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Chapter 4: Problem 19

If \(U\) and \(V\) are positive constants, find all critical points of $$ F(t)=U e^{t}+V e^{-t} $$

Short Answer

Expert verified
The critical point is \( t = \frac{1}{2} \ln\left(\frac{V}{U}\right) \).

Step by step solution

01

Find the Derivative

The first step is to find the first derivative of the function \( F(t) = U e^{t} + V e^{-t} \). To do this, apply the derivative rules for exponential functions. The derivative of \( e^{t} \) is \( e^{t} \), and the derivative of \( e^{-t} \) is \( -e^{-t} \). Therefore, the derivative is:\[ F'(t) = U e^{t} - V e^{-t} \]
02

Set the Derivative to Zero

To find the critical points, set the derivative \( F'(t) \) equal to zero and solve for \( t \):\[ U e^{t} - V e^{-t} = 0 \]
03

Solve the Equation

Rearrange the equation to find a solution for \( t \). First, move \( V e^{-t} \) to the right side:\[ U e^{t} = V e^{-t} \]Multiply both sides by \( e^{t} \) to eliminate the negative exponent:\[ U e^{2t} = V \]Now solve for \( e^{2t} \):\[ e^{2t} = \frac{V}{U} \]Take the natural logarithm of both sides:\[ 2t = \ln\left(\frac{V}{U}\right) \]Finally, solve for \( t \):\[ t = \frac{1}{2} \ln\left(\frac{V}{U}\right) \]
04

Verify the Critical Point

The critical point found is \( t = \frac{1}{2} \ln\left(\frac{V}{U}\right) \). Since \( U \) and \( V \) are positive constants, \( \ln\left(\frac{V}{U}\right) \) is a real number, and hence \( \frac{1}{2} \ln\left(\frac{V}{U}\right) \) is a valid critical point for the function \( F(t) \).

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Key Concepts

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

Critical Points
Critical points are essential in understanding the behavior of functions. They are the values of the variable where the function's derivative is zero, which marks the points where the function may change from increasing to decreasing or vice versa. To find a critical point:
  • Calculate the function's first derivative.
  • Set the derivative equal to zero to identify critical values.
  • Solve the equation for the variable.
In our problem, the critical point for the function \(F(t) = U e^{t} + V e^{-t}\) was determined by finding when its derivative \(F'(t) = U e^{t} - V e^{-t}\) was zero. Solving this equation led to finding the critical point \(t = \frac{1}{2} \ln\left(\frac{V}{U}\right)\). This indicates a potential peak or trough for the function.
Exponential Functions
Exponential functions involve variables in the exponent and have the general form \(a e^{bx}\), where \(a\) and \(b\) are constants. Exponential growth or decay occurs based on the sign of the exponent. Exponential functions have some unique properties:
  • They start slow but increase quickly.
  • Their derivative at any point is proportional to their current value.
  • They never touch zero, always positive for all real numbers if \(a > 0\).
In the exercise, \(U e^{t}\) represents exponential growth while \(V e^{-t}\) represents exponential decay. As \(t\) increases, \(U e^{t}\) grows larger while \(V e^{-t}\) approaches zero. Understanding this helps to comprehend the overall behavior of the function \(F(t)\).
Derivatives
Derivatives are a core concept in calculus, representing the rate of change or the slope of a function at a particular point. For exponential functions like \(e^{t}\), the derivative is straightforward:
  • The derivative of \(e^{t}\) is \(e^{t}\) itself, indicating constant growth.
  • For \(e^{-t}\), the derivative is \(-e^{-t}\), indicating a slowly diminishing function.
The derivative \(F'(t) = U e^{t} - V e^{-t}\) reveals how the function \(F(t)\) changes over time.
  • If the derivative is positive, \(F(t)\) is increasing.
  • If it is negative, \(F(t)\) is decreasing.
  • If zero, the function is momentarily flat, indicating a possible extrema, like a peak or a valley.
Utilizing derivatives allows us to pinpoint critical points and analyze the behavior of complex functions effortlessly.

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Most popular questions from this chapter

A grapefruit is tossed straight up with an initial velocity of \(50 \mathrm{ft} / \mathrm{sec}\). The grapefruit is 5 feet above the ground when it is released. Its height at time \(t\) is given by $$y=-16 t^{2}+50 t+5$$ How high does it go before returning to the ground?

The quantity of a drug in the bloodstream \(t\) hours after a tablet is swallowed is given, in \(\mathrm{mg}\), by $$q(t)=20\left(e^{-t}-e^{-2 t}\right)$$ (a) How much of the drug is in the bloodstream at time \(t=0 ?\) (b) When is the maximum quantity of drug in the bloodstream? What is that maximum? (c) In the long run, what happens to the quantity?

At a price of \(\$ 8\) per ticket, a musical theater group can fill every seat in the theater, which has a capacity of \(1500 .\) For every additional dollar charged, the number of people buying tickets decreases by \(75 .\) What ticket price maximizes revenue?

Show that a demand equation \(q=k / p^{r}\), where \(r\) is a positive constant, gives constant elasticity \(E=r\).

A company manufactures only one product. The quantity, \(q\), of this product produced per month depends on the amount of capital, \(K\), invested (i.e., the number of machines the company owns, the size of its building, and so on) and the amount of labor, \(L\), available each month. We assume that \(q\) can be expressed as a Cobb-Douglas production function: $$ q=c K^{\alpha} L^{\beta} $$ where \(c, \alpha, \beta\) are positive constants, with \(0<\alpha<1\) and \(0<\beta<1 .\) In this problem we will see how the Russian government could use a Cobb-Douglas function to estimate how many people a newly privatized industry might employ. A company in such an industry has only a small amount of capital available to it and needs to use all of it, so \(K\) is fixed. Suppose \(L\) is measured in man-hours per month, and that each man-hour costs the company \(w\) rubles (a ruble is the unit of Russian currency). Suppose the company has no other costs besides labor, and that each unit of the good can be sold for a fixed price of \(p\) rubles. How many man-hours of labor per month should the company use in order to maximize its profit?
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