Explanation The Directional Derivative is defined as the rate of change by which the given function is changing at a given point in the given direction. Given: A function f ( x , y , z ) = sin ( x y + z ) and the direction is v = j + k Formula Used: 1. First find the gradient of a function by differentiating the given function with respect to each variable in the function. 2. Evaluate the gradient at the given point 3. Find the length of the vector 4. Normalize the vector by dividing each component of vector by the length of the vector. 5. Directional Derivative is the product of gradient and normalized vector Calculation: The given function is f ( x , y , z ) = sin ( x y + z ) First find the gradient of the function f ( x , y , z ) = sin ( x y + z ) Gradient is given as: ∇ f = ( ∂ f ∂ x , ∂ f ∂ y , ∂ f ∂ x ) ∂ f ∂ x = ∂ f ∂ x ( s i n ( x y + z ) ) = ∂ f ∂ u ( s i n u ) ∂ ∂ x ( x y + z ) = ( c o s u ) . y = y c o s ( x y + z ) ∂ f ∂ y = ∂ f ∂ y ( s i n ( x y + z ) ) = ∂ f ∂ u ( s i n u ) ∂ ∂ y ( x y + z ) = ( c o s u ) . x = x c o s ( x y + z ) ∂ f ∂ z = ∂ f ∂ z ( s i n ( x y + z ) ) = ∂ f ∂ u ( s i n u ) ∂ ∂ z ( x y + z ) = ( c o s u ) .1 = c o s ( x y + z ) Thus, the gradient of the function is ∇ f = ( y c o s ( x y + z ) , x c o s ( x y + z ) , c o s ( x y + z ) ) Now, finding the value of the gradient at the point P = ( 0 , 0 , 0 ) ∇ ( sin ( x y + z ) ) | ( x , y , z ) = ( 0 , 0 , 0 ) = ( 0...

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Author: Rogawski

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Chapter 14, Problem 36CRE

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The directional derivative for the function $f (x, y, z) = \sin (x y + z)$ at point $P = (0, 0, 0)$ in the direction of vector $v = j + k$

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Chapter 14, Problem 35CRE

Chapter 14, Problem 37CRE

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Chapter 14 Solutions

CALCULUS: EARLY TRANS 4TH ED W/ ACCESS

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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The directional derivative for the function f ( x , y , z ) = sin ( x y + z ) at point P = ( 0 , 0 , 0 ) in the direction of vector v = j + k

The directional derivative for the function f(x, y, z) = sin(xy + z) at point P = (0, 0, 0) in the direction of vector v = j + k

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Chapter 14 Solutions

The directional derivative for the function $f (x, y, z) = \sin (x y + z)$ at point $P = (0, 0, 0)$ in the direction of vector $v = j + k$