Explanation Chain Rule: “Suppose that z = f ( x , y ) is a differentiable function of x and y , where x = g ( s , t ) and y = h ( s , t ) are both differentiable functions of s and t . Then ∂ z ∂ s = ∂ z ∂ x ⋅ ∂ x ∂ s + ∂ z ∂ y ⋅ ∂ y ∂ s and ∂ z ∂ t = ∂ z ∂ x ⋅ ∂ x ∂ t + ∂ z ∂ y ⋅ ∂ y ∂ t ”. Calculation: The function is, z = f ( x , y ) . x = r cos θ and y = r sin θ . Take the partial derivative of x and y with respect to r , ∂ x ∂ r = ∂ ∂ r ( r cos θ ) = cos θ ∂ y ∂ r = ∂ ∂ r ( r sin θ ) = sin θ Thus, the partial derivates, ∂ x ∂ r = cos θ and ∂ y ∂ r = sin θ . Take the partial derivative of x and y with respect to θ , ∂ x ∂ θ = ∂ ∂ θ ( r cos θ ) = − r sin θ [ ∵ ∂ ∂ x ( cos x ) = − sin x ] ∂ y ∂ θ = ∂ ∂ θ ( r sin θ ) = r cos θ [ ∵ ∂ ∂ x ( sin x ) = cos x ] Thus, the partial derivates, ∂ x ∂ θ = − r sin θ and ∂ y ∂ θ = r cos θ . Use chain rule and find the vaue of ∂ z ∂ r , ∂ z ∂ r = ∂ z ∂ x ⋅ ∂ x ∂ r + ∂ z ∂ y ⋅ ∂ y ∂ r = ∂ z ∂ x . cos θ + ∂ z ∂ y . sin θ Therefore, the value, ∂ z ∂ r is ∂ z ∂ x . cos θ + ∂ z ∂ y . sin θ . Similarly find the value of ∂ z ∂ θ . ∂ z ∂ θ = ∂ z ∂ x ⋅ ∂ x ∂ θ + ∂ z ∂ y ⋅ ∂ y ∂ θ = ∂ z ∂ x . ( − r sin θ ) + ∂ z ∂ y . ( r cos θ ) Therefore, the value, ∂ z ∂ θ is ∂ z ∂ x . ( − r sin θ ) + ∂ z ∂ y . ( r cos θ ) . Obtain the value of ∂ 2 z ∂ r 2 as follows, ∂ ∂ r [ ∂ z ∂ x . cos θ + ∂ z ∂ y . sin θ ] = { cos θ ( ∂ 2 z ∂ x 2 ⋅ cos θ + ∂ 2 z ∂ y ∂ x sin θ ) + sin θ ( ∂ 2 z ∂ y 2 ⋅ sin θ + ∂ 2 z ∂ x ∂ y cos θ ) } = { cos 2 θ ∂ 2 z ∂ x 2 + cos θ sin θ ∂ 2 z ∂ y ∂ x + sin 2 θ ∂ 2 z ∂ y 2 + cos θ sin θ ∂ 2 z ∂ x ∂ y } = { cos 2 θ ∂ 2 z ∂ x 2 + 2 cos θ sin θ ∂ 2 z ∂ y ∂ x + sin 2 θ ∂ 2 z ∂ y 2 } Obtain the value of ∂ 2 z ∂ θ 2 as follows, ∂ ∂ θ [ ∂ z ∂ x

Essential Calculus: Early Transcendentals; MAC 2311 Sequence| MAC 2281Sequence USF (Essential Calculus)

2nd Edition

ISBN: 9781285101552

Author: James Stewart

Publisher: Cenage

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Question

Chapter 11.5, Problem 45E

To determine

To show: The equation (∂2z∂x2)+(∂2z∂y2)=(∂2z∂r2)+1r2(∂2z∂θ2)+1r.∂z∂r , where x=rcosθ and y=rsinθ .

Expert Solution & Answer

Chapter 11 Solutions

Essential Calculus: Early Transcendentals; MAC 2311 Sequence| MAC 2281Sequence USF (Essential Calculus)

Ch. 11.1 - Let g(x, y) = cos(x + 2y). (a) Evaluate g(2, 1)....Ch. 11.1 - Let F(x,y)=1+4y2. (a) Evaluate F(3,1). (b) Find...Ch. 11.1 - Let f(x,y,z)=x+y+z+ln(4x2y2z2). (a) Evaluate f(1,...Ch. 11.1 - Let g(x,y,z)=x3y2z10xyz. (a) Evaluate g(1, 2, 3)....Ch. 11.1 - Find and sketch the domain of the function....Ch. 11.1 - Find and sketch the domain of the function....Ch. 11.1 - Find and sketch the domain of the function. 15....Ch. 11.1 - Find and sketch the domain of the function....Ch. 11.1 - Find and sketch the domain of the function. 19....Ch. 11.1 - Find and sketch the domain of the function. f(x,...

Knowledge Booster

The equation ( ∂ 2 z ∂ x 2 ) + ( ∂ 2 z ∂ y 2 ) = ( ∂ 2 z ∂ r 2 ) + 1 r 2 ( ∂ 2 z ∂ θ 2 ) + 1 r . ∂ z ∂ r , where x = r cos θ and y = r sin θ .

Solution Summary: The author explains the equation z=f(x,y), where x=rmathrmcostheta

To show: The equation (∂2z∂x2)+(∂2z∂y2)=(∂2z∂r2)+1r2(∂2z∂θ2)+1r.∂z∂r , where x=rcosθ and y=rsinθ .

Chapter 11 Solutions