Modeling the Dynamics of Life: Calculus and Probability for Life Scientists
Modeling the Dynamics of Life: Calculus and Probability for Life Scientists
3rd Edition
ISBN: 9780840064189
Author: Frederick R. Adler
Publisher: Cengage Learning
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Chapter 2.10, Problem 19E
To determine

To calculate: The derivative of the function c(z)=ezsin(z) also the value of the function and slope at the points 0,π2 and π . Sketch and describe the graph of the function on the domain 0z2π .

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Answer to Problem 19E

The derivative of the function is c'(z)=ez(coszsinz) , value and slope of the function at the points0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=eπ2,c'(π2)=eπ2,c(π)=0 and c'(π)=eπ. The graph of the function is provided below,

  Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.10, Problem 19E , additional homework tip  1

The behavior of the function is interpreted as, the function first increases reaches the maximum at z=π2 then decreases becomes zero at z=π , then becomes negative and again zero at z=2π .

Explanation of Solution

Given information:

The function c(z)=ezsin(z) and the points 0,π2 and π .

Formula used:

The slope of a function at a given point is the derivative of the function at that point.

Product rule of differentiation ddxf(x)g(x)=f(x)ddxg(x)+g(x)ddxf(x) .

Calculation:

Consider the function c(z)=ezsin(z) .

Differentiate the function with respect to z ,

Recall product rule of differentiation ddxf(x)g(x)=f(x)ddxg(x)+g(x)ddxf(x) .

Apply it,

  ddzc(z)=ddz(e zsinz)=ezddz(sinz)+sinzddz(e z)=ezcoszezsinz=ez(coszsinz)

Therefore, derivative of the function of the function is c'(z)=ez(coszsinz) .

Recall that the slope of a function at a given point is the derivative of the function at that point.

Evaluate the value of the function and slope at the points 0,π2 and π .

At the point z=0 ,

Substitute z=0 to evaluate the value of the function c(z)=ezsin(z) ,

  c(0)=e0sin(0)=0

Substitute z=0 in c'(z)=ez(coszsinz) to evaluate the slope of the function c(z)=ezsin(z) ,

  c'(0)=e(0)(cos(0)sin(0))=1

At the point z=π2 ,

Substitute z=π2 to evaluate the value of the function c(z)=ezsin(z) ,

  c(π2)=e( π 2 )sin(π2)=e( π 2 )1=e( π 2 )

Substitute z=π2 in c'(z)=ez(coszsinz) to evaluate the slope of the function c(z)=ezsin(z) ,

  c'(π2)=e( π 2 )(cos( π 2 )sin( π 2 ))=e( π 2 )(01)=e( π 2 )

At the point z=π ,

Substitute z=π to evaluate the value of the function c(z)=ezsin(z) ,

  c(π)=e(π)sin((π))=e(π)0=0

Substitute z=π in c'(z)=ez(coszsinz) to evaluate the slope of the function b(y)=y2+3cos(y) ,

  c'(π)=e(π)(cos(π)sin(π))=e(π)(10)=e(π)

The graph of the function is provided below,

  Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.10, Problem 19E , additional homework tip  2

The behavior of the function is interpreted as, the functionfirst increases reaches the maximum at z=π2 then decreases becomes zero at z=π , then becomes negative and again zero at z=2π .

Thus, derivative of the function is c'(z)=ez(coszsinz), value and slope of the function at the points 0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=eπ2,c'(π2)=eπ2,c(π)=0 and c'(π)=eπ .

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Chapter 2 Solutions

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists

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