a. Explanation Given: The position y of the mass after t seconds is y = y 0 cos ( t k m ) . Result used: Chain Rule If h is differentiable at θ and g is differentiable at h ( x ) , then the composite function F = g ∘ h defined by F ( θ ) = g ( h ( θ ) ) is differentiable at θ and F ′ is given by the product F ′ ( θ ) = g ′ ( h ( θ ) ) ⋅ h ′ ( θ ) (1) Calculation: Obtain the first derivative of y. y ′ = d d t ( y ) = d d t ( y 0 cos ( t k m ) ) y ′ = y 0 d d t ( cos ( t k m ) ) (2) Obtain the derivative d d t ( cos ( t k m ) ) by using the chain rule as shown in equation (1). Let h ( t ) = t k m and g ( u ) = cos u where u = h ( t ) . d d t ( cos ( t k m ) ) = g ′ ( h ( t ) ) ⋅ h ′ ( t ) (3) The derivative g ′ ( h ( x ) ) is computed as follows, g ′ ( h ( x ) ) = g ′ ( u ) = d d u ( g ( u ) ) = d d u ( cos u ) = − sin u Substitute u = t k m in the above equation, g ′ ( h ( x ) ) = − sin ( t k m ) Thus, the derivative g ′ ( h ( x ) ) is g ′ ( h ( x ) ) = − sin ( t k m ) . The derivative of h ( t ) is computed as follows, h ′ ( t ) = d d t ( t k m ) = k m ( 1 t 1 − 1 ) = k m Thus, the derivative of h ( t ) is h ′ ( t ) = k m b. To determine To verify: The second derivative of d 2 y d t 2 = − k m y .

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ISBN: 9781323178522

Author: Briggs

Publisher: PEARSON

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Chapter 3.7, Problem 85E

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To find: The second derivatives of y .

To determine

To verify: The second derivative of d2ydt2=−kmy .

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Chapter 3.7, Problem 84E

Chapter 3.7, Problem 86E

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The second derivatives of y .

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To find: The second derivatives of y .

To verify: The second derivative of d2ydt2=−kmy .

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Chapter 3 Solutions