The derivative of the function c ( z ) = e − z sin ( z ) also the value of the function and slope at the points 0 , π 2 and π . Sketch and describe the graph of the function on the domain 0 ≤ z ≤ 2 π .

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Answer 1

Question

Chapter 2.10, Problem 19E

To determine

To calculate: The derivative of the function c(z)=e−zsin(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 0≤z≤2π .

Expert Solution & Answer

Answer to Problem 19E

The derivative of the function is c'(z)=e−z(cosz−sinz) , value and slope of the function at the points0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=e−π2,c'(π2)=−e−π2,c(π)=0 and c'(π)=−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 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)=e−zsin(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)=e−zsin(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)=e−zddz(sinz)+sinzddz(e −z)=e−zcosz−e−zsinz=e−z(cosz−sinz)

Therefore, derivative of the function of the function is c'(z)=e−z(cosz−sinz) .

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)=e−zsin(z) ,

c(0)=e−0sin(0)=0

Substitute z=0 in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function c(z)=e−zsin(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)=e−zsin(z) ,

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

Substitute z=π2 in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function c(z)=e−zsin(z) ,

c'(π2)=e−( π 2 )(cos( π 2 )−sin( π 2 ))=e−( π 2 )(0−1)=−e−( π 2 )

At the point z=π ,

Substitute z=π to evaluate the value of the function c(z)=e−zsin(z) ,

c(π)=e−(π)sin((π))=e−(π)⋅0=0

Substitute z=π in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function b(y)=y2+3cos(y) ,

c'(π)=e−(π)(cos(π)−sin(π))=e−(π)(−1−0)=−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)=e−z(cosz−sinz), value and slope of the function at the points 0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=e−π2,c'(π2)=−e−π2,c(π)=0 and c'(π)=−e−π .

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Answer 2

Question

Chapter 2.10, Problem 19E

To determine

To calculate: The derivative of the function c(z)=e−zsin(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 0≤z≤2π .

Expert Solution & Answer

Answer to Problem 19E

The derivative of the function is c'(z)=e−z(cosz−sinz) , value and slope of the function at the points0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=e−π2,c'(π2)=−e−π2,c(π)=0 and c'(π)=−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 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)=e−zsin(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)=e−zsin(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)=e−zddz(sinz)+sinzddz(e −z)=e−zcosz−e−zsinz=e−z(cosz−sinz)

Therefore, derivative of the function of the function is c'(z)=e−z(cosz−sinz) .

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)=e−zsin(z) ,

c(0)=e−0sin(0)=0

Substitute z=0 in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function c(z)=e−zsin(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)=e−zsin(z) ,

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

Substitute z=π2 in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function c(z)=e−zsin(z) ,

c'(π2)=e−( π 2 )(cos( π 2 )−sin( π 2 ))=e−( π 2 )(0−1)=−e−( π 2 )

At the point z=π ,

Substitute z=π to evaluate the value of the function c(z)=e−zsin(z) ,

c(π)=e−(π)sin((π))=e−(π)⋅0=0

Substitute z=π in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function b(y)=y2+3cos(y) ,

c'(π)=e−(π)(cos(π)−sin(π))=e−(π)(−1−0)=−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)=e−z(cosz−sinz), value and slope of the function at the points 0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=e−π2,c'(π2)=−e−π2,c(π)=0 and c'(π)=−e−π .

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Answer 3

Question

Chapter 2.10, Problem 19E

To determine

To calculate: The derivative of the function c(z)=e−zsin(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 0≤z≤2π .

Expert Solution & Answer

Answer to Problem 19E

The derivative of the function is c'(z)=e−z(cosz−sinz) , value and slope of the function at the points0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=e−π2,c'(π2)=−e−π2,c(π)=0 and c'(π)=−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 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)=e−zsin(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)=e−zsin(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)=e−zddz(sinz)+sinzddz(e −z)=e−zcosz−e−zsinz=e−z(cosz−sinz)

Therefore, derivative of the function of the function is c'(z)=e−z(cosz−sinz) .

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)=e−zsin(z) ,

c(0)=e−0sin(0)=0

Substitute z=0 in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function c(z)=e−zsin(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)=e−zsin(z) ,

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

Substitute z=π2 in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function c(z)=e−zsin(z) ,

c'(π2)=e−( π 2 )(cos( π 2 )−sin( π 2 ))=e−( π 2 )(0−1)=−e−( π 2 )

At the point z=π ,

Substitute z=π to evaluate the value of the function c(z)=e−zsin(z) ,

c(π)=e−(π)sin((π))=e−(π)⋅0=0

Substitute z=π in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function b(y)=y2+3cos(y) ,

c'(π)=e−(π)(cos(π)−sin(π))=e−(π)(−1−0)=−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)=e−z(cosz−sinz), value and slope of the function at the points 0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=e−π2,c'(π2)=−e−π2,c(π)=0 and c'(π)=−e−π .

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Answer 4

Question

Chapter 2.10, Problem 19E

To determine

To calculate: The derivative of the function c(z)=e−zsin(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 0≤z≤2π .

Expert Solution & Answer

Answer to Problem 19E

The derivative of the function is c'(z)=e−z(cosz−sinz) , value and slope of the function at the points0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=e−π2,c'(π2)=−e−π2,c(π)=0 and c'(π)=−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 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)=e−zsin(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)=e−zsin(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)=e−zddz(sinz)+sinzddz(e −z)=e−zcosz−e−zsinz=e−z(cosz−sinz)

Therefore, derivative of the function of the function is c'(z)=e−z(cosz−sinz) .

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)=e−zsin(z) ,

c(0)=e−0sin(0)=0

Substitute z=0 in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function c(z)=e−zsin(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)=e−zsin(z) ,

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

Substitute z=π2 in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function c(z)=e−zsin(z) ,

c'(π2)=e−( π 2 )(cos( π 2 )−sin( π 2 ))=e−( π 2 )(0−1)=−e−( π 2 )

At the point z=π ,

Substitute z=π to evaluate the value of the function c(z)=e−zsin(z) ,

c(π)=e−(π)sin((π))=e−(π)⋅0=0

Substitute z=π in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function b(y)=y2+3cos(y) ,

c'(π)=e−(π)(cos(π)−sin(π))=e−(π)(−1−0)=−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)=e−z(cosz−sinz), value and slope of the function at the points 0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=e−π2,c'(π2)=−e−π2,c(π)=0 and c'(π)=−e−π .

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 2.10, Problem 19E

To determine

To calculate: The derivative of the function c(z)=e−zsin(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 0≤z≤2π .

Expert Solution & Answer

Answer to Problem 19E

The derivative of the function is c'(z)=e−z(cosz−sinz) , value and slope of the function at the points0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=e−π2,c'(π2)=−e−π2,c(π)=0 and c'(π)=−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 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)=e−zsin(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)=e−zsin(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)=e−zddz(sinz)+sinzddz(e −z)=e−zcosz−e−zsinz=e−z(cosz−sinz)

Therefore, derivative of the function of the function is c'(z)=e−z(cosz−sinz) .

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)=e−zsin(z) ,

c(0)=e−0sin(0)=0

Substitute z=0 in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function c(z)=e−zsin(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)=e−zsin(z) ,

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

Substitute z=π2 in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function c(z)=e−zsin(z) ,

c'(π2)=e−( π 2 )(cos( π 2 )−sin( π 2 ))=e−( π 2 )(0−1)=−e−( π 2 )

At the point z=π ,

Substitute z=π to evaluate the value of the function c(z)=e−zsin(z) ,

c(π)=e−(π)sin((π))=e−(π)⋅0=0

Substitute z=π in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function b(y)=y2+3cos(y) ,

c'(π)=e−(π)(cos(π)−sin(π))=e−(π)(−1−0)=−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)=e−z(cosz−sinz), value and slope of the function at the points 0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=e−π2,c'(π2)=−e−π2,c(π)=0 and c'(π)=−e−π .

Answer 6

Chapter 2.10, Problem 19E

To determine

To calculate: The derivative of the function c(z)=e−zsin(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 0≤z≤2π .

Expert Solution & Answer

Answer to Problem 19E

The derivative of the function is c'(z)=e−z(cosz−sinz) , value and slope of the function at the points0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=e−π2,c'(π2)=−e−π2,c(π)=0 and c'(π)=−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 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)=e−zsin(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)=e−zsin(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)=e−zddz(sinz)+sinzddz(e −z)=e−zcosz−e−zsinz=e−z(cosz−sinz)

Therefore, derivative of the function of the function is c'(z)=e−z(cosz−sinz) .

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)=e−zsin(z) ,

c(0)=e−0sin(0)=0

Substitute z=0 in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function c(z)=e−zsin(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)=e−zsin(z) ,

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

Substitute z=π2 in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function c(z)=e−zsin(z) ,

c'(π2)=e−( π 2 )(cos( π 2 )−sin( π 2 ))=e−( π 2 )(0−1)=−e−( π 2 )

At the point z=π ,

Substitute z=π to evaluate the value of the function c(z)=e−zsin(z) ,

c(π)=e−(π)sin((π))=e−(π)⋅0=0

Substitute z=π in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function b(y)=y2+3cos(y) ,

c'(π)=e−(π)(cos(π)−sin(π))=e−(π)(−1−0)=−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)=e−z(cosz−sinz), value and slope of the function at the points 0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=e−π2,c'(π2)=−e−π2,c(π)=0 and c'(π)=−e−π .

Answer 7

Chapter 2.10, Problem 19E

To determine

To calculate: The derivative of the function c(z)=e−zsin(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 0≤z≤2π .

Answer 8

Expert Solution & Answer

Answer to Problem 19E

The derivative of the function is c'(z)=e−z(cosz−sinz) , value and slope of the function at the points0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=e−π2,c'(π2)=−e−π2,c(π)=0 and c'(π)=−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 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 .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

The function c(z)=e−zsin(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)=e−zsin(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)=e−zddz(sinz)+sinzddz(e −z)=e−zcosz−e−zsinz=e−z(cosz−sinz)

Therefore, derivative of the function of the function is c'(z)=e−z(cosz−sinz) .

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)=e−zsin(z) ,

c(0)=e−0sin(0)=0

Substitute z=0 in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function c(z)=e−zsin(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)=e−zsin(z) ,

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

Substitute z=π2 in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function c(z)=e−zsin(z) ,

c'(π2)=e−( π 2 )(cos( π 2 )−sin( π 2 ))=e−( π 2 )(0−1)=−e−( π 2 )

At the point z=π ,

Substitute z=π to evaluate the value of the function c(z)=e−zsin(z) ,

c(π)=e−(π)sin((π))=e−(π)⋅0=0

Substitute z=π in c'(z)=e−z(cosz−sinz) to evaluate the slope of the function b(y)=y2+3cos(y) ,

c'(π)=e−(π)(cos(π)−sin(π))=e−(π)(−1−0)=−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)=e−z(cosz−sinz), value and slope of the function at the points 0,π2 and π are, c(0)=0,c'(0)=1,c(π2)=e−π2,c'(π2)=−e−π2,c(π)=0 and c'(π)=−e−π .

Answer 12

Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E

The derivative of the function c ( z ) = e − z sin ( z ) also the value of the function and slope at the points 0 , π 2 and π . Sketch and describe the graph of the function on the domain 0 ≤ z ≤ 2 π .

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To calculate: The derivative of the function c(z)=e−zsin(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 0≤z≤2π .

Answer to Problem 19E

Explanation of Solution

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