Problem 17 - Calculus Volume 1

Question

Draw and label diagrams to help solve the related-rates problems. The volume of a cube decreases at a rate of \(10 \mathrm{~m} / \mathrm{sec}\). Find the rate at which the side of the cube changes when the side of the cube is \(2 \mathrm{~m}\).

Step-by-Step Solution

Answer

The side of the cube decreases at \(-\frac{5}{6}\) m/sec when the side length is 2 m.

Step 1: Understand the Problem

First, recognize that the problem asks for the rate of change of the side of a cube. We are given that the volume decreases at a constant rate, and we have the side length of the cube at the moment of interest.

Step 2: Write the Formula for Volume of a Cube

Recall the formula for the volume of a cube, which is given by \[ V = s^3 \]where \( V \) is the volume and \( s \) is the side of the cube.

Step 3: Differentiate the Volume with Respect to Time

To find how the side length changes with time, differentiate the volume formula with respect to time \( t \). Apply the chain rule to get:\[ \frac{dV}{dt} = 3s^2 \frac{ds}{dt} \]

Step 4: Substitute the Given Values

We are given \( \frac{dV}{dt} = -10 \) m/sec (the negative sign indicates a decrease) and \( s = 2 \) m. Substitute these values into the differentiated equation:\[ -10 = 3(2)^2 \frac{ds}{dt} \]

Step 5: Solve for the Rate of Change of the Side

Solve the equation from Step 4 for \( \frac{ds}{dt} \):\[ -10 = 12 \frac{ds}{dt} \]\[ \frac{ds}{dt} = \frac{-10}{12} = -\frac{5}{6} \]Thus, the rate at which the side of the cube changes is \[ \frac{ds}{dt} = -\frac{5}{6} \] m/sec.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that helps us understand how things change. It involves finding the derivative of a function, which represents the rate at which a quantity changes with respect to another.
In this problem, we need to find the rate of change of the side of a cube. We are given that the volume of the cube decreases at a known rate. By differentiating the volume formula with respect to time, we can relate this to how the side length changes.
To differentiate the volume of a cube, we apply the chain rule, which allows us to differentiate composite functions. This rule is helpful when a variable indirectly influences another through another variable, like how time affects both volume and side length in this case.
Through differentiation, we can express a physical scenario in mathematical terms, enabling us to make accurate predictions about dynamic systems.

Volume of a Cube

The volume of a cube, a simple yet vital concept, is calculated as the cube of its side length. The formula for the volume is:

• \[ V = s^3 \]

Where \( V \) represents the volume, and \( s \) is the length of the side.
Understanding this formula is crucial because it sets the stage for analyzing how changes in the cube impact its volume.
In the context of related rates, knowing how volume and side length interact helps us determine how changes in one affect the other. A decrease in volume implies a change in side length, given the fixed relationship in the formula.
Without understanding how to calculate the volume, it would be difficult to approach any change-related questions regarding the shape, size, or any property of the cube.

Rate of Change

The rate of change is a crucial concept when discussing related rates and calculus in general. It tells us how quickly or slowly something changes over time or relative to another variable.
In our exercise, the rate at which the cube's volume decreases is given, and our goal is to determine the rate of change of its side length.

• The rate of change of volume, \( \frac{dV}{dt} \), was provided as \(-10\) m/sec.
• We then calculated the rate of change of the side, \( \frac{ds}{dt} \), using the differentiated formula.

Understanding this concept allows us to make predictions or reach useful conclusions in dynamic situations, such as predicting how dimensions alter as a function of time in this scenario.
By computing the rate of change of the side of the cube, we deduced that it changes at a rate of \( -\frac{5}{6} \) m/sec when the side is 2 meters. This tells us how quickly the cube shrinks at that particular moment.