Explanation Partial derivative is defined as the derivative of a function having more than one variable with respect to one variable, while the other variables being treated as constant. Let us consider that there is a function f of two variables x and y , then the partial derivatives of the function f x and f y are also the functions of the two variables. The partial derivatives of f x and f y are referred to as the second partial derivatives. The notations for second order partial derivatives are as follows: ( f x ) x = f x x = ∂ ∂ x ( ∂ f ∂ x ) = ∂ 2 f ∂ x 2 ( f x ) y = f x y = ∂ ∂ y ( ∂ f ∂ x ) = ∂ 2 f ∂ y ∂ x ( f y ) x = f y x = ∂ ∂ x ( ∂ f ∂ y ) = ∂ 2 f ∂ x ∂ y And ( f y ) y = f y y = ∂ ∂ y ( ∂ f ∂ y ) = ∂ 2 f ∂ y 2 Given: A two variable function u ( t , x ) = s i n ( α c t + β ) s i n ( α x ) and wave equation as ∂ 2 u ∂ t 2 = c 2 ∂ 2 u ∂ x 2 Formula Used: 1. In order to find f x ( x , y ) , treat the variable y as a constant and then differentiate the function f with respect to the variable x . 2. In order to find f y ( x , y ) , treat the variable x as a constant and then differentiate the function f with respect to the variable y . Calculation: The given function is u ( t , x ) = s i n ( α c t + β ) s i n ( α x ) Partial Derivative of the function with respect to t is given as follows ∂ u ∂ t = ∂ ∂ t u ( t , x ) = ∂ ∂ t s i n ( α c t + β ) s i n ( α x ) Taking out the constant and then applying the Chain Rule of derivation: d f ( u ) d x = d f d u · d u d x ∂ u ∂ t = ∂ ∂ t u ( t , x ) = s i n ( α x ) ∂ ∂ t s i n ( α c t + β ) Putting f = s i n ( u ) and u = α c t + β ∂ u ∂ t = ∂ ∂ t u ( t , x ) = s i n ( α x ) ∂ ∂ u s i n ( u ) ∂ ∂ t ( α c t + β ) ---- Equation 1 Solving ∂ ∂ u ( s i n ( u ) ) , we get ∂ ∂ u ( s i n ( u ) ) = c o s ( u ) And solving ∂ ∂ t ( α c t + β ) , we get the following ∂ ∂ t ( α c t + β ) = α c Substituting the above values in Equation 1, we get, ∂ u ∂ t = s i n ( α x ) c o s ( u ) α c ∂ u ∂ t = s i n ( α x ) c o s ( α c t + β ) α c ∂ u ∂ t = α c . s i n ( α x ) c o s ( α c t + β ) Again finding partial derivative of α c s i n ( α x ) c o s ( α c t + β ) , we get ∂ 2 u ∂ t 2 = ∂ ∂ t α c . s i n ( α x ) c o s ( α c t + β ) Taking out the constant and then applying the Chain Rule of derivation: d f ( u ) d x = d f d u · d u d x ∂ 2 u ∂ t 2 = α c . s i n ( α x ) ∂ ∂ t c o s ( α c t + β ) Putting f = c o s ( u ) and u = α c t + β ∂ 2 u ∂ t 2 = α c . s i n ( α x ) ∂ ∂ u c o s ( u ) ∂ ∂ t ( α c t + β ) ---- Equation 2 Solving ∂ ∂ u ( c o s ( u ) ) , we get ∂ ∂ u ( c o s ( u ) ) = - s i n ( u ) And solving ∂ ∂ t ( α c t + β ) , we get the following ∂ ∂ t ( α c t + β ) = α c Substituting the above values in Equation 2, we get ∂ 2 u ∂ t 2 = α c ...

4th Edition

ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 15, Problem 22CRE

To determine

That for the constants $α$ , $β$ and $c = 0$ , the function $u (t, x) = s i n (α c t + β) s i n (α x)$ satisfies the wave equation $\frac{\partial^{2} u}{\partial t^{2}} = c^{2} \frac{\partial^{2} u}{\partial x^{2}}$

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Chapter 15, Problem 21CRE

Chapter 15, Problem 23CRE

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Chapter 15 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

Ch. 15.1 - Prob. 1PQ Ch. 15.1 - Prob. 2PQ Ch. 15.1 - Prob. 3PQ Ch. 15.1 - Prob. 4PQ Ch. 15.1 - Prob. 5PQ Ch. 15.1 - Prob. 1E Ch. 15.1 - Prob. 2E Ch. 15.1 - Prob. 3E Ch. 15.1 - Prob. 4E Ch. 15.1 - Prob. 5E

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That for the constants α , β and c = 0 , the function u ( t , x ) = s i n ( α c t + β ) s i n ( α x ) satisfies the wave equation ∂ 2 u ∂ t 2 = c 2 ∂ 2 u ∂ x 2

Solution Summary: The author explains that the function u ( t, x ) = s i n satisfies the wave equation 2

That for the constants $α$ , $β$ and $c = 0$ , the function $u (t, x) = s i n (α c t + β) s i n (α x)$ satisfies the wave equation $\frac{\partial^{2} u}{\partial t^{2}} = c^{2} \frac{\partial^{2} u}{\partial x^{2}}$

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Chapter 15 Solutions

That for the constants α , β and c = 0 , the function u ( t , x ) = s i n ( α c t + β ) s i n ( α x ) satisfies the wave equation ∂ 2 u ∂ t 2 = c 2 ∂ 2 u ∂ x 2

Solution Summary: The author explains that the function u ( t, x ) = s i n satisfies the wave equation 2

That for the constants α,β and c=0, the function u(t, x) = sin(αct + β)sin(αx) satisfies the wave equation ∂2u∂t2=c2∂2u∂x2

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Chapter 15 Solutions

That for the constants $α$ , $β$ and $c = 0$ , the function $u (t, x) = s i n (α c t + β) s i n (α x)$ satisfies the wave equation $\frac{\partial^{2} u}{\partial t^{2}} = c^{2} \frac{\partial^{2} u}{\partial x^{2}}$