mm}\) is \(-\frac{1}{48} \, \text{mm}/\text{second}\). The negative sign indicates that the edge length is decreasing. Thus, the rate is \(\boxed{-\frac{1}{48} \, \text{mm}/\text{second}}\).

Examples

Understanding the rate of change is essential in engineering and manufacturing. For instance, if a thermal expansion process causes a cube-shaped component to shrink uniformly as it cools, an engineer needs to predict how quickly each side length will shorten. By knowing the total surface area's rate of decrease, they can use calculus to determine the exact speed at which the dimensions are changing, ensuring that the part remains within specified tolerances during the cooling cycle.