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Find the exact coordinates of the point at which the following curve is steepest: $$ y=\frac{50}{1+6 e^{-2 t}} \quad \text { for } t \geq 0 . $$

Short Answer

Expert verified
The steepest point on the curve is at \((\ln(6)/2, 25)\).

Step by step solution

01

Understand the Expression

The given equation is a sigmoid function, \( y = \frac{50}{1 + 6e^{-2t}} \), where \( y \) is a function of \( t \). We need to find out where this curve is steepest. To do so, we'll have to calculate the derivative and find its maximum.
02

Calculate the First Derivative

To find the steepest point, we first need the derivative of \( y \) with respect to \( t \). Let \( u = 1 + 6e^{-2t} \), then \( y = \frac{50}{u} \). The derivative \( \frac{dy}{dt} \) = \( \frac{-50 du/dt}{u^2} \). Since \( du/dt = -12e^{-2t} \), then \( \frac{dy}{dt} = \frac{600e^{-2t}}{(1+6e^{-2t})^2} \).
03

Differentiate the Derivative (Find the Second Derivative)

To find where the rate of change of the derivative itself (how steep it is) is maximum, we need to find the second derivative \( \frac{d^2y}{dt^2} \) and set it to zero to find critical points. Differentiate \( \frac{dy}{dt} = \frac{600e^{-2t}}{(1+6e^{-2t})^2} \) using the quotient rule. This results in a more complex expression.
04

Solve for Critical Points

To find the steepest point of the curve, solve for \( \frac{d^2y}{dt^2} = 0 \). This requires simplifying and balancing the resulting derivative equation. You find that the critical point occurs when \( t = \ln(6)/(2) \).
05

Calculate the Exact Coordinates

At the steepest point, substitute \( t = \ln(6)/2 \) back into the original equation to find \( y \). This gives \( y = \frac{50}{1+6e^{-2(\ln(6)/2)}} = \frac{50}{1+1} = 25 \). Therefore, the coordinates at which the curve is steepest are: \( (\ln(6)/2, 25) \).

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Key Concepts

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

Derivative
In calculus, understanding derivatives is fundamental to analyzing the behavior of functions. The derivative of a function, represented as \( \frac{dy}{dt} \), describes the rate at which the function's value changes with respect to a change in the input variable. In simple terms, it's like measuring how fast something is moving or changing. For the sigmoid function \( y = \frac{50}{1 + 6e^{-2t}} \), calculating the derivative helps us understand how steep the curve is at any given point \( t \). The steeper the curve, the greater the derivative. This is crucial for finding points of maximum steepness or incline on the graph of the function, which is precisely what our problem requires.
Sigmoid Function
A sigmoid function is a type of mathematical function that produces an S-shaped curve. It's commonly used in fields like machine learning and neural networks because of its smooth and bounded nature. The specific sigmoid function we are looking at is \( y = \frac{50}{1 + 6e^{-2t}} \), which is defined for \( t \geq 0 \).
These functions are so named because their graphs resemble a stretched "S" shape. This unique shape occurs because, as \( t \) approaches infinity, the function approaches a maximum value, and as \( t \) approaches negative infinity, the function approaches a minimum value. For our purpose, the interest is in determining where the function's slope is the greatest, which corresponds to its steepest point.
Critical Points
Critical points are important in calculus because they offer insights about the function's extremities, such as maximum, minimum, or saddle points. To find a critical point, you typically look where the first derivative equals zero or is undefined. In our task, however, we seek where the rate of change of the derivative itself is maximized, which involves setting the second derivative to zero.
For the sigmoid function in question, \( y = \frac{50}{1 + 6e^{-2t}} \), we determine the critical point by solving \( \frac{d^2y}{dt^2} = 0 \). This step ensures we're identifying the steepest part of the curve. The result, \( t = \ln(6)/2 \), reflects where this change reaches its peak, and thus, where the graph is steepest.
Second Derivative
The second derivative of a function, denoted \( \frac{d^2y}{dt^2} \), provides information about the concavity of the function and the behavior of its first derivative. It tells us how the rate of change of the function is itself changing. This can indicate whether a function is curving up or down, much like how acceleration measures how velocity changes over time.
In the search for the steepest point of our sigmoid curve, we calculate the second derivative to identify where the first derivative itself is changing most rapidly. By setting \( \frac{d^2y}{dt^2} = 0 \), and resolving this equation, we pinpoint the exact \( t \)-value where the steepness is maximized. This step is crucial for accurate analysis and helps in discovering the peak rate of increase, leading us to our exact coordinates of interest on the graph.

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Most popular questions from this chapter

The elasticity of a good is \(E=0.5 .\) What is the effect on the quantity demanded of: (a) A \(3 \%\) price increase? (b) A \(3 \%\) price decrease?

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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Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither. \(f(x)=\frac{x^{3}}{6}+\frac{x^{2}}{4}-x+2\)

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