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Habitat fragmentation and species conservation The size of a class-structured population is modeled in Section $8.8 .$ In certain situations the long-term per capita growth rate of the population is given by$$r=\frac{1}{2}(1+\sqrt{1+8 s})$$where $s$ is the annual survival probability of juveniles.(a) Approximate the growth rate with a Taylor polynomial of degree one (linear approximation) centered at $0 .$(b) Approximate the growth rate with a Taylor polynomial of degree two centered at $0 .$

(a) Linear approximation $\rightarrow T_{1}(s)=1+2(x)$(b) Quadratic Approximation $\rightarrow y=1+2 x-\frac{1}{2} x^{2}$

Calculus 1 / AB

Chapter 3

Derivatives

Section 8

Linear Approximations and Taylor Polynomials

Missouri State University

University of Michigan - Ann Arbor

Boston College

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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Habitat fragmentation and …

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this question gives us scenario involving species preservation and asked us to determine an expression from a rate of change of growth rate with respect to a change in habitat area recall. That rate of change is the same things. The derivative. In other words, D r. Over D. A is the same thing as d R. Over Diaz. Times d asked over d A. You can see the d s is cancel. Therefore, this is equivalent to de over de s times 1/2 times one plus one plus A s. I'm just writing exactly. The equation was given the problem just with a negative, just with the positive expert fraction instead of a radical because it makes it easier to read. This is equivalent to 1/2 times 1/2 is 1/4. We're now taking the derivative. As you can see, our expert, it becomes negative because we're subtracting one times asked prime of a this simplifies to be d r over d A is to us prime of a over square root of one plus eight s

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