(a) To sketch the slope field of the differential equation given in the range − 1 ≤ t ≤ 3 , − 1 ≤ y ≤ 3 by observing that the isoclines of slope c is in the line t − y = c , so the segments have slope c at points on the line y = t − c

Question

y=f'(x) 1 8 The function f is defined on the closed interval [0,8]. The graph of its derivative f' is shown above. How many relative minima are there for f(x)? O 2 6 4 00

Answer 1

Question

Chapter 9.3, Problem 5E

To determine

(a)

To sketch the slope field of the differential equation given in the range $- 1 \leq t \leq 3, - 1 \leq y \leq 3$ by observing that the isoclines of slope $c$ is in the line $t - y = c$ , so the segments have slope $c$ at points on the line $y = t - c$

To determine

(b)

To show that $y = t - 1 + C e^{- t}$ is a solution for all $C$ . Since $\lim_{t \to \infty} e^{- t} = 0$ these solutions approach the particular solution $y = t - 1$ as $t \to \infty$ . Explain how this behavior is reflected in your slope field

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Students have asked these similar questions

y=f'(x) 1 8 The function f is defined on the closed interval [0,8]. The graph of its derivative f' is shown above. How many relative minima are there for f(x)? O 2 6 4 00

60! 5!.7!.15!.33!

• • Let > be a potential for the vector field F = (−2 y³, −6 xy² − 4 z³, −12 yz² + 4 2). Then the value of sin((-1.63, 2.06, 0.57) – (0,0,0)) is - 0.336 -0.931 -0.587 0.440 0.902 0.607 -0.609 0.146

Answer 2

Question

Chapter 9.3, Problem 5E

To determine

(a)

To sketch the slope field of the differential equation given in the range $- 1 \leq t \leq 3, - 1 \leq y \leq 3$ by observing that the isoclines of slope $c$ is in the line $t - y = c$ , so the segments have slope $c$ at points on the line $y = t - c$

To determine

(b)

To show that $y = t - 1 + C e^{- t}$ is a solution for all $C$ . Since $\lim_{t \to \infty} e^{- t} = 0$ these solutions approach the particular solution $y = t - 1$ as $t \to \infty$ . Explain how this behavior is reflected in your slope field

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 3

Question

Chapter 9.3, Problem 5E

To determine

(a)

To sketch the slope field of the differential equation given in the range $- 1 \leq t \leq 3, - 1 \leq y \leq 3$ by observing that the isoclines of slope $c$ is in the line $t - y = c$ , so the segments have slope $c$ at points on the line $y = t - c$

To determine

(b)

To show that $y = t - 1 + C e^{- t}$ is a solution for all $C$ . Since $\lim_{t \to \infty} e^{- t} = 0$ these solutions approach the particular solution $y = t - 1$ as $t \to \infty$ . Explain how this behavior is reflected in your slope field

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 4

Question

Chapter 9.3, Problem 5E

To determine

(a)

To sketch the slope field of the differential equation given in the range $- 1 \leq t \leq 3, - 1 \leq y \leq 3$ by observing that the isoclines of slope $c$ is in the line $t - y = c$ , so the segments have slope $c$ at points on the line $y = t - c$

To determine

(b)

To show that $y = t - 1 + C e^{- t}$ is a solution for all $C$ . Since $\lim_{t \to \infty} e^{- t} = 0$ these solutions approach the particular solution $y = t - 1$ as $t \to \infty$ . Explain how this behavior is reflected in your slope field

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 5

Chapter 9.3, Problem 5E

To determine

(a)

To sketch the slope field of the differential equation given in the range $- 1 \leq t \leq 3, - 1 \leq y \leq 3$ by observing that the isoclines of slope $c$ is in the line $t - y = c$ , so the segments have slope $c$ at points on the line $y = t - c$

To determine

(b)

To show that $y = t - 1 + C e^{- t}$ is a solution for all $C$ . Since $\lim_{t \to \infty} e^{- t} = 0$ these solutions approach the particular solution $y = t - 1$ as $t \to \infty$ . Explain how this behavior is reflected in your slope field

Answer 6

Chapter 9.3, Problem 5E

To determine

(a)

To sketch the slope field of the differential equation given in the range $- 1 \leq t \leq 3, - 1 \leq y \leq 3$ by observing that the isoclines of slope $c$ is in the line $t - y = c$ , so the segments have slope $c$ at points on the line $y = t - c$

Answer 7

Chapter 9.3, Problem 5E

To determine

(a)

To sketch the slope field of the differential equation given in the range $- 1 \leq t \leq 3, - 1 \leq y \leq 3$ by observing that the isoclines of slope $c$ is in the line $t - y = c$ , so the segments have slope $c$ at points on the line $y = t - c$

Answer 8

To determine

(b)

To show that $y = t - 1 + C e^{- t}$ is a solution for all $C$ . Since $\lim_{t \to \infty} e^{- t} = 0$ these solutions approach the particular solution $y = t - 1$ as $t \to \infty$ . Explain how this behavior is reflected in your slope field

Answer 9

To determine

(b)

To show that $y = t - 1 + C e^{- t}$ is a solution for all $C$ . Since $\lim_{t \to \infty} e^{- t} = 0$ these solutions approach the particular solution $y = t - 1$ as $t \to \infty$ . Explain how this behavior is reflected in your slope field

Answer 10

Expert Solution & Answer

Answer 11

Expert Solution & Answer

Answer 12

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 13

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 14

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 15

Trending nowThis is a popular solution!

Answer 16

Trending nowThis is a popular solution!

Answer 17

Trending nowThis is a popular solution!

Answer 18

Trending now

Answer 19

This is a popular solution!

Answer 20

See solution

Check out sample textbook solution

Answer 21

Ch. 9.1 - Prob. 1PQ Ch. 9.1 - Prob. 2PQ Ch. 9.1 - Prob. 3PQ Ch. 9.1 - Prob. 4PQ Ch. 9.1 - Prob. 5PQ Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E

(a) To sketch the slope field of the differential equation given in the range − 1 ≤ t ≤ 3 , − 1 ≤ y ≤ 3 by observing that the isoclines of slope c is in the line t − y = c , so the segments have slope c at points on the line y = t − c

(a) To sketch the slope field of the differential equation given in the range −1≤t≤3,−1≤y≤3 by observing that the isoclines of slope c is in the line t−y=c, so the segments have slope c at points on the line y=t−c

(b) To show that y=t−1+Ce−t is a solution for all C. Since limt→∞e−t=0 these solutions approach the particular solution y=t−1 as t→∞. Explain how this behavior is reflected in your slope field

Chapter 9 Solutions

(a)

To sketch the slope field of the differential equation given in the range $- 1 \leq t \leq 3, - 1 \leq y \leq 3$ by observing that the isoclines of slope $c$ is in the line $t - y = c$ , so the segments have slope $c$ at points on the line $y = t - c$

(b)

To show that $y = t - 1 + C e^{- t}$ is a solution for all $C$ . Since $\lim_{t \to \infty} e^{- t} = 0$ these solutions approach the particular solution $y = t - 1$ as $t \to \infty$ . Explain how this behavior is reflected in your slope field