ANSWER on Question No59206, Math / Differential Equations
QUESTION 6
If
A → = 5 t 2 i → + t j → − t 3 k → and B → = i → sin t − j → cos t . \overrightarrow {A} = 5 t ^ {2} \overrightarrow {i} + t \overrightarrow {j} - t ^ {3} \overrightarrow {k} \quad \text{and} \quad \overrightarrow {B} = \overrightarrow {i} \sin t - \overrightarrow {j} \cos t. A = 5 t 2 i + t j − t 3 k and B = i sin t − j cos t .
Evaluate
d d t ( A → × B → ) \frac {d}{d t} \left(\overrightarrow {A} \times \overrightarrow {B}\right) d t d ( A × B )
SOLUTION
By the definition
A → × B → = ∥ i → j → k → A x A y A z B x B y B z ∥ = i → ( A y B z − A z B y ) − j → ( A x B z − A z B x ) + k → ( A x B y − A y B x ) \overrightarrow {A} \times \overrightarrow {B} = \left\| \begin{array}{ccc} \overrightarrow {i} & \overrightarrow {j} & \overrightarrow {k} \\ A _ {x} & A _ {y} & A _ {z} \\ B _ {x} & B _ {y} & B _ {z} \end{array} \right\| = \overrightarrow {i} \left(A _ {y} B _ {z} - A _ {z} B _ {y}\right) - \overrightarrow {j} \left(A _ {x} B _ {z} - A _ {z} B _ {x}\right) + \overrightarrow {k} \left(A _ {x} B _ {y} - A _ {y} B _ {x}\right) A × B = ∥ ∥ i A x B x j A y B y k A z B z ∥ ∥ = i ( A y B z − A z B y ) − j ( A x B z − A z B x ) + k ( A x B y − A y B x )
In our case
A → = 5 t 2 ⏟ A x i → + t ⏟ A y j → + − t 3 ⏟ A z k → and B → = i → sin t ⏟ B x + j → − cos t ⏟ B y + 0 ⏟ B z k → \overrightarrow {A} = \underbrace {5 t ^ {2}} _ {A _ {x}} \overrightarrow {i} + \underbrace {t} _ {A _ {y}} \overrightarrow {j} + \underbrace {- t ^ {3}} _ {A _ {z}} \overrightarrow {k} \quad \text{and} \quad \overrightarrow {B} = \overrightarrow {i} \underbrace {\sin t} _ {B _ {x}} + \overrightarrow {j} \underbrace {- \cos t} _ {B _ {y}} + \underbrace {0} _ {B _ {z}} \overrightarrow {k} A = A x 5 t 2 i + A y t j + A z − t 3 k and B = i B x sin t + j B y − cos t + B z 0 k A → × B → = i → ( t ∗ 0 − ( − t 3 ) ∗ ( − cos t ) ) − j → ( ( 5 t 2 ) ∗ 0 − ( − t 3 ) ∗ sin t ) + k → ( ( 5 t 2 ) ∗ ( − cos t ) − t ∗ sin t ) = = − t 3 cos t i → − t 3 ∗ sin t j → − ( 5 t 2 cos t + t ∗ sin t ) k → \begin{array}{l} \overrightarrow {A} \times \overrightarrow {B} = \overrightarrow {i} (t * 0 - (- t ^ {3}) * (- \cos t)) - \overrightarrow {j} ((5 t ^ {2}) * 0 - (- t ^ {3}) * \sin t) + \overrightarrow {k} ((5 t ^ {2}) * (- \cos t) - t * \sin t) = \\ = - t ^ {3} \cos t \overrightarrow {i} - t ^ {3} * \sin t \overrightarrow {j} - (5 t ^ {2} \cos t + t * \sin t) \overrightarrow {k} \\ \end{array} A × B = i ( t ∗ 0 − ( − t 3 ) ∗ ( − cos t )) − j (( 5 t 2 ) ∗ 0 − ( − t 3 ) ∗ sin t ) + k (( 5 t 2 ) ∗ ( − cos t ) − t ∗ sin t ) = = − t 3 cos t i − t 3 ∗ sin t j − ( 5 t 2 cos t + t ∗ sin t ) k d d t ( A → × B → ) = d d t ( − t 3 cos t i → − t 3 ∗ sin t j → − ( 5 t 2 cos t + t ∗ sin t ) k → ) = = d d t ( − t 3 cos t ) i → − d d t ( t 3 ∗ sin t ) j → − d d t ( 5 t 2 cos t + t ∗ sin t ) k → = = ( − 3 t 2 cos t + ( − t 3 ) ∗ ( − sin t ) ) i → − ( 3 t 2 ∗ sin t + t 3 ∗ cos t ) j → − − ( 5 ∗ 2 t cos t + 5 t 2 ∗ ( − sin t ) + sin t + t ∗ cos t ) k → = = ( t 3 sin t − 3 t 2 cos t ) i → − ( 3 t 2 sin t + t 3 cos t ) j → − ( 11 t cos t − 5 t 2 sin t + sin t ) k → \begin{array}{l} \frac {d}{d t} \left(\overrightarrow {A} \times \overrightarrow {B}\right) = \frac {d}{d t} \left(- t ^ {3} \cos t \overrightarrow {i} - t ^ {3} * \sin t \overrightarrow {j} - \left(5 t ^ {2} \cos t + t * \sin t\right) \overrightarrow {k}\right) = \\ = \frac {d}{d t} \left(- t ^ {3} \cos t\right) \overrightarrow {i} - \frac {d}{d t} \left(t ^ {3} * \sin t\right) \overrightarrow {j} - \frac {d}{d t} \left(5 t ^ {2} \cos t + t * \sin t\right) \overrightarrow {k} = \\ = \left(- 3 t ^ {2} \cos t + (- t ^ {3}) * (- \sin t)\right) \overrightarrow {i} - \left(3 t ^ {2} * \sin t + t ^ {3} * \cos t\right) \overrightarrow {j} - \\ - \left(5 * 2 t \cos t + 5 t ^ {2} * (- \sin t) + \sin t + t * \cos t\right) \overrightarrow {k} = \\ = \left(t ^ {3} \sin t - 3 t ^ {2} \cos t\right) \overrightarrow {i} - \left(3 t ^ {2} \sin t + t ^ {3} \cos t\right) \overrightarrow {j} - \left(11 t \cos t - 5 t ^ {2} \sin t + \sin t\right) \overrightarrow {k} \\ \end{array} d t d ( A × B ) = d t d ( − t 3 cos t i − t 3 ∗ sin t j − ( 5 t 2 cos t + t ∗ sin t ) k ) = = d t d ( − t 3 cos t ) i − d t d ( t 3 ∗ sin t ) j − d t d ( 5 t 2 cos t + t ∗ sin t ) k = = ( − 3 t 2 cos t + ( − t 3 ) ∗ ( − sin t ) ) i − ( 3 t 2 ∗ sin t + t 3 ∗ cos t ) j − − ( 5 ∗ 2 t cos t + 5 t 2 ∗ ( − sin t ) + sin t + t ∗ cos t ) k = = ( t 3 sin t − 3 t 2 cos t ) i − ( 3 t 2 sin t + t 3 cos t ) j − ( 11 t cos t − 5 t 2 sin t + sin t ) k
ANSWER
d d t ( A → × B → ) = ( t 3 sin t − 3 t 2 cos t ) i → − ( 3 t 2 sin t + t 3 cos t ) j → − ( 11 t cos t − 5 t 2 sin t + sin t ) k → \frac {d}{d t} \left(\overrightarrow {A} \times \overrightarrow {B}\right) = \left(t ^ {3} \sin t - 3 t ^ {2} \cos t\right) \overrightarrow {i} - \left(3 t ^ {2} \sin t + t ^ {3} \cos t\right) \overrightarrow {j} - \left(11 t \cos t - 5 t ^ {2} \sin t + \sin t\right) \overrightarrow {k} d t d ( A × B ) = ( t 3 sin t − 3 t 2 cos t ) i − ( 3 t 2 sin t + t 3 cos t ) j − ( 11 t cos t − 5 t 2 sin t + sin t ) k
QUESTION 7
If
A → = 5 t 2 i → + t j → − t 3 k → and B → = i → sin t − j → cos t . \overrightarrow {A} = 5 t ^ {2} \overrightarrow {i} + t \overrightarrow {j} - t ^ {3} \overrightarrow {k} \quad \text{and} \quad \overrightarrow {B} = \overrightarrow {i} \sin t - \overrightarrow {j} \cos t. A = 5 t 2 i + t j − t 3 k and B = i sin t − j cos t .
Evaluate
d d t ( A → ⋅ A → ) \frac {d}{d t} \left(\overrightarrow {A} \cdot \overrightarrow {A}\right) d t d ( A ⋅ A )
SOLUTION
By the definition
A → ⋅ A → = A x ∗ A x + A y ∗ A y + A z ∗ A z \overrightarrow {A} \cdot \overrightarrow {A} = A _ {x} * A _ {x} + A _ {y} * A _ {y} + A _ {z} * A _ {z} A ⋅ A = A x ∗ A x + A y ∗ A y + A z ∗ A z
In our case
A → = 5 t 2 ⏟ A x i → + t ⏟ A y j → + − t 3 ⏟ A z k → \overrightarrow {A} = \underbrace {5 t ^ {2}} _ {A _ {x}} \overrightarrow {i} + \underbrace {t} _ {A _ {y}} \overrightarrow {j} + \underbrace {- t ^ {3}} _ {A _ {z}} \overrightarrow {k} A = A x 5 t 2 i + A y t j + A z − t 3 k A → ⋅ A → = 5 t 2 ∗ 5 t 2 + t ∗ t + ( − t 3 ) ∗ ( − t 3 ) = 25 t 4 + t 2 + t 6 \overrightarrow {A} \cdot \overrightarrow {A} = 5 t ^ {2} * 5 t ^ {2} + t * t + (- t ^ {3}) * (- t ^ {3}) = 2 5 t ^ {4} + t ^ {2} + t ^ {6} A ⋅ A = 5 t 2 ∗ 5 t 2 + t ∗ t + ( − t 3 ) ∗ ( − t 3 ) = 25 t 4 + t 2 + t 6 d d t ( A → ⋅ A → ) = d d t ( 25 t 4 + t 2 + t 6 ) = 25 ∗ 4 t 3 + 2 t + 6 t 5 = 100 t 3 + 2 t + 6 t 5 \frac {d}{d t} \left(\overrightarrow {A} \cdot \overrightarrow {A}\right) = \frac {d}{d t} \left(2 5 t ^ {4} + t ^ {2} + t ^ {6}\right) = 2 5 * 4 t ^ {3} + 2 t + 6 t ^ {5} = 1 0 0 t ^ {3} + 2 t + 6 t ^ {5} d t d ( A ⋅ A ) = d t d ( 25 t 4 + t 2 + t 6 ) = 25 ∗ 4 t 3 + 2 t + 6 t 5 = 100 t 3 + 2 t + 6 t 5
ANSWER
d d t ( A → ⋅ A → ) = 6 t 5 + 100 t 3 + 2 t \frac {d}{d t} \left(\overrightarrow {A} \cdot \overrightarrow {A}\right) = 6 t ^ {5} + 1 0 0 t ^ {3} + 2 t d t d ( A ⋅ A ) = 6 t 5 + 100 t 3 + 2 t
QUESTION 8
If
A → = sin u i → + cos u j → + u k → B → = cos u i → − sin u j → − 3 k → and C → = 2 i → + 3 j → − k → \overrightarrow {A} = \sin u \overrightarrow {i} + \cos u \overrightarrow {j} + u \overrightarrow {k} \quad \overrightarrow {B} = \cos u \overrightarrow {i} - \sin u \overrightarrow {j} - 3 \overrightarrow {k} \quad \text{and} \quad \overrightarrow {C} = 2 \overrightarrow {i} + 3 \overrightarrow {j} - \overrightarrow {k} A = sin u i + cos u j + u k B = cos u i − sin u j − 3 k and C = 2 i + 3 j − k
Evaluate
d d u ( A → × ( B → × C → ) ) at u = 0 \frac {d}{d u} \left(\overrightarrow {A} \times \left(\overrightarrow {B} \times \overrightarrow {C}\right)\right) \quad \text{at} \quad u = 0 d u d ( A × ( B × C ) ) at u = 0
SOLUTION
By the definition
B → × C → = ∥ i → j → k → cos u − sin u − 3 2 3 − 1 ∥ = \overrightarrow {B} \times \overrightarrow {C} = \left\| \begin{array}{ccc} \overrightarrow {i} & \overrightarrow {j} & \overrightarrow {k} \\ \cos u & - \sin u & - 3 \\ 2 & 3 & - 1 \end{array} \right\| = B × C = ∥ ∥ i cos u 2 j − sin u 3 k − 3 − 1 ∥ ∥ = = i → ( ( − sin u ) ∗ ( − 1 ) − ( 3 ) ∗ ( − 3 ) ) − j → ( cos u ∗ ( − 1 ) − 2 ∗ ( − 3 ) ) + k → ( cos u ∗ ( 3 ) − 2 ∗ ( − sin u ) ) = = i → ( sin u + 9 ) − j → ( 6 − cos u ) + k → ( 3 cos u + 2 sin u ) = = i → ( sin u + 9 ) + j → ( cos u − 6 ) + k → ( 3 cos u + 2 sin u ) \begin{array}{l}
= \overrightarrow {i} ((- \sin u) * (- 1) - (3) * (- 3)) - \overrightarrow {j} (\cos u * (- 1) - 2 * (- 3)) + \overrightarrow {k} (\cos u * (3) - 2 * (- \sin u)) = \\
= \overrightarrow {i} (\sin u + 9) - \overrightarrow {j} (6 - \cos u) + \overrightarrow {k} (3 \cos u + 2 \sin u) = \\
= \overrightarrow {i} (\sin u + 9) + \overrightarrow {j} (\cos u - 6) + \overrightarrow {k} (3 \cos u + 2 \sin u) \\
\end{array} = i (( − sin u ) ∗ ( − 1 ) − ( 3 ) ∗ ( − 3 )) − j ( cos u ∗ ( − 1 ) − 2 ∗ ( − 3 )) + k ( cos u ∗ ( 3 ) − 2 ∗ ( − sin u )) = = i ( sin u + 9 ) − j ( 6 − cos u ) + k ( 3 cos u + 2 sin u ) = = i ( sin u + 9 ) + j ( cos u − 6 ) + k ( 3 cos u + 2 sin u ) A → × ( B → × C → ) = ∥ i → j → k → sin u cos u u sin u + 9 cos u − 6 3 cos u + 2 sin u ∥ = \overrightarrow {A} \times \left(\overrightarrow {B} \times \overrightarrow {C}\right) = \left\| \begin{array}{ccc} \overrightarrow {i} & \overrightarrow {j} & \overrightarrow {k} \\ \sin u & \cos u & u \\ \sin u + 9 & \cos u - 6 & 3 \cos u + 2 \sin u \end{array} \right\| = A × ( B × C ) = ∥ ∥ i sin u sin u + 9 j cos u cos u − 6 k u 3 cos u + 2 sin u ∥ ∥ = = i → ( cos u ∗ ( 3 cos u + 2 sin u ) − u ∗ ( cos u − 6 ) ) − j → ( sin u ∗ ( 3 cos u + 2 sin u ) − u ∗ ( sin u + 9 ) ) + = \overrightarrow {i} (\cos u * (3 \cos u + 2 \sin u) - u * (\cos u - 6)) - \overrightarrow {j} (\sin u * (3 \cos u + 2 \sin u) - u * (\sin u + 9)) + = i ( cos u ∗ ( 3 cos u + 2 sin u ) − u ∗ ( cos u − 6 )) − j ( sin u ∗ ( 3 cos u + 2 sin u ) − u ∗ ( sin u + 9 )) + + k → ( sin u ∗ ( cos u − 6 ) − cos u ∗ ( sin u + 9 ) ) = + \overrightarrow {k} (\sin u * (\cos u - 6) - \cos u * (\sin u + 9)) = + k ( sin u ∗ ( cos u − 6 ) − cos u ∗ ( sin u + 9 )) = = i → ( 3 cos 2 u + 2 cos u sin u − u cos u + 6 u ) − j → ( 3 sin u cos u + 2 sin 2 u − u sin u − 9 u ) + = \overrightarrow {i} (3 \cos^ {2} u + 2 \cos u \sin u - u \cos u + 6 u) - \overrightarrow {j} (3 \sin u \cos u + 2 \sin^ {2} u - u \sin u - 9 u) + = i ( 3 cos 2 u + 2 cos u sin u − u cos u + 6 u ) − j ( 3 sin u cos u + 2 sin 2 u − u sin u − 9 u ) + + k → ( sin u cos u − 6 sin u − cos u sin u − 9 cos u ) = + \overrightarrow {k} (\sin u \cos u - 6 \sin u - \cos u \sin u - 9 \cos u) = + k ( sin u cos u − 6 sin u − cos u sin u − 9 cos u ) = = i → ( 3 cos 2 u + 2 cos u sin u − u cos u + 6 u ) − j → ( 3 sin u cos u + 2 sin 2 u − u sin u − 9 u ) − = \overrightarrow {i} (3 \cos^ {2} u + 2 \cos u \sin u - u \cos u + 6 u) - \overrightarrow {j} (3 \sin u \cos u + 2 \sin^ {2} u - u \sin u - 9 u) - = i ( 3 cos 2 u + 2 cos u sin u − u cos u + 6 u ) − j ( 3 sin u cos u + 2 sin 2 u − u sin u − 9 u ) − − k → ( 6 sin u + 9 cos u ) - \overrightarrow {k} (6 \sin u + 9 \cos u) − k ( 6 sin u + 9 cos u ) d d u ( A → × ( B → × C → ) ) = \frac {d}{d u} \left(\overrightarrow {A} \times \left(\overrightarrow {B} \times \overrightarrow {C}\right)\right) = d u d ( A × ( B × C ) ) = = i → d d u ( 3 cos 2 u + 2 cos u sin u − u cos u + 6 u ) − j → d d u ( 3 sin u cos u + 2 sin 2 u − u sin u − 9 u ) − = \overrightarrow {i} \frac {d}{d u} \left(3 \cos^ {2} u + 2 \cos u \sin u - u \cos u + 6 u\right) - \overrightarrow {j} \frac {d}{d u} \left(3 \sin u \cos u + 2 \sin^ {2} u - u \sin u - 9 u\right) - = i d u d ( 3 cos 2 u + 2 cos u sin u − u cos u + 6 u ) − j d u d ( 3 sin u cos u + 2 sin 2 u − u sin u − 9 u ) − − k → d d u ( 6 sin u + 9 cos u ) = - \overrightarrow {k} \frac {d}{d u} (6 \sin u + 9 \cos u) = − k d u d ( 6 sin u + 9 cos u ) = = i → ( 3 ∗ 2 cos u ( − sin u ) + 2 ( − sin u ) sin u + 2 cos u ∗ cos u − cos u − u ∗ ( − sin u ) + 6 ) − = \overrightarrow {i} (3 * 2 \cos u (- \sin u) + 2 (- \sin u) \sin u + 2 \cos u * \cos u - \cos u - u * (- \sin u) + 6) - = i ( 3 ∗ 2 cos u ( − sin u ) + 2 ( − sin u ) sin u + 2 cos u ∗ cos u − cos u − u ∗ ( − sin u ) + 6 ) − − j → ( 3 cos u cos u + 3 sin u ∗ ( − sin u ) + 2 ∗ 2 sin u ∗ cos u − sin u − u cos u − 9 ) − - \overrightarrow {j} (3 \cos u \cos u + 3 \sin u * (- \sin u) + 2 * 2 \sin u * \cos u - \sin u - u \cos u - 9) - − j ( 3 cos u cos u + 3 sin u ∗ ( − sin u ) + 2 ∗ 2 sin u ∗ cos u − sin u − u cos u − 9 ) − − k → ( 6 cos u + 9 ( − sin u ) ) = - \overrightarrow {k} (6 \cos u + 9 (- \sin u)) = − k ( 6 cos u + 9 ( − sin u )) = = i → ( − 6 cos u sin u − 2 sin 2 u + 2 cos 2 u − cos u + u sin u + 6 ) − = \overrightarrow {i} (- 6 \cos u \sin u - 2 \sin^ {2} u + 2 \cos^ {2} u - \cos u + u \sin u + 6) - = i ( − 6 cos u sin u − 2 sin 2 u + 2 cos 2 u − cos u + u sin u + 6 ) − − j → ( 3 cos 2 u − 3 sin 2 u + 4 sin u cos u − sin u − u cos u − 9 ) − k → ( 6 cos u − 9 sin u ) - \overrightarrow {j} (3 \cos^ {2} u - 3 \sin^ {2} u + 4 \sin u \cos u - \sin u - u \cos u - 9) - \overrightarrow {k} (6 \cos u - 9 \sin u) − j ( 3 cos 2 u − 3 sin 2 u + 4 sin u cos u − sin u − u cos u − 9 ) − k ( 6 cos u − 9 sin u ) d d u ( A → × ( B → × C → ) ) ∣ u = 0 = \left. \frac {d}{d u} \left(\overrightarrow {A} \times \left(\overrightarrow {B} \times \overrightarrow {C}\right)\right) \right| _ {u = 0} = d u d ( A × ( B × C ) ) ∣ ∣ u = 0 = = ( i → ( − 6 cos u sin u − 2 sin 2 u + 2 cos 2 u − cos u + u sin u + 6 ) − = \left(\overrightarrow {i} (- 6 \cos u \sin u - 2 \sin^ {2} u + 2 \cos^ {2} u - \cos u + u \sin u + 6) - \right. = ( i ( − 6 cos u sin u − 2 sin 2 u + 2 cos 2 u − cos u + u sin u + 6 ) − − j → ( 3 cos 2 u − 3 sin 2 u + 4 sin u cos u − sin u − u cos u − 9 ) − k → ( 6 cos u − 9 sin u ) ) ∣ u = 0 = - \overrightarrow {j} (3 \cos^ {2} u - 3 \sin^ {2} u + 4 \sin u \cos u - \sin u - u \cos u - 9) - \overrightarrow {k} (6 \cos u - 9 \sin u) \Big) \Big | _ {u = 0} = − j ( 3 cos 2 u − 3 sin 2 u + 4 sin u cos u − sin u − u cos u − 9 ) − k ( 6 cos u − 9 sin u ) ) ∣ ∣ u = 0 = = i → ( − 6 cos 0 sin 0 − 2 sin 2 0 + 2 cos 2 0 − cos 0 + 0 ∗ sin 0 + 6 ) − = \overrightarrow {i} (- 6 \cos 0 \sin 0 - 2 \sin^ {2} 0 + 2 \cos^ {2} 0 - \cos 0 + 0 * \sin 0 + 6) - = i ( − 6 cos 0 sin 0 − 2 sin 2 0 + 2 cos 2 0 − cos 0 + 0 ∗ sin 0 + 6 ) − − j → ( 3 cos 2 0 − 3 sin 2 0 + 4 sin 0 cos 0 − sin 0 − 0 ∗ cos 0 − 9 ) − k → ( 6 cos 0 − 9 sin 0 ) = - \overrightarrow {j} (3 \cos^ {2} 0 - 3 \sin^ {2} 0 + 4 \sin 0 \cos 0 - \sin 0 - 0 * \cos 0 - 9) - \overrightarrow {k} (6 \cos 0 - 9 \sin 0) = − j ( 3 cos 2 0 − 3 sin 2 0 + 4 sin 0 cos 0 − sin 0 − 0 ∗ cos 0 − 9 ) − k ( 6 cos 0 − 9 sin 0 ) = = i → ( 2 − 1 + 6 ) − j → ( 3 − 9 ) − k → ( 6 ) = 7 i → + 6 j → − 6 k → = \overrightarrow {i} (2 - 1 + 6) - \overrightarrow {j} (3 - 9) - \overrightarrow {k} (6) = 7 \overrightarrow {i} + 6 \overrightarrow {j} - 6 \overrightarrow {k} = i ( 2 − 1 + 6 ) − j ( 3 − 9 ) − k ( 6 ) = 7 i + 6 j − 6 k
ANSWER
d d u ( A → × ( B → × C → ) ) ∣ u = 0 = 7 i → + 6 j → − 6 k → \left. \frac {d}{d u} \left(\overrightarrow {A} \times \left(\overrightarrow {B} \times \overrightarrow {C}\right)\right) \right| _ {u = 0} = 7 \overrightarrow {i} + 6 \overrightarrow {j} - 6 \overrightarrow {k} d u d ( A × ( B × C ) ) ∣ ∣ u = 0 = 7 i + 6 j − 6 k
QUESTION 9
Let
A → = x 2 y z i → − 2 x z 3 j → − x z 2 k → and B → = 4 z i → + y j → + 4 x 2 k → \overrightarrow {A} = x ^ {2} y z \overrightarrow {i} - 2 x z ^ {3} \overrightarrow {j} - x z ^ {2} \overrightarrow {k} \quad \text{and} \quad \overrightarrow {B} = 4 z \overrightarrow {i} + y \overrightarrow {j} + 4 x ^ {2} \overrightarrow {k} A = x 2 yz i − 2 x z 3 j − x z 2 k and B = 4 z i + y j + 4 x 2 k
Find
∂ 2 ∂ x ∂ y ( A → × B → ) at ( 1 , 0 , − 2 ) \frac {\partial^ {2}}{\partial x \partial y} \left(\overrightarrow {A} \times \overrightarrow {B}\right) \quad \text{at} \quad (1, 0, - 2) ∂ x ∂ y ∂ 2 ( A × B ) at ( 1 , 0 , − 2 ) SOLUTION
By the definition
A → × B → = ∥ i → j → k → A x A y A z B x B y B z ∥ = i → ( A y B z − A z B y ) − j → ( A x B z − A z B x ) + k → ( A x B y − A y B x ) \overrightarrow {A} \times \overrightarrow {B} = \left\| \begin{array}{c c c} \overrightarrow {i} & \overrightarrow {j} & \overrightarrow {k} \\ A _ {x} & A _ {y} & A _ {z} \\ B _ {x} & B _ {y} & B _ {z} \end{array} \right\| = \overrightarrow {i} \left(A _ {y} B _ {z} - A _ {z} B _ {y}\right) - \overrightarrow {j} \left(A _ {x} B _ {z} - A _ {z} B _ {x}\right) + \overrightarrow {k} \left(A _ {x} B _ {y} - A _ {y} B _ {x}\right) A × B = ∥ ∥ i A x B x j A y B y k A z B z ∥ ∥ = i ( A y B z − A z B y ) − j ( A x B z − A z B x ) + k ( A x B y − A y B x )
In our case
A → = x 2 y z ⏟ A x i → + ( − 2 x z 3 ) ⏟ A y j → + ( − x z 2 ) ⏟ A z k → and B → = 4 z ⏟ B x i → + y ⏟ B y j → + 4 x 2 ⏟ B z k → \overrightarrow {A} = \underbrace {x ^ {2} y z} _ {A _ {x}} \overrightarrow {i} + \underbrace {(- 2 x z ^ {3})} _ {A _ {y}} \overrightarrow {j} + \underbrace {(- x z ^ {2})} _ {A _ {z}} \overrightarrow {k} \quad \text{and} \quad \overrightarrow {B} = \underbrace {4 z} _ {B _ {x}} \overrightarrow {i} + \underbrace {y} _ {B _ {y}} \overrightarrow {j} + \underbrace {4 x ^ {2}} _ {B _ {z}} \overrightarrow {k} A = A x x 2 yz i + A y ( − 2 x z 3 ) j + A z ( − x z 2 ) k and B = B x 4 z i + B y y j + B z 4 x 2 k A → × B → = i → ( − 2 x z 3 ∗ 4 x 2 − ( − x z 2 ) y ) − j → ( x 2 y z ∗ 4 x 2 − ( − x z 2 ) ∗ 4 z ) + k → ( x 2 y z ∗ y − ( − 2 x z 3 ) ∗ 4 z ) = = i → ( x y z 2 − 8 x 3 z 3 ) − j → ( 4 x 4 y z + 4 x z 3 ) + k → ( x 2 y 2 z + 8 x z 4 ) \begin{array}{l} \overrightarrow {A} \times \overrightarrow {B} = \overrightarrow {i} \left(- 2 x z ^ {3} * 4 x ^ {2} - (- x z ^ {2}) y\right) - \overrightarrow {j} \left(x ^ {2} y z * 4 x ^ {2} - (- x z ^ {2}) * 4 z\right) + \overrightarrow {k} \left(x ^ {2} y z * y - (- 2 x z ^ {3}) * 4 z\right) = \\ = \overrightarrow {i} \left(x y z ^ {2} - 8 x ^ {3} z ^ {3}\right) - \overrightarrow {j} \left(4 x ^ {4} y z + 4 x z ^ {3}\right) + \overrightarrow {k} \left(x ^ {2} y ^ {2} z + 8 x z ^ {4}\right) \\ \end{array} A × B = i ( − 2 x z 3 ∗ 4 x 2 − ( − x z 2 ) y ) − j ( x 2 yz ∗ 4 x 2 − ( − x z 2 ) ∗ 4 z ) + k ( x 2 yz ∗ y − ( − 2 x z 3 ) ∗ 4 z ) = = i ( x y z 2 − 8 x 3 z 3 ) − j ( 4 x 4 yz + 4 x z 3 ) + k ( x 2 y 2 z + 8 x z 4 ) ∂ 2 ∂ x ∂ y ( A → × B → ) = ∂ 2 ∂ x ∂ y ( i → ( x y z 2 − 8 x 3 z 3 ) − j → ( 4 x 4 y z + 4 x z 3 ) + k → ( x 2 y 2 z + 8 x z 4 ) ) = = i → ∂ 2 ∂ x ∂ y ( x y z 2 − 8 x 3 z 3 ) − j → ∂ 2 ∂ x ∂ y ( 4 x 4 y z + 4 x z 3 ) + k → ∂ 2 ∂ x ∂ y ( x 2 y 2 z + 8 x z 4 ) = = i → ∂ ∂ x ( x z 2 ) − j → ∂ ∂ x ( 4 x 4 z ) + k → ∂ ∂ x ( 2 x 2 y z ) = i → z 2 − j → 4 ∗ 4 x 3 z + k → 2 ∗ 2 x y z ∂ 2 ∂ x ∂ y ( A → × B → ) ∣ ( 1 , 0 , − 2 ) = ( i → z 2 − j → 4 ∗ 4 x 3 z + k → 2 ∗ 2 x y z ) ∣ ( 1 , 0 , − 2 ) = = i → ∗ ( − 2 ) 2 − j → ∗ 16 ∗ 1 3 ∗ ( − 2 ) + k → ∗ 4 ∗ 1 ∗ 0 ∗ ( − 2 ) = 4 i → + 32 j → ∂ 2 ∂ x ∂ y ( A → × B → ) ∣ ( 1 , 0 , − 2 ) = 4 i → + 32 j → \begin{array}{l} \frac {\partial^ {2}}{\partial x \partial y} \left(\overrightarrow {A} \times \overrightarrow {B}\right) = \frac {\partial^ {2}}{\partial x \partial y} \left(\overrightarrow {i} \left(x y z ^ {2} - 8 x ^ {3} z ^ {3}\right) - \overrightarrow {j} \left(4 x ^ {4} y z + 4 x z ^ {3}\right) + \overrightarrow {k} \left(x ^ {2} y ^ {2} z + 8 x z ^ {4}\right)\right) = \\ = \overrightarrow {i} \frac {\partial^ {2}}{\partial x \partial y} \left(x y z ^ {2} - 8 x ^ {3} z ^ {3}\right) - \overrightarrow {j} \frac {\partial^ {2}}{\partial x \partial y} \left(4 x ^ {4} y z + 4 x z ^ {3}\right) + \overrightarrow {k} \frac {\partial^ {2}}{\partial x \partial y} \left(x ^ {2} y ^ {2} z + 8 x z ^ {4}\right) = \\ = \overrightarrow {i} \frac {\partial}{\partial x} \left(x z ^ {2}\right) - \overrightarrow {j} \frac {\partial}{\partial x} \left(4 x ^ {4} z\right) + \overrightarrow {k} \frac {\partial}{\partial x} \left(2 x ^ {2} y z\right) = \overrightarrow {i} z ^ {2} - \overrightarrow {j} 4 * 4 x ^ {3} z + \overrightarrow {k} 2 * 2 x y z \\ \left. \frac {\partial^ {2}}{\partial x \partial y} \left(\overrightarrow {A} \times \overrightarrow {B}\right) \right| _ {(1, 0, - 2)} = \left. \left(\overrightarrow {i} z ^ {2} - \overrightarrow {j} 4 * 4 x ^ {3} z + \overrightarrow {k} 2 * 2 x y z\right) \right| _ {(1, 0, - 2)} = \\ = \overrightarrow {i} * (- 2) ^ {2} - \overrightarrow {j} * 16 * 1 ^ {3} * (- 2) + \overrightarrow {k} * 4 * 1 * 0 * (- 2) = 4 \overrightarrow {i} + 32 \overrightarrow {j} \\ \left. \frac {\partial^ {2}}{\partial x \partial y} \left(\overrightarrow {A} \times \overrightarrow {B}\right) \right| _ {(1, 0, - 2)} = 4 \overrightarrow {i} + 32 \overrightarrow {j} \\ \end{array} ∂ x ∂ y ∂ 2 ( A × B ) = ∂ x ∂ y ∂ 2 ( i ( x y z 2 − 8 x 3 z 3 ) − j ( 4 x 4 yz + 4 x z 3 ) + k ( x 2 y 2 z + 8 x z 4 ) ) = = i ∂ x ∂ y ∂ 2 ( x y z 2 − 8 x 3 z 3 ) − j ∂ x ∂ y ∂ 2 ( 4 x 4 yz + 4 x z 3 ) + k ∂ x ∂ y ∂ 2 ( x 2 y 2 z + 8 x z 4 ) = = i ∂ x ∂ ( x z 2 ) − j ∂ x ∂ ( 4 x 4 z ) + k ∂ x ∂ ( 2 x 2 yz ) = i z 2 − j 4 ∗ 4 x 3 z + k 2 ∗ 2 x yz ∂ x ∂ y ∂ 2 ( A × B ) ∣ ∣ ( 1 , 0 , − 2 ) = ( i z 2 − j 4 ∗ 4 x 3 z + k 2 ∗ 2 x yz ) ∣ ∣ ( 1 , 0 , − 2 ) = = i ∗ ( − 2 ) 2 − j ∗ 16 ∗ 1 3 ∗ ( − 2 ) + k ∗ 4 ∗ 1 ∗ 0 ∗ ( − 2 ) = 4 i + 32 j ∂ x ∂ y ∂ 2 ( A × B ) ∣ ∣ ( 1 , 0 , − 2 ) = 4 i + 32 j ANSWER
∂ 2 ∂ x ∂ y ( A → × B → ) ∣ ( 1 , 0 , − 2 ) = 4 i → + 32 j → \left. \frac {\partial^ {2}}{\partial x \partial y} \left(\overrightarrow {A} \times \overrightarrow {B}\right) \right| _ {(1, 0, - 2)} = 4 \overrightarrow {i} + 32 \overrightarrow {j} ∂ x ∂ y ∂ 2 ( A × B ) ∣ ∣ ( 1 , 0 , − 2 ) = 4 i + 32 j QUESTION 10
Solve
d 2 A d t 2 − 4 d A d t − 5 A = 0 \frac {d ^ {2} A}{d t ^ {2}} - 4 \frac {d A}{d t} - 5 A = 0 d t 2 d 2 A − 4 d t d A − 5 A = 0
SOLUTION
d 2 A d t 2 − 4 d A d t − 5 A = 0 \frac {d ^ {2} A}{d t ^ {2}} - 4 \frac {d A}{d t} - 5 A = 0 d t 2 d 2 A − 4 d t d A − 5 A = 0
It is ordinary, homogeneous second order differential equation with constant coefficients. The solution of this equation is found in the form (this is known from the theory of differential equations)
A ( t ) = e λ t ⟺ d A d t = λ e λ t ⟺ d 2 A d t 2 = λ 2 e λ t A (t) = e ^ {\lambda t} \Longleftrightarrow \frac {d A}{d t} = \lambda e ^ {\lambda t} \Longleftrightarrow \frac {d ^ {2} A}{d t ^ {2}} = \lambda^ {2} e ^ {\lambda t} A ( t ) = e λ t ⟺ d t d A = λ e λ t ⟺ d t 2 d 2 A = λ 2 e λ t d 2 A d t 2 − 4 d A d t − 5 A = 0 ⟺ λ 2 e λ t − 4 λ e λ t − 5 e λ t = 0 ⟺ e λ t ( λ 2 − 4 λ − 5 ) = 0 \frac {d ^ {2} A}{d t ^ {2}} - 4 \frac {d A}{d t} - 5 A = 0 \Longleftrightarrow \lambda^ {2} e ^ {\lambda t} - 4 \lambda e ^ {\lambda t} - 5 e ^ {\lambda t} = 0 \Longleftrightarrow e ^ {\lambda t} \big (\lambda^ {2} - 4 \lambda - 5 \big) = 0 d t 2 d 2 A − 4 d t d A − 5 A = 0 ⟺ λ 2 e λ t − 4 λ e λ t − 5 e λ t = 0 ⟺ e λ t ( λ 2 − 4 λ − 5 ) = 0 λ 2 − 4 λ − 5 = 0 ⟺ { λ 1 = 5 λ 2 = − 1 \lambda^ {2} - 4 \lambda - 5 = 0 \Longleftrightarrow \left\{ \begin{array}{l l} & \lambda_ {1} = 5 \\ & \lambda_ {2} = - 1 \end{array} \right. λ 2 − 4 λ − 5 = 0 ⟺ { λ 1 = 5 λ 2 = − 1 A ( t ) = C 1 e λ 1 t + C 2 e λ 2 t = C 1 e 5 t + C 2 e − t A (t) = C _ {1} e ^ {\lambda_ {1} t} + C _ {2} e ^ {\lambda_ {2} t} = C _ {1} e ^ {5 t} + C _ {2} e ^ {- t} A ( t ) = C 1 e λ 1 t + C 2 e λ 2 t = C 1 e 5 t + C 2 e − t
ANSWER
A ( t ) = C 1 e 5 t + C 2 e − t A (t) = C _ {1} e ^ {5 t} + C _ {2} e ^ {- t} A ( t ) = C 1 e 5 t + C 2 e − t
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#340153 on Dec 2023