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Chapter 1: Problem 56

Find a formula for the described function and state its domain. Express the surface area of a cube as a function of its volume.

Short Answer

Expert verified
The function is \( S(V) = 6V^{2/3} \) with domain \( V > 0 \).

Step by step solution

01

Understand the problem

We need to express the surface area of a cube as a function of its volume, meaning we need to find a relationship between these two properties.
02

Recall cube properties

For a cube, the surface area \( S \) is given by \( S = 6a^2 \) and the volume \( V \) is given by \( V = a^3 \), where \( a \) is the length of a side of the cube.
03

Solve for side length in terms of volume

To express the surface area in terms of volume, solve for the side length \( a \) in terms of volume: \( a = V^{1/3} \).
04

Substitute side length into surface area formula

Substitute \( a = V^{1/3} \) into the surface area formula \( S = 6a^2 \):\[S = 6(V^{1/3})^2 = 6V^{2/3}.\]
05

Determine the domain

The volume \( V \) must be positive since it's the volume of a cube. Thus, the domain of the function is \( V > 0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Composition
Function composition is a crucial concept in calculus, especially when expressing one function in terms of another. It allows us to connect different mathematical properties to find a new relationship. When composing functions, we essentially plug one function into another.

In our exercise, we wanted to express the surface area of a cube as a function of its volume. This involves composing these two functions: the surface area and the volume of a cube. By expressing the side length of the cube in terms of its volume and substituting this expression back into the surface area formula, we've created a new function.
  • Volume function: \( V = a^3 \).
  • Surface area function in terms of side length: \( S = 6a^2 \).
Composing these functions involves finding the side length from the volume equation and then substituting into the surface area function. This gives us one function expressed in terms of another, showing how calculus can bridge different mathematical relationships seamlessly.
Volume of a Cube
The volume of a cube is a fundamental geometric concept. It represents the amount of space enclosed within the cube. For a cube, all sides are equal in length, and its volume \( V \) is calculated by:
  • \( V = a^3 \)
where \( a \) is the length of a side.

Understanding and calculating the volume helps in various fields such as engineering and biology, where managing space efficiently is crucial. Notably, in the life sciences, determining volumes can be vital for understanding cell sizes or container capacities.
Surface Area of a Cube
The surface area of a cube measures the total area covered by the six identical square faces that make up the cube. For a cube, each face has the same surface area. It's calculated as:
  • \( S = 6a^2 \)
where \( a \) is the length of a side.

Calculating the surface area is important in practical situations, such as determining the amount of material needed to cover a box or understanding heat exchange processes in biological systems where surface contact is critical. By expressing the surface area in terms of volume, we streamline these calculations, benefiting applications across various domains.
Domain Determination
Determining the domain of a function is essential to understand the values for which our function is valid or meaningful. In mathematics, the domain refers to all the possible input values (or "x" values) that will output real and meaningful results for a function.

In our exercise, when determining the domain of the function \( S = 6V^{2/3} \) (the surface area, expressed in terms of volume), we realize that the volume must always be positive. This is because a negative or zero volume does not make physical sense in real-world scenarios.
  • The domain of our function is thus \( V > 0 \).

Understanding domain is crucial when applying calculus to real-life scenarios, ensuring our mathematical models align with physical realities.

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Most popular questions from this chapter

\(\begin{array}{ll}{\text { Use the table to evaluate each expression. }} \\\ {\text { (a) } f(g(1))} & {\text { (b) } g(f(1))} & {\text { (c) } f(f(1))} \\\ {\text { (d) } g(g(1))} & {\text { (e) }(g \circ f)(3)} & {\text { (f) }(f \circ g)(6)}\end{array}\) \(\begin{array}{|c|c|c|c|c|c|c|}\hline x & {1} & {2} & {3} & {4} & {5} & {6} \\\ \hline f(x) & {3} & {1} & {4} & {2} & {2} & {5} \\ \hline g(x) & {6} & {3} & {2} & {1} & {2} & {3} \\ \hline\end{array}\)

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Dialysis time Hemodialysis is a process by which a machine is used to filter urea and other waste products from an individual's blood when the kidneys fail. The concentration of urea in the blood is often modeled as exponential decay. If \(K\) is the mass transfer coefficient \((\) in \(\mathrm{mL} / \mathrm{min}), c(t)\) is the urea concentration in the blood at time \(t(\mathrm{in} \mathrm{mg} / \mathrm{mL})\) and \(V\) is the blood volume, then \(c(t)=c_{0} e^{-K t / V}\) where \(c_{0}\) is the initial concentration at time \(t=0.\) (a) How long should a patient be put on dialysis to reduce the blood urea concentration from an initial value of 1.65 \(\mathrm{mg} / \mathrm{mL}\) to 0.60 \(\mathrm{mg} / \mathrm{mL}\) , given that \(K=340 \mathrm{mL} / \mathrm{min}\) and \(V=32,941 \mathrm{mL} ?\) (b) Derive a general formula for the dialysis time \(T\) in terms of the initial urea concentration \(c_{0}\) and the target urea concentration \(c(T)\) .

A function is given by a table of values, a graph, a formula, or a verbal description. Determine whether it is one-to-one. \(g(x)=\cos x\)

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