show that w ( t ) = ( sin ( 3t+4 ) , sin ( 3t-2 ) , cos ( 3t ) ) satisfies the differential equation w'' ( t ) = -9w ( t ) .

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Answer 1

Question

Chapter 14.2, Problem 64E

To determine

show that $w (t) = (sin (3t+4), sin (3t-2), cos (3t))$ satisfies the differential equation $w'' (t) = -9w (t) .$

To determine

To prove: $w'' (t) = -9w (t) .$

Proof:

$w (t) = (sin (3t+4), sin (3t-2), cos (3t))$

Differentiating the above equation we get,

$w' (t) =3cos (3t+4), 3cos (3t-2), (-3sin (3t))$

Differentiating the derivative vector to ﬁnd the second derivative

$w'' (t) = (-3×3 sin (3t+4)), (-3×3 sin (3t-2)), (-3×3 cos (3t))$

On rearranging we get,

$w ″ (t) = - 9 (\sin (3 t + 4), \sin (3 t - 2), \cos (3 t))$

$w'' (t) = -9w (t) .$

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Answer 2

Question

Chapter 14.2, Problem 64E

To determine

show that $w (t) = (sin (3t+4), sin (3t-2), cos (3t))$ satisfies the differential equation $w'' (t) = -9w (t) .$

To determine

To prove: $w'' (t) = -9w (t) .$

Proof:

$w (t) = (sin (3t+4), sin (3t-2), cos (3t))$

Differentiating the above equation we get,

$w' (t) =3cos (3t+4), 3cos (3t-2), (-3sin (3t))$

Differentiating the derivative vector to ﬁnd the second derivative

$w'' (t) = (-3×3 sin (3t+4)), (-3×3 sin (3t-2)), (-3×3 cos (3t))$

On rearranging we get,

$w ″ (t) = - 9 (\sin (3 t + 4), \sin (3 t - 2), \cos (3 t))$

$w'' (t) = -9w (t) .$

Expert Solution & Answer

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See solution

Answer 3

Question

Chapter 14.2, Problem 64E

To determine

show that $w (t) = (sin (3t+4), sin (3t-2), cos (3t))$ satisfies the differential equation $w'' (t) = -9w (t) .$

To determine

To prove: $w'' (t) = -9w (t) .$

Proof:

$w (t) = (sin (3t+4), sin (3t-2), cos (3t))$

Differentiating the above equation we get,

$w' (t) =3cos (3t+4), 3cos (3t-2), (-3sin (3t))$

Differentiating the derivative vector to ﬁnd the second derivative

$w'' (t) = (-3×3 sin (3t+4)), (-3×3 sin (3t-2)), (-3×3 cos (3t))$

On rearranging we get,

$w ″ (t) = - 9 (\sin (3 t + 4), \sin (3 t - 2), \cos (3 t))$

$w'' (t) = -9w (t) .$

Expert Solution & Answer

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See solution

Answer 4

Question

Chapter 14.2, Problem 64E

To determine

show that $w (t) = (sin (3t+4), sin (3t-2), cos (3t))$ satisfies the differential equation $w'' (t) = -9w (t) .$

To determine

To prove: $w'' (t) = -9w (t) .$

Proof:

$w (t) = (sin (3t+4), sin (3t-2), cos (3t))$

Differentiating the above equation we get,

$w' (t) =3cos (3t+4), 3cos (3t-2), (-3sin (3t))$

Differentiating the derivative vector to ﬁnd the second derivative

$w'' (t) = (-3×3 sin (3t+4)), (-3×3 sin (3t-2)), (-3×3 cos (3t))$

On rearranging we get,

$w ″ (t) = - 9 (\sin (3 t + 4), \sin (3 t - 2), \cos (3 t))$

$w'' (t) = -9w (t) .$

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Chapter 14.2, Problem 64E

To determine

show that $w (t) = (sin (3t+4), sin (3t-2), cos (3t))$ satisfies the differential equation $w'' (t) = -9w (t) .$

To determine

To prove: $w'' (t) = -9w (t) .$

Proof:

$w (t) = (sin (3t+4), sin (3t-2), cos (3t))$

Differentiating the above equation we get,

$w' (t) =3cos (3t+4), 3cos (3t-2), (-3sin (3t))$

Differentiating the derivative vector to ﬁnd the second derivative

$w'' (t) = (-3×3 sin (3t+4)), (-3×3 sin (3t-2)), (-3×3 cos (3t))$

On rearranging we get,

$w ″ (t) = - 9 (\sin (3 t + 4), \sin (3 t - 2), \cos (3 t))$

$w'' (t) = -9w (t) .$

Answer 6

Chapter 14.2, Problem 64E

To determine

show that $w (t) = (sin (3t+4), sin (3t-2), cos (3t))$ satisfies the differential equation $w'' (t) = -9w (t) .$

Answer 7

Chapter 14.2, Problem 64E

To determine

show that $w (t) = (sin (3t+4), sin (3t-2), cos (3t))$ satisfies the differential equation $w'' (t) = -9w (t) .$

Answer 8

To determine

To prove: $w'' (t) = -9w (t) .$

Proof:

$w (t) = (sin (3t+4), sin (3t-2), cos (3t))$

Differentiating the above equation we get,

$w' (t) =3cos (3t+4), 3cos (3t-2), (-3sin (3t))$

Differentiating the derivative vector to ﬁnd the second derivative

$w'' (t) = (-3×3 sin (3t+4)), (-3×3 sin (3t-2)), (-3×3 cos (3t))$

On rearranging we get,

$w ″ (t) = - 9 (\sin (3 t + 4), \sin (3 t - 2), \cos (3 t))$

$w'' (t) = -9w (t) .$

Answer 9

To determine

To prove: $w'' (t) = -9w (t) .$

Proof:

$w (t) = (sin (3t+4), sin (3t-2), cos (3t))$

Differentiating the above equation we get,

$w' (t) =3cos (3t+4), 3cos (3t-2), (-3sin (3t))$

Differentiating the derivative vector to ﬁnd the second derivative

$w'' (t) = (-3×3 sin (3t+4)), (-3×3 sin (3t-2)), (-3×3 cos (3t))$

On rearranging we get,

$w ″ (t) = - 9 (\sin (3 t + 4), \sin (3 t - 2), \cos (3 t))$

$w'' (t) = -9w (t) .$

Answer 10

Expert Solution & Answer

Answer 11

Expert Solution & Answer

Answer 12

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Answer 13

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 6PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E

show that w ( t ) = ( sin ( 3t+4 ) , sin ( 3t-2 ) , cos ( 3t ) ) satisfies the differential equation w'' ( t ) = -9w ( t ) .

Solution Summary: The general solution is given as Given: w ( t ) = ( sin ( 3t+4), sin (3t-2 ), cos (3t) satisfies the differential equation

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show that $w (t) = (sin (3t+4), sin (3t-2), cos (3t))$ satisfies the differential equation $w'' (t) = -9w (t) .$

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Chapter 14 Solutions

show that w ( t ) = ( sin ( 3t+4 ) , sin ( 3t-2 ) , cos ( 3t ) ) satisfies the differential equation w'' ( t ) = -9w ( t ) .

Solution Summary: The general solution is given as Given: w ( t ) = ( sin ( 3t+4), sin (3t-2 ), cos (3t) satisfies the differential equation

Concept explainers

Videos

show that w(t) = (sin (3t+4), sin (3t-2), cos (3t)) satisfies the differential equation w''(t) = -9w(t).

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Chapter 14 Solutions

show that $w (t) = (sin (3t+4), sin (3t-2), cos (3t))$ satisfies the differential equation $w'' (t) = -9w (t) .$