Answer on Question #52628 - Math - Vector Calculus
1) Verify that the vector fields are conservation by comparing cross derivatives, then potential functions for the, by taking anti-derivatives
A) f = ( z , 1 , x ) f = (z, 1, x) f = ( z , 1 , x )
B) f = ( y 2 , 2 x y + e z , y e z ) f = (y^{2}, 2xy + e^{z}, ye^{z}) f = ( y 2 , 2 x y + e z , y e z )
C) f = ( cos z , 2 y , − x sin z ) f = (\cos z, 2y, -x \sin z) f = ( cos z , 2 y , − x sin z )
Solution
Curl is given by
∇ × f = ∣ i ⃗ j ⃗ k ⃗ ∂ ∂ x ∂ ∂ y ∂ ∂ z f x f y f z ∣ = [ ∂ f z ∂ y − ∂ f y ∂ z ] i ⃗ + [ ∂ f x ∂ z − ∂ f z ∂ x ] j ⃗ + [ ∂ f y ∂ x − ∂ f x ∂ y ] k ⃗ \nabla \times f = \left| \begin{array}{ccc} \vec {i} & \vec {j} & \vec {k} \\ \frac {\partial}{\partial x} & \frac {\partial}{\partial y} & \frac {\partial}{\partial z} \\ f _ {x} & f _ {y} & f _ {z} \end{array} \right| = \left[ \frac {\partial f _ {z}}{\partial y} - \frac {\partial f _ {y}}{\partial z} \right] \vec {i} + \left[ \frac {\partial f _ {x}}{\partial z} - \frac {\partial f _ {z}}{\partial x} \right] \vec {j} + \left[ \frac {\partial f _ {y}}{\partial x} - \frac {\partial f _ {x}}{\partial y} \right] \vec {k} ∇ × f = ∣ ∣ i ∂ x ∂ f x j ∂ y ∂ f y k ∂ z ∂ f z ∣ ∣ = [ ∂ y ∂ f z − ∂ z ∂ f y ] i + [ ∂ z ∂ f x − ∂ x ∂ f z ] j + [ ∂ x ∂ f y − ∂ y ∂ f x ] k
A) Since
∇ × f = ∣ i ⃗ j ⃗ k ⃗ ∂ ∂ x ∂ ∂ y ∂ ∂ z z 1 x ∣ = [ ∂ x ∂ y − ∂ 1 ∂ z ] i ⃗ + [ ∂ z ∂ z − ∂ x ∂ x ] j ⃗ + [ ∂ 1 ∂ x − ∂ z ∂ y ] k ⃗ = = [ 0 − 0 ] i ⃗ + [ 1 − 1 ] j ⃗ + [ 0 − 0 ] k ⃗ = 0 ⃗ , \begin{array}{l}
\nabla \times f = \left| \begin{array}{ccc} \vec {i} & \vec {j} & \vec {k} \\ \frac {\partial}{\partial x} & \frac {\partial}{\partial y} & \frac {\partial}{\partial z} \\ z & 1 & x \end{array} \right| = \left[ \frac {\partial x}{\partial y} - \frac {\partial 1}{\partial z} \right] \vec {i} + \left[ \frac {\partial z}{\partial z} - \frac {\partial x}{\partial x} \right] \vec {j} + \left[ \frac {\partial 1}{\partial x} - \frac {\partial z}{\partial y} \right] \vec {k} = \\
= [ 0 - 0 ] \vec {i} + [ 1 - 1 ] \vec {j} + [ 0 - 0 ] \vec {k} = \vec {0},
\end{array} ∇ × f = ∣ ∣ i ∂ x ∂ z j ∂ y ∂ 1 k ∂ z ∂ x ∣ ∣ = [ ∂ y ∂ x − ∂ z ∂ 1 ] i + [ ∂ z ∂ z − ∂ x ∂ x ] j + [ ∂ x ∂ 1 − ∂ y ∂ z ] k = = [ 0 − 0 ] i + [ 1 − 1 ] j + [ 0 − 0 ] k = 0 ,
the vector field f f f ( f = ∇ V ) (f = \nabla V) ( f = ∇ V )
V = x z + y + C V = xz + y + C V = x z + y + C C C C
∇ V = ( ∂ ( x z + y + C ) ∂ x ; ∂ ( x z + y + C ) ∂ y ; ∂ ( x z + y + C ) ∂ z ) = ( z , 1 , x ) . \nabla V = \left(\frac {\partial (x z + y + C)}{\partial x}; \frac {\partial (x z + y + C)}{\partial y}; \frac {\partial (x z + y + C)}{\partial z}\right) = (z, 1, x). ∇ V = ( ∂ x ∂ ( x z + y + C ) ; ∂ y ∂ ( x z + y + C ) ; ∂ z ∂ ( x z + y + C ) ) = ( z , 1 , x ) .
To find function V V V
∂ V ∂ x = f x , ∂ V ∂ y = f y , ∂ V ∂ z = f z ; that is, \frac {\partial V}{\partial x} = f _ {x}, \frac {\partial V}{\partial y} = f _ {y}, \frac {\partial V}{\partial z} = f _ {z}; \text{ that is,} ∂ x ∂ V = f x , ∂ y ∂ V = f y , ∂ z ∂ V = f z ; that is, ∂ V ∂ x = z , ∂ V ∂ y = 1 , ∂ V ∂ z = x ; \frac {\partial V}{\partial x} = z, \frac {\partial V}{\partial y} = 1, \frac {\partial V}{\partial z} = x; ∂ x ∂ V = z , ∂ y ∂ V = 1 , ∂ z ∂ V = x ;
For ∂ V ∂ x = z \frac{\partial V}{\partial x} = z ∂ x ∂ V = z x x x V = z x + g ( y , z ) V = zx + g(y,z) V = z x + g ( y , z ) y y y
∂ V ∂ y = ∂ ∂ y ( z x + g ( y , z ) ) = ∂ g ( y , z ) ∂ y . \frac {\partial V}{\partial y} = \frac {\partial}{\partial y} (z x + g (y, z)) = \frac {\partial g (y , z)}{\partial y}. ∂ y ∂ V = ∂ y ∂ ( z x + g ( y , z )) = ∂ y ∂ g ( y , z ) .
On the other hand, taking into account the system, ∂ V ∂ y = 1 \frac{\partial V}{\partial y} = 1 ∂ y ∂ V = 1
Equating right-hand sides of two formulas gives ∂ g ( y , z ) ∂ y = 1 \frac{\partial g(y,z)}{\partial y} = 1 ∂ y ∂ g ( y , z ) = 1
Integrating both sides with respect to y y y g ( y , z ) = y + C ( z ) g(y,z) = y + C(z) g ( y , z ) = y + C ( z )
V = z x + g ( y , z ) = z x + y + C ( z ) . V = z x + g (y, z) = z x + y + C (z). V = z x + g ( y , z ) = z x + y + C ( z ) .
Taking the partial derivative of the both sides with respect to z z z
∂ V ∂ z = ∂ ∂ z ( z x + y + C ( z ) ) = x + C ′ ( z ) . \frac {\partial V}{\partial z} = \frac {\partial}{\partial z} (z x + y + C (z)) = x + C ^ {\prime} (z). ∂ z ∂ V = ∂ z ∂ ( z x + y + C ( z )) = x + C ′ ( z ) .
On the other hand, taking into account the system, ∂ V ∂ z = x \frac{\partial V}{\partial z} = x ∂ z ∂ V = x
Equating right-hand sides of two formulas gives x + C ′ ( z ) = x x + C'(z) = x x + C ′ ( z ) = x C ′ ( z ) = 0 C'(z) = 0 C ′ ( z ) = 0 z z z C ( z ) = C C(z) = C C ( z ) = C C C C
Thus, V = z x + g ( y , z ) = z x + y + C ( z ) = z x + y + C V = zx + g(y,z) = zx + y + C(z) = zx + y + C V = z x + g ( y , z ) = z x + y + C ( z ) = z x + y + C
B) Since
∇ × f = ∣ i ⃗ j ⃗ k ⃗ ∂ ∂ x ∂ ∂ y ∂ ∂ z y 2 2 x y + e z y e z ∣ = = [ ∂ ( y e z ) ∂ y − ∂ ( 2 x y + e z ) ∂ z ] i ⃗ + [ ∂ ( y 2 ) ∂ z − ∂ ( y e z ) ∂ x ] j ⃗ + [ ∂ ( 2 x y + e z ) ∂ x − ∂ ( y 2 ) ∂ y ] k ⃗ = [ e z − e z ] i ⃗ + [ 0 − 0 ] j ⃗ + [ 2 y − 2 y ] k ⃗ = 0 ⃗ \begin{array}{l} \nabla \times f = \left| \begin{array}{c c c} \vec {i} & \vec {j} & \vec {k} \\ \frac {\partial}{\partial x} & \frac {\partial}{\partial y} & \frac {\partial}{\partial z} \\ y ^ {2} & 2 x y + e ^ {z} & y e ^ {z} \end{array} \right| = \\ = \left[ \frac {\partial (y e ^ {z})}{\partial y} - \frac {\partial (2 x y + e ^ {z})}{\partial z} \right] \vec {i} + \left[ \frac {\partial (y ^ {2})}{\partial z} - \frac {\partial (y e ^ {z})}{\partial x} \right] \vec {j} \\ + \left[ \frac {\partial (2 x y + e ^ {z})}{\partial x} - \frac {\partial (y ^ {2})}{\partial y} \right] \vec {k} \\ = [ e ^ {z} - e ^ {z} ] \vec {i} + [ 0 - 0 ] \vec {j} + [ 2 y - 2 y ] \vec {k} = \vec {0} \\ \end{array} ∇ × f = ∣ ∣ i ∂ x ∂ y 2 j ∂ y ∂ 2 x y + e z k ∂ z ∂ y e z ∣ ∣ = = [ ∂ y ∂ ( y e z ) − ∂ z ∂ ( 2 x y + e z ) ] i + [ ∂ z ∂ ( y 2 ) − ∂ x ∂ ( y e z ) ] j + [ ∂ x ∂ ( 2 x y + e z ) − ∂ y ∂ ( y 2 ) ] k = [ e z − e z ] i + [ 0 − 0 ] j + [ 2 y − 2 y ] k = 0
the vector field f f f ( f = ∇ V ) (f = \nabla V) ( f = ∇ V )
V = x y 2 + y e z + C V = xy^{2} + ye^{z} + C V = x y 2 + y e z + C C C C
because
∇ V = ( ∂ ( x y 2 + y e z + C ) ∂ x ; ∂ ( x y 2 + y e z + C ) ∂ y ; ∂ ( x y 2 + y e z + C ) ∂ z ) = ( y 2 , e z , y e z ) . \nabla V = \left(\frac {\partial (x y ^ {2} + y e ^ {z} + C)}{\partial x}; \frac {\partial (x y ^ {2} + y e ^ {z} + C)}{\partial y}; \frac {\partial (x y ^ {2} + y e ^ {z} + C)}{\partial z}\right) = (y ^ {2}, e ^ {z}, y e ^ {z}). ∇ V = ( ∂ x ∂ ( x y 2 + y e z + C ) ; ∂ y ∂ ( x y 2 + y e z + C ) ; ∂ z ∂ ( x y 2 + y e z + C ) ) = ( y 2 , e z , y e z ) .
To find function V V V
∂ V ∂ x = f x , ∂ V ∂ y = f y , ∂ V ∂ z = f z \frac{\partial V}{\partial x} = f_x, \frac{\partial V}{\partial y} = f_y, \frac{\partial V}{\partial z} = f_z ∂ x ∂ V = f x , ∂ y ∂ V = f y , ∂ z ∂ V = f z
∂ V ∂ x = y 2 , ∂ V ∂ y = 2 x y + e z , ∂ V ∂ z = y e z ; \frac {\partial V}{\partial x} = y ^ {2}, \frac {\partial V}{\partial y} = 2 x y + e ^ {z}, \frac {\partial V}{\partial z} = y e ^ {z}; ∂ x ∂ V = y 2 , ∂ y ∂ V = 2 x y + e z , ∂ z ∂ V = y e z ;
For ∂ V ∂ x = y 2 \frac{\partial V}{\partial x} = y^2 ∂ x ∂ V = y 2 x x x V = x y 2 + g ( y , z ) V = xy^2 + g(y, z) V = x y 2 + g ( y , z ) y y y
∂ V ∂ y = ∂ ∂ y ( x y 2 + g ( y , z ) ) = 2 x y + ∂ g ( y , z ) ∂ y . \frac {\partial V}{\partial y} = \frac {\partial}{\partial y} (x y ^ {2} + g (y, z)) = 2 x y + \frac {\partial g (y , z)}{\partial y}. ∂ y ∂ V = ∂ y ∂ ( x y 2 + g ( y , z )) = 2 x y + ∂ y ∂ g ( y , z ) .
On the other hand, taking into account the system, ∂ V ∂ y = 2 x y + e z \frac{\partial V}{\partial y} = 2xy + e^z ∂ y ∂ V = 2 x y + e z
Equating right-hand sides of two formulas gives ∂ g ( y , z ) ∂ y = e z \frac{\partial g(y,z)}{\partial y} = e^z ∂ y ∂ g ( y , z ) = e z
Integrating both sides with respect to y y y g ( y , z ) = y e z + C ( z ) g(y,z) = ye^{z} + C(z) g ( y , z ) = y e z + C ( z )
V = x y 2 + g ( y , z ) = x y 2 + y e z + C ( z ) . V = x y ^ {2} + g (y, z) = x y ^ {2} + y e ^ {z} + C (z). V = x y 2 + g ( y , z ) = x y 2 + y e z + C ( z ) .
Taking the partial derivative of the both sides with respect to z z z
∂ V ∂ z = ∂ ∂ z ( x y 2 + y e z + C ( z ) ) = y e z + C ′ ( z ) . \frac {\partial V}{\partial z} = \frac {\partial}{\partial z} (x y ^ {2} + y e ^ {z} + C (z)) = y e ^ {z} + C ^ {\prime} (z). ∂ z ∂ V = ∂ z ∂ ( x y 2 + y e z + C ( z )) = y e z + C ′ ( z ) .
On the other hand, taking into account the system, ∂ V ∂ z = y e z \frac{\partial V}{\partial z} = ye^z ∂ z ∂ V = y e z
Equating right-hand sides of two formulas gives y e z + C ′ ( z ) = y e z ye^{z} + C^{\prime}(z) = ye^{z} y e z + C ′ ( z ) = y e z C ′ ( z ) = 0 C^{\prime}(z) = 0 C ′ ( z ) = 0 z z z C ( z ) = C C(z) = C C ( z ) = C C C C
Thus, V = x y 2 + g ( y , z ) = x y 2 + y e z + C ( z ) = x y 2 + y e z + C V = xy^{2} + g(y,z) = xy^{2} + ye^{z} + C(z) = xy^{2} + ye^{z} + C V = x y 2 + g ( y , z ) = x y 2 + y e z + C ( z ) = x y 2 + y e z + C
C) Since
∇ × f = ∣ i ⃗ j ⃗ k ⃗ ∂ ∂ x ∂ ∂ y ∂ ∂ z cos z 2 y − x sin z ∣ = = [ ∂ ( − x sin z ) ∂ y − ∂ ( 2 y ) ∂ z ] i ⃗ + [ ∂ ( cos z ) ∂ z − ∂ ( − x sin z ) ∂ x ] j ⃗ + [ ∂ ( 2 y ) ∂ x − ∂ ( cos z ) ∂ y ] k ⃗ = [ 0 − 0 ] i ⃗ + [ − sin z − ( − sin z ) ] j ⃗ + [ 0 − 0 ] k ⃗ = 0 ⃗ , \begin{array}{l} \nabla \times f = \left| \begin{array}{c c c} \vec {i} & \vec {j} & \vec {k} \\ \frac {\partial}{\partial x} & \frac {\partial}{\partial y} & \frac {\partial}{\partial z} \\ \cos z & 2 y & - x \sin z \end{array} \right| = \\ = \left[ \frac {\partial (- x \sin z)}{\partial y} - \frac {\partial (2 y)}{\partial z} \right] \vec {i} + \left[ \frac {\partial (\cos z)}{\partial z} - \frac {\partial (- x \sin z)}{\partial x} \right] \vec {j} \\ + \left[ \frac {\partial (2 y)}{\partial x} - \frac {\partial (\cos z)}{\partial y} \right] \vec {k} \\ = [ 0 - 0 ] \vec {i} + [ - \sin z - (- \sin z) ] \vec {j} + [ 0 - 0 ] \vec {k} = \vec {0}, \\ \end{array} ∇ × f = ∣ ∣ i ∂ x ∂ cos z j ∂ y ∂ 2 y k ∂ z ∂ − x sin z ∣ ∣ = = [ ∂ y ∂ ( − x s i n z ) − ∂ z ∂ ( 2 y ) ] i + [ ∂ z ∂ ( c o s z ) − ∂ x ∂ ( − x s i n z ) ] j + [ ∂ x ∂ ( 2 y ) − ∂ y ∂ ( c o s z ) ] k = [ 0 − 0 ] i + [ − sin z − ( − sin z )] j + [ 0 − 0 ] k = 0 ,
the vector field f f f ( f = ∇ V ) (f = \nabla V) ( f = ∇ V )
V = y 2 + x cos z + C V = y^{2} + x\cos z + C V = y 2 + x cos z + C C C C
∇ V = ( ∂ ( y 2 + x cos z + C ) ∂ x ; ∂ ( y 2 + x cos z + C ) ∂ y ; ∂ ( y 2 + x cos z + C ) ∂ z ) = ( cos z , 2 y , − x sin z ) . \nabla V = \left(\frac {\partial (y ^ {2} + x \cos z + C)}{\partial x}; \frac {\partial (y ^ {2} + x \cos z + C)}{\partial y}; \frac {\partial (y ^ {2} + x \cos z + C)}{\partial z}\right) = (\cos z, 2 y, - x \sin z). ∇ V = ( ∂ x ∂ ( y 2 + x cos z + C ) ; ∂ y ∂ ( y 2 + x cos z + C ) ; ∂ z ∂ ( y 2 + x cos z + C ) ) = ( cos z , 2 y , − x sin z ) .
To find function V V V
∂ V ∂ x = f x , ∂ V ∂ y = f y , ∂ V ∂ z = f z ; that is, \frac {\partial V}{\partial x} = f _ {x}, \frac {\partial V}{\partial y} = f _ {y}, \frac {\partial V}{\partial z} = f _ {z}; \text{ that is,} ∂ x ∂ V = f x , ∂ y ∂ V = f y , ∂ z ∂ V = f z ; that is, ∂ V ∂ x = cos z , ∂ V ∂ y = 2 y , ∂ V ∂ z = − x sin z ; \frac {\partial V}{\partial x} = \cos z, \frac {\partial V}{\partial y} = 2 y, \frac {\partial V}{\partial z} = - x \sin z; ∂ x ∂ V = cos z , ∂ y ∂ V = 2 y , ∂ z ∂ V = − x sin z ;
For ∂ V ∂ x = cos z \frac{\partial V}{\partial x} = \cos z ∂ x ∂ V = cos z x x x V = x cos z + g ( y , z ) V = x\cos z + g(y,z) V = x cos z + g ( y , z ) y y y
∂ V ∂ y = ∂ ∂ y ( x cos z + g ( y , z ) ) = ∂ g ( y , z ) ∂ y . \frac {\partial V}{\partial y} = \frac {\partial}{\partial y} (x \cos z + g (y, z)) = \frac {\partial g (y , z)}{\partial y}. ∂ y ∂ V = ∂ y ∂ ( x cos z + g ( y , z )) = ∂ y ∂ g ( y , z ) .
On the other hand, taking into account the system, ∂ V ∂ y = 2 y \frac{\partial V}{\partial y} = 2y ∂ y ∂ V = 2 y
Equating right-hand sides of two formulas gives ∂ g ( y , z ) ∂ y = 2 y \frac{\partial g(y,z)}{\partial y} = 2y ∂ y ∂ g ( y , z ) = 2 y
Integrating both sides with respect to y y y g ( y , z ) = y 2 + C ( z ) g(y,z) = y^{2} + C(z) g ( y , z ) = y 2 + C ( z )
V = x cos z + g ( y , z ) = x cos z + y 2 + C ( z ) . V = x \cos z + g (y, z) = x \cos z + y ^ {2} + C (z). V = x cos z + g ( y , z ) = x cos z + y 2 + C ( z ) .
Taking the partial derivative of the both sides with respect to z z z
∂ V ∂ z = ∂ ∂ z ( x cos z + y 2 + C ( z ) ) = − x sin z + C ′ ( z ) . \frac {\partial V}{\partial z} = \frac {\partial}{\partial z} (x \cos z + y ^ {2} + C (z)) = - x \sin z + C ^ {\prime} (z). ∂ z ∂ V = ∂ z ∂ ( x cos z + y 2 + C ( z )) = − x sin z + C ′ ( z ) .
On the other hand, taking into account the system, ∂ V ∂ z = − x sin z \frac{\partial V}{\partial z} = -x\sin z ∂ z ∂ V = − x sin z
Equating right-hand sides of two formulas gives − x sin z + C ′ ( z ) = − x sin z -x\sin z + C'(z) = -x\sin z − x sin z + C ′ ( z ) = − x sin z C ′ ( z ) = 0 C'(z) = 0 C ′ ( z ) = 0 z z z C ( z ) = C C(z) = C C ( z ) = C C C C
Thus, V = x cos z + g ( y , z ) = x cos z + y 2 + C ( z ) = x cos z + y 2 + C V = x\cos z + g(y,z) = x\cos z + y^{2} + C(z) = x\cos z + y^{2} + C V = x cos z + g ( y , z ) = x cos z + y 2 + C ( z ) = x cos z + y 2 + C
Answer:
A) x z + y + C xz + y + C x z + y + C
B) x y 2 + y e z + C xy^{2} + ye^{z} + C x y 2 + y e z + C
C) x cos z + y 2 + C x\cos z + y^2 + C x cos z + y 2 + C
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