Explanation Given: The kinetic energy is K = 1 2 m v 2 . Here, m is mass and v is velocity. Proof: Compute the partial derivative of K with respect to m . K m = 1 2 v 2 ∂ ∂ m ( m ) = 1 2 v 2 ( 1 ) = 1 2 v 2 Thus, K m = 1 2 v 2 . Compute the partial derivative of K with respect to v . K v = 1 2 m ∂ ∂ v ( v 2 ) = 1 2 m ( 2 v ) = m v Hence, K v = m v . Again, compute the partial derivative of K v = m v with respect to v

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

10th Edition

ISBN: 9781305657861

Author: Soo T. Tan

Publisher: Cengage Learning

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Chapter 8.2, Problem 64E

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To show: The derivatives ∂K∂m⋅∂2K∂v2 is equal to Kinetic energy (K) .

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Chapter 8.2, Problem 63E

Chapter 8.2, Problem 66E

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Heat transfer Fourier’s Law of heat transfer (or heat conduction) states that the heat flow vector F at a point is proportional to the negative gradient of the temperature; that is, F = -k∇T, which means that heat energy flows from hot regions to cold regions. The constant k > 0 is called the conductivity, which has metric units of J/(m-s-K). A temperature function for a region D is given. Find the net outward heat flux ∫∫S F ⋅ n dS = -k∫∫S ∇T ⋅ n dS across the boundary S of D. In some cases, it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume k = 1. T(x, y, z) = 100 + x + 2y + z;D = {(x, y, z): 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1}

Heat transfer Fourier’s Law of heat transfer (or heat conduction) states that the heat flow vector F at a point is proportional to the negative gradient of the temperature; that is, F = -k∇T, which means that heat energy flows from hot regions to cold regions. The constant k > 0 is called the conductivity, which has metric units of J/(m-s-K). A temperature function for a region D is given. Find the net outward heat flux ∫∫S F ⋅ n dS = -k∫∫S ∇T ⋅ n dS across the boundary S of D. In some cases, it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume k = 1. T(x, y, z) = 100 + e-z;D = {(x, y, z): 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1}

Heat transfer Fourier’s Law of heat transfer (or heat conduction) states that the heat flow vector F at a point is proportional to the negative gradient of the temperature; that is, F = -k∇T, which means that heat energy flows from hot regions to cold regions. The constant k > 0 is called the conductivity, which has metric units of J/(m-s-K). A temperature function for a region D is given. Find the net outward heat flux ∫∫S F ⋅ n dS = -k∫∫S ∇T ⋅ n dS across the boundary S of D. In some cases, it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume k = 1. T(x, y, z) = 100 + x2 + y2 + z2;;D = {(x, y, z): 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1}

Chapter 8 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

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The derivatives ∂ K ∂ m ⋅ ∂ 2 K ∂ v 2 is equal to Kinetic energy ( K ) .

Solution Summary: The author explains how the derivatives partial mcdot and K are equal to Kinetic energy.

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Chapter 8 Solutions