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Chapter 7: Problem 2

Write a formula for the indicated rate of change. $$ S(c, k)=c\left(39^{k}\right) ; \frac{d c}{d k} $$

Short Answer

Expert verified
\( \frac{d c}{d k} = 0 \) because \( c \) is independent of \( k \).

Step by step solution

01

Understand the Function

The given function is \( S(c, k) = c(39^k) \). Here, \( S \) is a function of two variables \( c \) and \( k \). The problem asks for the rate of change of \( c \) with respect to \( k \), which implies finding \( \frac{d c}{d k} \).
02

Identify the Related Rates

Since \( c \) is a parameter (a multiplication factor in the context of the function \( S \)), we consider how \( c \) can change with respect to \( k \). However, \( c \) is not dependent on \( k \) within the expression of \( S \).
03

Analyze Dependency

Examine if there is an implicit relation that defines \( c \) as a function of \( k \). Without additional information or given functions linking \( c \) and \( k \), \( c \) is independent of \( k \), implying no direct change in \( c \) with changes in \( k \).
04

Conclusion on Rate of Change

Since there is no dependency of \( c \) on \( k \) presented in the function \( S(c, k) = c(39^k) \), \( \frac{d c}{d k} = 0 \). This is because the derivative of a constant with respect to any other variable is zero.

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Key Concepts

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

Partial Derivatives
In calculus, partial derivatives are an essential tool for dealing with functions of multiple variables. A partial derivative focuses on how a function changes as only one of the input variables changes, while all other variables are held constant. For instance, if we have a function \( f(x, y) \), the partial derivative of \( f \) with respect to \( x \) considers how \( f \) changes with variations in \( x \), treating \( y \) as a constant.

Partial derivatives are denoted as \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). This is different from the standard derivative, which is used when a function depends on a single variable.
  • To compute a partial derivative, differentiate with respect to one variable while treating the others as constants.
  • Partial derivatives help us understand the slope of a surface defined by a function of two or more variables.
  • They are foundational in fields like physics, engineering, and economics, where functions often depend on multiple factors.
Functions of Multiple Variables
Functions of multiple variables are mathematical expressions that depend on more than one input. These functions are essential in many real-world applications since they model situations where multiple factors influence a result. For example, in the given function \( S(c, k) = c(39^k) \), both \( c \) and \( k \) are variables, with \( c \) being a constant factor for any change in \( k \).

  • Such functions are commonly denoted as \( f(x_1, x_2, ..., x_n) \).
  • They allow for the exploration of how different variables interact and contribute to the outcome of a function.
  • In the case of \( S(c, k) \), understanding that \( c \) does not depend on \( k \) is crucial for determining rate changes, making it a vital part of analysis.
Understanding these interactions through visualization and analysis helps in predicting behavior and optimizing results in various fields.
Rate of Change
The rate of change is a significant concept in calculus that describes how one quantity varies relative to another. In the context of functions with multiple variables, it often helps in analyzing how a dependent variable changes with respect to changes in independent variables.

For the function \( S(c, k) = c(39^k) \), we investigate the rate of change of \( c \) with respect to \( k \), denoted as \( \frac{d c}{d k} \). In this scenario, \( c \) is constant with respect to \( k \), meaning its rate of change is zero. This outcome occurs because, unless explicitly defined by an equation, \( c \) and \( k \) are independent variables in this function.
  • In simpler terms, if changing \( k \) doesn't affect \( c \), then \( \frac{d c}{d k} = 0 \).
  • This concept is crucial in understanding the dynamics and sensitivity of systems described by mathematical functions.
  • Analyzing rates of change helps in modeling and predicting how variations in one parameter can impact overall behavior, important in fields like economics and natural sciences.

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Most popular questions from this chapter

Future Value The future value \(F(t, r)\) of an investment of \(\$ 1000\) after \(t\) years in an account for which the interest rate is \(100 r \%\) compounded continuously is given by the function \(F(t, r)=1000 e^{r t}\) dollars. a. Find and interpret \(F(10, r)\). b. Find and interpret \(\left.\frac{\partial F}{\partial r}\right|_{(t, r)=(10,0.072)}\). c. Explain how the rate of change in part \(b\) is related to a graph of the cross-sectional function in part \(a\). Illustrate graphically.

Draw the contour curves \(P(s, u)=40, P(s, u)=60,\) and \(P(s, u)=100\) for the function \(P(s, u)=38.6 s-2 s u+\) \(13 u+0.99 u^{2}\) for \(0 \leq u \leq 5\).

Sketch the contour curve indicated in the activity. Sketch the tangent line indicated and calculate its slope. $$ \begin{aligned} &f(s, t)=s \ln (2 t)+e^{-1.34 t}\\\ &f(s, t)=2 \text { for } 1 \leq t \leq 20\\\ &\text { tangent line at } t=10 \end{aligned} $$

What must be true about the partial derivatives of a function with two input variables at a relative maximum? Explain from a graphical viewpoint why this is true.

a. Calculate the output associated with the given input values. b. Approximate the change needed in one input variable to compensate for the given change in the other input variable. $$ \begin{array}{l} f(h, s)=0.00091 s\left[0.103\left(2.5^{b}\right)+1\right] \quad \text { when } \quad h=3.5, \\ s=1148, \text { and } \Delta h=-0.5 \end{array} $$
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