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Graphs of logistic function (Figures 2 and 3) look suspiciously similar to the graph of the hyperbolic tangent function (Figure 3.11.3). Explain the similarity by showing that the logistic function given by Equation 7 can be written as $ P(t) = \frac {1}{2} M[ 1 + \tanh(\frac {1}{2} k(t -c))] $where $ c = (\ln A)/k. $ Thus the logistic function is really just a shifted hyperbolic tangent.

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$$\frac{K}{2} \cdot \frac{2}{1+e^{-2 u}}=\frac{K}{1+e^{-2 u}}=\frac{K}{1+A e^{-k t}}=P(t)$$

Calculus 2 / BC

Chapter 9

Differential Equations

Section 4

Models for Population Growth

University of Michigan - Ann Arbor

University of Nottingham

Boston College

Lectures

13:37

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

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Graphs of logistic functio…

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Match the logistic equatio…

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Find the value of $t$ for …

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Show that for any constant…

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Okay. And this question rest to relate the hyperbolic tangent function to the logistic function. So Pft equals one half mm. Times one plus 10 H. One half. Okay T minus C. And I'm going to evaluate this to show that's the same as the equation for the logistic function. So we know that hyperbolic tan equals hyperbolic sine over hyperbolic cosign. Which can be evaluated as E. To the X minus E. To the negative X. Divided by E. To the X. Plus E. To the negative X. So we're gonna plug in that one half K. T minus C. In there to get one half. M. One yes. Eat the one half. Okay T minus E. My C. E. To the negative one half. Okay. Team on the sea divided by the E. Two and one half K. Timur. C plus E. To the negative one half. Okay. Team. RC Going to rewrite this one as a fraction so that we can combine like terms. So this becomes 1/2. Um We have Eat the 1/2. Okay do you wanna see plus E. To the negative one half? Okay. Team I see Plus E. To the 1/2. Okay T minus C minus E. To the negative one half K. T. You wanna see all over E. to the 1/2 k. She want a C minus E. To the negative one half. K. T. M. I. C. And then we can see that these cancel out and there's gonna be two of these ease. They'll cancel it with the one half. So we got em times E. To the 1/2. Okay. Team on the sea Divided by E. to the 1/2. Okay. T. M. I C minus E. To the negative. Actually this plus -1 F. K. Timor Sea. And now to simplify this bit of multiply bye. E. to the negative 1/2 K. Team on a. C. Over. Eat a negative 1/2 k. Team on the sea. two can't allow some of these exponents. So this gives us em we have a one in the numerator because those exponents cancel out. We have a one in the denominator and then plus E. To the negative K. Team on the sea equals mm over one plus eat a negative K. T. E. Two to the K. C. Now C. Equals Ellen of a over K. So this is gonna simplify to be I am over one plus a eat a negative K. T. Which equals pft. Uh huh.

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