Problem 39
Question
In Exercises \(37-42,\) write a differential formula that estimates the given change in volume or surface area. The change in the surface area \(S=6 x^{2}\) of a cube when the edge lengths change from \(x_{0}\) to \(x_{0}+d x\)
Step-by-Step Solution
Verified Answer
The change in surface area is approximately \( 12x_0 \, dx \).
1Step 1: Identify the Original Surface Area Formula
The surface area formula for a cube is given by \( S = 6x^2 \), where \( x \) represents the edge length of the cube.
2Step 2: Differentiate the Surface Area Formula
To find the rate of change of the surface area with respect to the edge length, we differentiate the formula \( S = 6x^2 \) with respect to \( x \). This yields \( \frac{dS}{dx} = 12x \).
3Step 3: Apply the Differential Formula
The differential formula can be used to estimate the change \( dS \) in surface area as the edge length changes. Using the derivative from Step 2, we write \( dS = \frac{dS}{dx} \cdot dx = 12x \, dx \).
4Step 4: Evaluate at the Initial Condition
Substitute the initial edge length \( x = x_0 \) into the differential expression: \( dS = 12x_0 \cdot dx \). This represents the estimated change in surface area when the edge length changes from \( x_0 \) to \( x_0 + dx \).
Key Concepts
Surface Area of a CubeRate of ChangeDifferentiationApplications of Derivatives
Surface Area of a Cube
Understanding the surface area of a cube is fundamental in geometry. A cube is a three-dimensional shape with all sides equal. Because all faces of a cube are squares, the formula for the surface area of a cube is six times the area of one square face. Mathematically, this is written as \( S = 6x^2 \), where \( x \) is the edge length of the cube. This equation highlights that the surface area depends on the square of the edge length. Therefore, any change in the edge length results in a quadratic change in the surface area, making this relationship non-linear.
Rate of Change
When analyzing how a quantity changes, the concept of rate of change becomes crucial. It tells us how one quantity varies in relation to another. In differential calculus, the rate of change is often represented by a derivative. In our problem, the quantity of interest is the surface area \( S \) of the cube, which changes as the edge length \( x \) changes. The rate of change of the surface area with respect to the edge length is represented as \( \frac{dS}{dx} \). This gives insight into how fast the surface area is increasing or decreasing for small changes in the cube's edge length.
Differentiation
Differentiation is a key concept in calculus used to determine the rate of change of a function. It involves finding the derivative of a function, which is the function's slope or gradient. In the context of the given exercise, the differentiation of the surface area formula \( S = 6x^2 \) with respect to \( x \) gives us the derivative \( \frac{dS}{dx} = 12x \). This result tells us how the surface area changes as the edge length undergoes a tiny change. Differentiation helps us understand the sensitivity of one variable in relation to changes in another variable, which is why it's a cornerstone concept in calculus.
Applications of Derivatives
Derivatives have broad applications across various fields. In this problem, the derivative tells us how the surface area changes as the cube's size changes. By using the derivative, we can form the differential equation \( dS = 12x \, dx \) to estimate the change in the surface area for a small change \( dx \) in edge length. Such applications are vital in engineering, physics, and economics, where understanding and predicting changes is essential. For instance, in engineering, knowing how material strength changes with modifications to dimensions is critical for construction and design purposes. Derivatives provide the tools needed for these predictive analyses.
Other exercises in this chapter
Problem 38
Find the derivatives of the functions in Exercises \(19-38\) $$ q=\cot \left(\frac{\sin t}{t}\right) $$
View solution Problem 38
Find the first and second derivatives of the functions $$ p=\frac{q^{2}+3}{(q-1)^{3}+(q+1)^{3}} $$
View solution Problem 39
In Exercises \(37-42,\) use implicit differentiation to find \(d y / d x\) and then \(d^{2} y / d x^{2} .\) $$ y^{2}=x^{2}+2 x $$
View solution Problem 39
In Exercises \(39-48,\) find \(d y / d t\) $$ y=\sin ^{2}(\pi t-2) $$
View solution