Section A brief on transformations

Transcription

1 Section 127 A brief on transformations consider the transformation another name for it: change of variable x = rcosθ The determinant y = rsinθ r y r is called the Jacobian determinant of x,y with respect to r,θ, and is denoted by r,θ Let us calculate this determinant: r r θ = θ cosθ sinθ x = xu,v Definition If y = yu,v θ θ r sinθ r cosθ = rcos2 θ + rsin 2 θ = rcos 2 θ + sin 2 θ = r is a transformation, then the determinant u,v = is called the Jacobian determinant of the functions x and y in terms of the variables u and v This Jacobian is a function of the variables u and v see the next example x = u + 3v 2 Example Find the Jacobian of the transformation y = u 2 uv Solution 1

2 u,v = = 1 6v 2u v u u 6v2u v u 12uv + 6v 2 this is a function of u and v Note Sometimes we are interested in a particular value of the Jacobian See the next example: Example In the previous example, what is the value of the Jacobian u,v at the point where u,v = 1, 1 Solution u,v u 12uv + 6v 2 u,v=1, 1 = 17 u,v=1, 1 Fact It is true that u,v = 1 u,v This is a useful formula when it is difficult or impossible to find the inverse transformation u = ux,y v = vx,y Example For the transformation x = u 3 + uv + v y = v 3 + uv + v calculate u,v at the point where u,v = 1, 1 Solution We are not able to calculate u,v in terms of u,v the inverse transformation therefore we 2

3 cannot calculate u,v directly So we use the formula u,v = 1 u,v By substituting u = 1 and v 1 we get x 1 and y 3 u,v = 2 2 u,v = u,v=1, = 3u 2 + v u + 1 v 3v 2 + u + 1 Definition For a 3 3 transformation = 12 u,v = 1 u,v x = xu,v,w y = yu,v,w z = zu,v,w the Jacobian determinant is defined by x,y,z u,v,w = w w w = 1 12 This quantity enjoys a similar identity as in the two dimensional case: x,y,z u,v,w = 1 u,v,w x,y,z Note In general if we have an n n system of equations such as: y 1 = f 1 x 1,,, x n y 2 = f 2 x 1,,, x n y n = f n x 1,,, x n 3

4 then its Jacobian determinant is defined by: y 1,,, y n x 1,,, x n = n n n 2 1 n 2 n n n the first row being the derivatives of y 1 4

5 Implicit Differentiation Suppose we have a system of m equations with m + n unknowns: F 1 x 1,, x n, y 1,, y m = 0 F 2 x 1,, x n, y 1,, y m = 0 F m x 1,, x n, y 1,, y m = 0 We consider the n number of variables the difference between m + n and m, ie the difference between the number of unknowns and the number of equations as independent variables For simplicity suppose that x 1,, x n are the independent variables, and so y 1,, y m are the dependent variables Then the first order partial derivatives of the dependent variables with respect to the independent variables are calculated through: i k F 1,,F m y 1,,x k,,y m F 1,,F m y 1,,y i,,y m x k is sat the i-th place in th numerator The denominator is the Jacobian with respect to the dependent variables Example Consider the system xy 2 + xzu + yv 2 = 3 x 3 yz + 2xv u 2 v 2 = 2 We must consider as many as 5 2 = 3 variables as independent ones Let us consider the variables u,v as the dependent variables and the variables x,y,z as the independent ones Find at the point with coordinates x,y,z,u,v = 1,1,1,1,1 Solution Set the first step is the naming of the equations: 5

6 Fx,y,z,u,v = xy 2 + xzu + yv 2 3 Gx,y,z,u,v = x 3 yz + 2xv u 2 v 2 2 Then, at the point with coordinates 1,1,1,1,1 we have: u,v = xz 2uv 2 2yv 2x 2u 2 v = = 4 u,y = xz 2xy + v 2 2uv 2 x 3 z = = 7 u,y u,v 7 4 Example Consider 4 variables which are tangled together through: u = x 2 + xy y 2 v = 2xy + y 2 i Consider x and y as functions of u,v Find at the point with x,y = 2, 1 ii Now consider x and v as functions of y and u Find at the point with x,y = 2, 1 Solution to part i We write the equations in the form: Fx,y,u,v = x 2 + xy y 2 u Gx,y,z,u,v = 2xy + y 2 v At the point with characteristics x = 2 and y 1 we have: 6

7 u,y 1 x 2y 0 2x + 2y = 2x + y x 2y 2y 2x + 2y = 1 7 Solution to part ii Now we are considering x,v as dependent variables So: u,v x,v = 2x + y 0 2y = 1 3 Note Consider an equation Fx,y = 0 which actually describes a curve in the plane defined implicitly such as x 3 y 2 2xy + 5 = 0 If y is considered as a function of x, then by applying the implicit differentiation formula we learned for the general case, we will have: Note Consider an equation Fx,y,z = 0 If z is considered as a function of x,y, then by applying the implicit differentiation formula we learned for the general case, we will have: Example section 127 exercise 1 Consider y as a function of x in the equation x 3 y 2 2xy + 5 = 0 Find dy dx 7

8 Solution First Method Put Fx,y = x 3 y 2 2xy + 5 dy dx 3x2 y 2 2y 2x 3 y 2x Second Method Differentiate both sides of x 3 y 2 2xy + 5 = 0 with respect to x considering y as a function of x this is an elementary Calculus subject: { {x 3 } {y 2 } + {x 3 }{y 2 } } { 2 {x} {y} + {x}{y} } = 0 { } { } {3x 2 }{y 2 } + {x 3 }{2yy } 2 {1}{y} + {x}{y } = 0 3x 2 y 2 2y + 2x 3 y 2xy = 0 y 3x2 y 2 2y 2x 3 y 2x Note As you see, the first method is much simpler Example section 127 exercise 5 The variable z is defined implicitly as a function of x and y through x 2 sinz ye z = 2x Find Solution First Method Put F = x 2 sinz ye z 2x = 0 2xsinz 2 x 2 cosz ye z = 2 2xsinz x 2 cosz ye z 8

9 Second Method Differentiate the equation x 2 sinz ye z = 2x with respect to x keeping in mind that the derivative of y with respect to x is zero because in this question it is assumed that x and y are independent and z is a function of them { {x 2 } {sinz} + {x 2 }{sinz} } { {y} {e z } + {y}{e z } } = {2x} { } { } {2x}{sinz} + {x 2 }{coszz } {0}{e z } + {y}{e z z } = 2 { } } {2x}{sinz} + {x 2 }{coszz } + {y}{e z z } = 2 2xsinz + x 2 cosz + ye z z = 2 z = 2 2xsinz x 2 cosz + ye z Example section 127 exercise 17 Find s if s = v x2 + y 2, and x and y are functions of u and v defined by u = x 2 y 2 v = x 2 y Solution We are assuming that s = sx,y x = xu,v y = yu,v s = s v + s = 2x + 2y Now to calculate and we need to apply the implicit differentiation technique because x and y are not given explicitly in terms of u and v: 9

10 Fx,y,u,v = x 2 y 2 u Gx,y,u,v = x 2 y v u,y 1 2y 0 1 2x 2y 2x 1 1 2x + 4xy = 1 2x 4xy x,u 2x 1 2x 0 2x 2y 2x 1 2x 2x + 4xy = 2x 2x 4xy Now by putting these into *, one gets: s = 2x v 1 2x 4xy 2x + 2y = 2x 4xy 2x + 2y 2x 4xy = x + y x 2xy 10