Explanation Given information: f ( x , y ) = 2 x + 3 y , r ( t ) = 〈 t 3 , t 2 〉 , t = − 2 . Calculation: Since the gradient of function is denoted by ∇ f = 〈 ∂ f ∂ x , ∂ f ∂ y 〉 . The given equation f ( x , y ) = 2 x + 3 y Differentiating f with respect to x we get, ∂ f ∂ x = ∂ ∂ x ( 2 x + 3 y ) ⇒ ∂ f ∂ x = 2 Differentiating f with respect to y we get, ∂ f ∂ y = ∂ ∂ y 2 x + 3 y ⇒ ∂ f ∂ y = 3 Hence ∇ f = 〈 ∂ f ∂ x , ∂ f ∂ y 〉 = 〈 2 , 3 〉 Now we have to calculate r ' ( t ) . Since r ( t ) = 〈 t 3 , t 2 〉 Differentiate r ( t ) with respect to t we get, r ' ( t ) = d d t 〈 t 3 , t 2 〉

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Chapter 14.5, Problem 10E

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ddtf(r(t)) at the value of t given.

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Chapter 14 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

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

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Tocalculate: d d t f ( r ( t ) ) at the value of t given.

Solution Summary: The author explains how the gradient of function is denoted by nabla f=langle

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Tocalculate:

ddtf(r(t)) at the value of t given.

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Chapter 14 Solutions

Tocalculate: d d t f ( r ( t ) ) at the value of t given.

Solution Summary: The author explains how the gradient of function is denoted by nabla f=langle

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Tocalculate: ddtf(r(t)) at the value of t given.

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Chapter 14 Solutions

Tocalculate:

ddtf(r(t)) at the value of t given.