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Chapter 7: Problem 6

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

Short Answer

Expert verified
Find \( s = \frac{2 - e^{-13.4}}{\ln(20)} \) and the slope \( m \) at \( t = 10 \) for the tangent line equation.

Step by step solution

01

Understand the Function

The given function is \( f(s, t) = s \ln(2t) + e^{-1.34t} \). We need to understand that this is a function of two variables, \( s \) and \( t \).
02

Establish the Contour Curve

We are tasked with finding the contour where \( f(s,t) = 2 \). This means solving \( s \ln(2t) + e^{-1.34t} = 2 \) for a given range \( 1 \leq t \leq 20 \).
03

Solve for \( s \) in terms of \( t \)

Re-arrange the contour equation to solve for \( s \):\[ s = \frac{2 - e^{-1.34t}}{\ln(2t)} \]for \( 1 \leq t \leq 20 \).
04

Determine the Tangent Line Equation at \( t = 10 \)

First, substitute \( t = 10 \) into \( s(t) = \frac{2 - e^{-1.34t}}{\ln(2t)} \) to find the value of \( s \). Then find the derivative of \( s \) with respect to \( t \) to determine the slope at \( t = 10 \).
05

Compute \( s \) at \( t = 10 \)

Substitute \( t = 10 \) into \( s(t) = \frac{2 - e^{-1.34t}}{\ln(2t)} \): \[ s = \frac{2 - e^{-13.4}}{\ln(20)} \]. This value of \( s \) will be where the tangent line intersects \( t = 10 \).
06

Differentiate to Find Slope

Differentiate \( s(t) = \frac{2 - e^{-1.34t}}{\ln(2t)} \) using the quotient rule to find \( \frac{ds}{dt} \).
07

Evaluate the Slope at \( t = 10 \)

Calculate the derivative at \( t = 10 \). This will give the slope of the tangent line at \( t = 10 \).
08

Write the Equation of the Tangent Line

Using the point-slope form of the line \( y - y_1 = m(x - x_1) \), where \( (t_1, s_1) \) is the point and \( m \) is the slope, we can write the equation of the tangent line at \( t = 10 \).

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Key Concepts

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

Contour Curves
In calculus, contour curves are a way to visualize a multivariable function. Imagine slicing through a 3D graph of the function at a constant height. Each slice creates a curve on a plane, much like a topographic map shows different elevations.
  • A contour curve represents all points where the function has the same value.
  • For a function of two variables, such as our function, these curves lie in a 2D plane.
Consider the activity's function: \(f(s, t) = s \ln(2t) + e^{-1.34t}\). Here, our task is to find a contour curve where \(f(s, t) = 2\), meaning every point \((s, t)\) on this curve results in the same function value of 2. By rearranging this equation for \(s\), you can express \(s\) as a function of \(t\), showing how \(s\) changes as \(t\) varies, keeping \(f(s, t) = 2\).
Tangent Line
Understanding tangent lines help us grasp how a function behaves at a specific point. Imagine a curve on a graph. The tangent line 'touches' the curve at that point. It is essentially a straight line that has the same slope as the curve at that point.
  • The slope of a tangent line tells us how steep the curve is at the point of tangency.
  • For the contour surface studied, the tangent line at \((t = 10, s)\) gives insights into how the contour behaves near that point.
For our exercise, we look at the function at \(t = 10\). Once we find the slope of the tangent line at this point, we use the point-slope form to write its equation, providing a linear approximation of the contour curve at \(t = 10\).
Partial Derivatives
Partial derivatives are fundamental in working with multivariable functions. They provide information on how a function changes as we vary each input variable separately, holding others constant.
  • The partial derivative with respect to \(s\) focuses on how the function changes as \(s\) changes alone.
  • The partial derivative with respect to \(t\) examines changes as \(t\) varies, with \(s\) constant.
In this scenario, we calculated \(\frac{ds}{dt}\) to find the slope of the tangent line. To do this, we applied the quotient rule, which is a method for differentiating functions that are ratios of other functions. By differentiating \(s(t)\), you understand how rapidly \(s\) changes with respect to \(t\), which directly relates to the steepness of the contour curve.
Multivariable Functions
Multivariable functions generalize calculus concepts to more than one variable. They serve a crucial role in many fields, including physics, economics, and engineering, allowing us to model complex systems that depend on multiple factors.
  • In a function of two variables, like \(f(s, t)\), both \(s\) and \(t\) are inputs that affect the output number \(f(s, t)\).
  • Such functions are often visualized as surfaces, with contour curves representing slices of constant output value.
Analyzing these types of functions involves evaluating how they change with respect to each variable individually or jointly. This understanding is essential for solving real-world problems where altering one or more variables can lead to significant outcome differences.

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Most popular questions from this chapter

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 \(0 \leq u \leq 5\).

Competitive Sales \(\quad\) Two vending machines sit side by side in a college dorm. One machine sells Coke products, and the other sells Pepsi products. Daily sales of Coke products, based on the prices of the products in the two machines, can be modeled as $$ \begin{aligned} S(c, p)=& 196.42 p-50.2 c^{2}+9.6 c \\ &+66.4-1.04 c p \text { cans } \end{aligned} $$ when Coke products cost \(c\) dollars and Pepsi products cost \(p\) dollars. a. Calculate the rate of change of the sale of Coke products with respect to the price of Coke products when Coke products cost \(\$ 0.75\) and Pepsi products cost \(\$ 1.25\) b. Calculate the rate of change of the sale of Coke products with respect to the price of Pepsi products when Coke products cost \(\$ 0.75\) and Pepsi products \(\operatorname{cost} \$ 1.25\) c. Calculate and interpret the two rates of change \(\left.\frac{\partial S}{\partial c}\right|_{(c, p)=(1.30,1.20)}\) and \(\left.\frac{\partial S}{\partial p}\right|_{(c, p)=(1.30,1.20)}\)

Price Index In \(1986,\) Cotterill developed a model for measuring the performance of supermarkets by considering their price index level. The price index level was an aggregate of 121 representative prices. The lowest-price supermarket was assigned a price index of 100 . According to the Cotterill study, the price index level of an independent supermarket can be modeled as $$ \begin{aligned} P(c, d, p, s)=& 109.168-0.730 s+0.027 s^{2} \\ &+0.002 d-0.041 p+0.175 c \end{aligned} $$ where the supermarket has \(s\) thousand square feet of sales space and is \(d\) miles from the warehouse, and the consumer base grew by \(p\) thousand people in 10 years and had a per capita income of \(c\) thousand dollars. Assume that the distance from a supermarket to its distribution warehouse is 100 miles and that the consumer base grew by 10,000 people in 10 years. (Source: \(\mathrm{P}\). G. Helmberger and \(\mathrm{J}\). P. Chavas, The Economics of Agricultural Prices, Upper Saddle River, NJ: Prentice-Hall, 1996 ) a. Write an equation for \(P(c, s)=P(c, 100,10, s)\). b. Draw the 110 contour curve for \(P(c, s)\) for supermarkets containing between 5000 and 25,000 square feet of sales space.

Peach Consumption The per capita consumption of peaches can be modeled as $$ C(p, i)=2 \ln i+2.7183^{-p}+4 \text { pounds } $$ where the price of peaches is \(\$(1.50+p)\) per pound and the person lives in a family with annual income \(\$ 10,000 i\). a. Use this model to calculate the rate of change of the per capita consumption of peaches with respect to yearly income when the yearly income is \(\$ 30,000\) and the price is \(\$ 1.70\) per pound. b. Use this model to calculate the rate of change of the per capita consumption of peaches with respect to price when the yearly income is \(\$ 30,000\) and the price is \(\$ 1.70\) per pound.

Payments The amount of a monthly payment on a loan with \(6 \%\) interest compounded monthly can be calculated as $$ m(A, t)=\frac{0.005 A}{1-0.9419^{t}} \text { dollars } $$ when the loan is for \(A\) dollars and is to be repaid over \(t\) years. a. What is the monthly payment for a loan of \(\$ 10,000\) to be repaid over a period of 5 years? b. Approximate the amount that could be borrowed withour increasing or decreasing the monthly payment determined in part \(a\) if the term of the loan is 4 years instead of \(5 .\)
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