Answer on Question #59910 – Math – Calculus
Question
If A = 5 t 2 + t j − t 3 k A = 5t^2 + tj - t^3k A = 5 t 2 + t j − t 3 k B = sinti-costj B = \text{sinti-costj} B = sinti-costj
Solution
d d t ( A ⃗ ⋅ B ⃗ ) = d A ⃗ d t ⋅ B ⃗ + A ⃗ ⋅ d B ⃗ d t = = ( d ( 5 t 2 ) d t i ^ + d t d t j ^ − d ( t 3 ) d t k ^ ) ⋅ ( sin t i ^ − cos t j ^ ) + ( 5 t 2 i ^ + t j ^ − t 3 k ^ ) ⋅ ( d ( sin t ) d t i ^ + d ( − cos t ) d t j ^ ) = = ( 10 t i ^ + j ^ − 3 t 2 k ^ ) ⋅ ( sin t i ^ − cos t j ^ ) + ( 5 t 2 i ^ + t j ^ − t 3 k ^ ) ⋅ ( cos t i ^ + sin t j ^ ) = 10 t sin t − cos t + ( ( − 3 t 2 ) ⋅ 0 + 5 t 2 cos t + t sin t + ( − t 3 ) ⋅ 0 ) = = 5 t 2 cos t + t sin t + 10 t sin t − cos t = \begin{aligned}
\frac{d}{dt} (\vec{A} \cdot \vec{B}) &= \frac{d\vec{A}}{dt} \cdot \vec{B} + \vec{A} \cdot \frac{d\vec{B}}{dt} = \\
&= \left( \frac{d(5t^2)}{dt} \hat{i} + \frac{dt}{dt} \hat{j} - \frac{d(t^3)}{dt} \hat{k} \right) \cdot (\sin t \hat{i} - \cos t \hat{j}) + \left( 5t^2 \hat{i} + t \hat{j} - t^3 \hat{k} \right) \cdot \left( \frac{d(\sin t)}{dt} \hat{i} + \frac{d(-\cos t)}{dt} \hat{j} \right) = \\
&= (10t \hat{i} + \hat{j} - 3t^2 \hat{k}) \cdot (\sin t \hat{i} - \cos t \hat{j}) + (5t^2 \hat{i} + t \hat{j} - t^3 \hat{k}) \cdot (\cos t \hat{i} + \sin t \hat{j}) \\
&= 10t \sin t - \cos t + \left( (-3t^2) \cdot 0 + 5t^2 \cos t + t \sin t + (-t^3) \cdot 0 \right) = \\
&= 5t^2 \cos t + t \sin t + 10t \sin t - \cos t =
\end{aligned} d t d ( A ⋅ B ) = d t d A ⋅ B + A ⋅ d t d B = = ( d t d ( 5 t 2 ) i ^ + d t d t j ^ − d t d ( t 3 ) k ^ ) ⋅ ( sin t i ^ − cos t j ^ ) + ( 5 t 2 i ^ + t j ^ − t 3 k ^ ) ⋅ ( d t d ( sin t ) i ^ + d t d ( − cos t ) j ^ ) = = ( 10 t i ^ + j ^ − 3 t 2 k ^ ) ⋅ ( sin t i ^ − cos t j ^ ) + ( 5 t 2 i ^ + t j ^ − t 3 k ^ ) ⋅ ( cos t i ^ + sin t j ^ ) = 10 t sin t − cos t + ( ( − 3 t 2 ) ⋅ 0 + 5 t 2 cos t + t sin t + ( − t 3 ) ⋅ 0 ) = = 5 t 2 cos t + t sin t + 10 t sin t − cos t = = ( 5 t 2 − 1 ) cos t + 11 t sin t . = (5t^2 - 1) \cos t + 11t \sin t. = ( 5 t 2 − 1 ) cos t + 11 t sin t .
Answer: ( 5 t 2 − 1 ) cos t + 11 t sin t (5t^2 - 1) \cos t + 11t \sin t ( 5 t 2 − 1 ) cos t + 11 t sin t
Question
If A = 5 t 2 + t j − t 3 k A = 5t^2 + tj - t^3k A = 5 t 2 + t j − t 3 k B = sinti-costj B = \text{sinti-costj} B = sinti-costj
Solution
d d t ( A ⃗ × B ⃗ ) = d A ⃗ d t × B ⃗ + A ⃗ × d B ⃗ d t = = ( d ( 5 t 2 ) d t i ^ + d ( t ) d t j ^ − d ( t 3 ) d t k ^ ) × ( sin t i ^ − cos t j ^ ) + ( 5 t 2 i ^ + t j ^ − t 3 k ^ ) × ( d ( sin t ) d t i ^ + d ( − cos t ) d t j ^ ) = ( 10 t i ^ + j ^ − 3 t 2 k ^ ) × ( sin t i ^ − cos t j ^ ) + ( 5 t 2 i ^ + t j ^ − t 3 k ^ ) × ( cos t i ^ + sin t j ^ ) = ∣ i ^ j ^ k ^ 10 t 1 − 3 t 2 sin t − cos t 0 ∣ + + ∣ i ^ j ^ k ^ 5 t 2 t − t 3 cos t sin t 0 ∣ = ( − 3 t 2 cos t i ^ − 3 t 2 sin t j ^ − ( 10 t cos t + sin t ) k ^ ) + ( t 3 sin t i ^ − t 3 cos t j ^ + ( 5 t 2 sin t − 11 t cos t − sin t ) k ^ ) + ( 5 t 2 sin t − t cos t ) k ^ ) = + ( 5 t 2 sin t − t cos t ) k ^ ) = ( t 3 sin t − 3 t 2 cos t ) i ^ − ( t 3 cos t + 3 t 2 sin t ) j ^ + ( 5 t 2 sin t − 11 t cos t − sin t ) k ^ ) . \begin{aligned}
\frac{d}{dt} (\vec{A} \times \vec{B}) &= \frac{d\vec{A}}{dt} \times \vec{B} + \vec{A} \times \frac{d\vec{B}}{dt} = \\
&= \left( \frac{d(5t^2)}{dt} \hat{i} + \frac{d(t)}{dt} \hat{j} - \frac{d(t^3)}{dt} \hat{k} \right) \times (\sin t \hat{i} - \cos t \hat{j}) + \left( 5t^2 \hat{i} + t \hat{j} - t^3 \hat{k} \right) \times \left( \frac{d(\sin t)}{dt} \hat{i} + \frac{d(-\cos t)}{dt} \hat{j} \right) = \\
(10t \hat{i} + \hat{j} - 3t^2 \hat{k}) \times (\sin t \hat{i} - \cos t \hat{j}) + (5t^2 \hat{i} + t \hat{j} - t^3 \hat{k}) \times (\cos t \hat{i} + \sin t \hat{j}) = \left| \begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
10t & 1 & -3t^2 \\
\sin t & -\cos t & 0
\end{array} \right| + \\
+ \left| \begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
5t^2 & t & -t^3 \\
\cos t & \sin t & 0
\end{array} \right| = (-3t^2 \cos t \hat{i} - 3t^2 \sin t \hat{j} - (10t \cos t + \sin t) \hat{k}) + (t^3 \sin t \hat{i} - t^3 \cos t \hat{j} + (5t^2 \sin t - 11t \cos t - \sin t) \hat{k}) + (5t^2 \sin t - t \cos t) \hat{k}) = \\
+ (5t^2 \sin t - t \cos t) \hat{k}) = (t^3 \sin t - 3t^2 \cos t) \hat{i} - (t^3 \cos t + 3t^2 \sin t) \hat{j} + (5t^2 \sin t - 11t \cos t - \sin t) \hat{k}).
\end{aligned} d t d ( A × B ) ( 10 t i ^ + j ^ − 3 t 2 k ^ ) × ( sin t i ^ − cos t j ^ ) + ( 5 t 2 i ^ + t j ^ − t 3 k ^ ) × ( cos t i ^ + sin t j ^ ) = ∣ ∣ i ^ 10 t sin t j ^ 1 − cos t k ^ − 3 t 2 0 ∣ ∣ + + ∣ ∣ i ^ 5 t 2 cos t j ^ t sin t k ^ − t 3 0 ∣ ∣ = ( − 3 t 2 cos t i ^ − 3 t 2 sin t j ^ − ( 10 t cos t + sin t ) k ^ ) + ( t 3 sin t i ^ − t 3 cos t j ^ + ( 5 t 2 sin t − 11 t cos t − sin t ) k ^ ) + ( 5 t 2 sin t − t cos t ) k ^ ) = + ( 5 t 2 sin t − t cos t ) k ^ ) = ( t 3 sin t − 3 t 2 cos t ) i ^ − ( t 3 cos t + 3 t 2 sin t ) j ^ + ( 5 t 2 sin t − 11 t cos t − sin t ) k ^ ) . = d t d A × B + A × d t d B = = ( d t d ( 5 t 2 ) i ^ + d t d ( t ) j ^ − d t d ( t 3 ) k ^ ) × ( sin t i ^ − cos t j ^ ) + ( 5 t 2 i ^ + t j ^ − t 3 k ^ ) × ( d t d ( sin t ) i ^ + d t d ( − cos t ) j ^ ) =
Answer: ( t 3 sin t − 3 t 2 cos t ) i ^ − ( t 3 cos t + 3 t 2 sin t ) j ^ + ( 5 t 2 sin t − 11 t cos t − sin t ) k ^ (t^3 \sin t - 3t^2 \cos t) \hat{i} - (t^3 \cos t + 3t^2 \sin t) \hat{j} + (5t^2 \sin t - 11t \cos t - \sin t) \hat{k} ( t 3 sin t − 3 t 2 cos t ) i ^ − ( t 3 cos t + 3 t 2 sin t ) j ^ + ( 5 t 2 sin t − 11 t cos t − sin t ) k ^
Question
If A = 5 t 2 + t j − t 3 k A = 5t^2 + tj - t^3k A = 5 t 2 + t j − t 3 k B = sinti-costj B = \text{sinti-costj} B = sinti-costj
Solution
d d t ( A ⃗ ⋅ A ⃗ ) = d A ⃗ d t ⋅ A ⃗ + A ⃗ ⋅ d A ⃗ d t = 2 A ⃗ ⋅ d A ⃗ d t = = 2 ( 5 t 2 i ^ + t j ^ − t 3 k ^ ) ⋅ ( d ( 5 t 2 ) d t i ^ + d ( t ) d t j ^ − d ( t 3 ) d t k ^ ) = 2 ( 5 t 2 i ^ + t j ^ − t 3 k ^ ) ⋅ ( 10 t i ^ + j ^ − 3 t 2 k ^ ) = = 2 ( 5 t 2 ⋅ 10 t + t ⋅ 1 − t 3 ⋅ ( − 3 t 2 ) ) = 2 ( 50 t 3 + t + 3 t 5 ) = 6 t 5 + 100 t 3 + 2 t . \begin{array}{l}
\frac{d}{dt} (\vec{A} \cdot \vec{A}) = \frac{d\vec{A}}{dt} \cdot \vec{A} + \vec{A} \cdot \frac{d\vec{A}}{dt} = 2\vec{A} \cdot \frac{d\vec{A}}{dt} = \\
= 2 (5t^2 \hat{i} + t \hat{j} - t^3 \hat{k}) \cdot \left( \frac{d(5t^2)}{dt} \hat{i} + \frac{d(t)}{dt} \hat{j} - \frac{d(t^3)}{dt} \hat{k} \right) = 2 (5t^2 \hat{i} + t \hat{j} - t^3 \hat{k}) \cdot (10t \hat{i} + \hat{j} - 3t^2 \hat{k}) = \\
= 2 (5t^2 \cdot 10t + t \cdot 1 - t^3 \cdot (-3t^2)) = 2(50t^3 + t + 3t^5) = 6t^5 + 100t^3 + 2t.
\end{array} d t d ( A ⋅ A ) = d t d A ⋅ A + A ⋅ d t d A = 2 A ⋅ d t d A = = 2 ( 5 t 2 i ^ + t j ^ − t 3 k ^ ) ⋅ ( d t d ( 5 t 2 ) i ^ + d t d ( t ) j ^ − d t d ( t 3 ) k ^ ) = 2 ( 5 t 2 i ^ + t j ^ − t 3 k ^ ) ⋅ ( 10 t i ^ + j ^ − 3 t 2 k ^ ) = = 2 ( 5 t 2 ⋅ 10 t + t ⋅ 1 − t 3 ⋅ ( − 3 t 2 )) = 2 ( 50 t 3 + t + 3 t 5 ) = 6 t 5 + 100 t 3 + 2 t .
Answer: 6 t 5 + 100 t 3 + 2 t 6t^5 + 100t^3 + 2t 6 t 5 + 100 t 3 + 2 t
Question
If A = sin u i + cos u j + u k A = \sin u_i + \cos u_j + uk A = sin u i + cos u j + u k B = cos u i − sin u j − 3 k B = \cos u_i - \sin u_j - 3k B = cos u i − sin u j − 3 k C = 2 i + 3 j − k C = 2i + 3j - k C = 2 i + 3 j − k ddu ( A × ( B × C ) ) \text{ddu}(A \times (B \times C)) ddu ( A × ( B × C )) u = 0 u = 0 u = 0
Solution
( B ⃗ × C ⃗ ) = ( cos u i ^ − sin u j ^ − 3 k ^ ) × ( 2 i ^ + 3 j ^ − k ^ ) = ∣ i ^ j ^ k ^ cos u − sin u − 3 2 3 − 1 ∣ = ( 9 + sin u ) i ^ + + ( − 6 + cos u ) j ^ + ( 3 cos u + 2 sin u ) k ^ ; \begin{array}{l}
\left(\vec{B} \times \vec{C}\right) = \left(\cos u \hat{i} - \sin u \hat{j} - 3 \hat{k}\right) \times \left(2 \hat{i} + 3 \hat{j} - \hat{k}\right) = \left| \begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
\cos u & -\sin u & -3 \\
2 & 3 & -1
\end{array} \right| = (9 + \sin u) \hat{i} + \\
+ (-6 + \cos u) \hat{j} + (3 \cos u + 2 \sin u) \hat{k};
\end{array} ( B × C ) = ( cos u i ^ − sin u j ^ − 3 k ^ ) × ( 2 i ^ + 3 j ^ − k ^ ) = ∣ ∣ i ^ cos u 2 j ^ − sin u 3 k ^ − 3 − 1 ∣ ∣ = ( 9 + sin u ) i ^ + + ( − 6 + cos u ) j ^ + ( 3 cos u + 2 sin u ) k ^ ; A ⃗ × ( B ⃗ × C ⃗ ) = ( sin u i ^ + cos u j ^ + u k ^ ) × ( ( 9 + sin u ) i ^ + ( − 6 + cos u ) j ^ + ( 3 cos u + 2 sin u ) k ^ ) = = ∣ i ^ j ^ k ^ sin u cos u u 9 + sin u − 6 + cos u 3 cos u + 2 sin u ∣ = ( 6 u − u cos u + 3 cos 2 u + 2 cos u sin u ) i ^ + + ( 9 u + u sin u − 2 sin 2 u − 3 cos u sin u ) j ^ + ( − 9 cos u − 6 sin u ) k ^ ; \begin{array}{l}
\vec{A} \times (\vec{B} \times \vec{C}) = (\sin u \hat{i} + \cos u \hat{j} + u \hat{k}) \times ((9 + \sin u) \hat{i} + (-6 + \cos u) \hat{j} + (3 \cos u + 2 \sin u) \hat{k}) = \\
= \left| \begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
\sin u & \cos u & u \\
9 + \sin u & -6 + \cos u & 3 \cos u + 2 \sin u
\end{array} \right| = (6u - u \cos u + 3 \cos^2 u + 2 \cos u \sin u) \hat{i} + \\
+ (9u + u \sin u - 2 \sin^2 u - 3 \cos u \sin u) \hat{j} + (-9 \cos u - 6 \sin u) \hat{k};
\end{array} A × ( B × C ) = ( sin u i ^ + cos u j ^ + u k ^ ) × (( 9 + sin u ) i ^ + ( − 6 + cos u ) j ^ + ( 3 cos u + 2 sin u ) k ^ ) = = ∣ ∣ i ^ sin u 9 + sin u j ^ cos u − 6 + cos u k ^ u 3 cos u + 2 sin u ∣ ∣ = ( 6 u − u cos u + 3 cos 2 u + 2 cos u sin u ) i ^ + + ( 9 u + u sin u − 2 sin 2 u − 3 cos u sin u ) j ^ + ( − 9 cos u − 6 sin u ) k ^ ; d d u A ⃗ × ( B ⃗ × C ⃗ ) = ( 6 − cos u + u sin u + 2 cos 2 u − 3 sin 2 u ) i ^ + ( 9 + sin u + u cos u − 3 cos 2 u − 2 sin 2 u ) j ^ + ( 9 sin u − 6 cos u ) k ^ ; \begin{array}{l}
\frac{d}{du} \vec{A} \times (\vec{B} \times \vec{C}) = (6 - \cos u + u \sin u + 2 \cos 2u - 3 \sin 2u) \hat{i} + (9 + \sin u + u \cos u - 3 \cos 2u - \\
2 \sin 2u) \hat{j} + (9 \sin u - 6 \cos u) \hat{k};
\end{array} d u d A × ( B × C ) = ( 6 − cos u + u sin u + 2 cos 2 u − 3 sin 2 u ) i ^ + ( 9 + sin u + u cos u − 3 cos 2 u − 2 sin 2 u ) j ^ + ( 9 sin u − 6 cos u ) k ^ ; ( d d u A ⃗ × ( B ⃗ × C ⃗ ) ) u = 0 = ( 6 − 1 + 0 + 2 − 0 ) i ^ + ( 9 + 0 + 0 − 3 − 0 ) j ^ + ( 0 − 6 ) k ^ = 7 i ^ + 6 j ^ − 6 k ^ . \left(\frac{d}{du} \vec{A} \times (\vec{B} \times \vec{C})\right)_{u=0} = (6 - 1 + 0 + 2 - 0) \hat{i} + (9 + 0 + 0 - 3 - 0) \hat{j} + (0 - 6) \hat{k} = 7 \hat{i} + 6 \hat{j} - 6 \hat{k}. ( d u d A × ( B × C ) ) u = 0 = ( 6 − 1 + 0 + 2 − 0 ) i ^ + ( 9 + 0 + 0 − 3 − 0 ) j ^ + ( 0 − 6 ) k ^ = 7 i ^ + 6 j ^ − 6 k ^ .
Answer: 7 i ^ + 6 j ^ − 6 k ^ 7 \hat{i} + 6 \hat{j} - 6 \hat{k} 7 i ^ + 6 j ^ − 6 k ^
Question
Let A = x 2 y 2 z 2 − 2 x 2 z 3 j ^ − x z 2 A = x^2 y^2 z^2 - 2x^2 z^3 \hat{j} - xz^2 A = x 2 y 2 z 2 − 2 x 2 z 3 j ^ − x z 2 B = 4 z 2 + y 2 z 3 + 4 x 2 z 3 k ^ B = 4z^2 + y^2 z^3 + 4x^2 z^3 \hat{k} B = 4 z 2 + y 2 z 3 + 4 x 2 z 3 k ^ ∂ 2 ∂ x ∂ y ( A × B ) \partial^2 \partial x \partial y (A \times B) ∂ 2 ∂ x ∂ y ( A × B ) ( 1 , 0 , − 2 ) (1, 0, -2) ( 1 , 0 , − 2 )
Solution
( A ⃗ × B ⃗ ) = ( x 2 y z i ^ − 2 x z 3 j ^ − x z 2 k ^ ) × ( 4 z i ^ + y j ^ + 4 x 2 k ^ ) = ∣ i ^ j ^ k ^ x 2 y z − 2 x z 3 − x z 2 4 z y 4 x 2 ∣ = = ( x y z 2 − 8 x 3 z 3 ) i ^ + ( − 4 x 4 y z − 4 x z 3 ) j ^ + ( x 2 y 2 z + 8 x z 4 ) k ^ ; \begin{array}{l}
(\vec{A} \times \vec{B}) = (x^2 y z \hat{i} - 2 x z^3 \hat{j} - x z^2 \hat{k}) \times (4 z \hat{i} + y \hat{j} + 4 x^2 \hat{k}) = \left| \begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
x^2 y z & -2 x z^3 & -x z^2 \\
4 z & y & 4 x^2
\end{array} \right| = \\
= (x y z^2 - 8 x^3 z^3) \hat{i} + (-4 x^4 y z - 4 x z^3) \hat{j} + (x^2 y^2 z + 8 x z^4) \hat{k};
\end{array} ( A × B ) = ( x 2 yz i ^ − 2 x z 3 j ^ − x z 2 k ^ ) × ( 4 z i ^ + y j ^ + 4 x 2 k ^ ) = ∣ ∣ i ^ x 2 yz 4 z j ^ − 2 x z 3 y k ^ − x z 2 4 x 2 ∣ ∣ = = ( x y z 2 − 8 x 3 z 3 ) i ^ + ( − 4 x 4 yz − 4 x z 3 ) j ^ + ( x 2 y 2 z + 8 x z 4 ) k ^ ; ∂ ∂ y ( A ⃗ × B ⃗ ) = ( x z 2 ) i ^ + ( − 4 x 4 z ) j ^ + ( 2 x 2 y z ) k ^ ; ∂ 2 ∂ x ∂ y ( A ⃗ × B ⃗ ) = z 2 i ^ − 16 x 3 z j ^ + 4 x y z k ^ ; \begin{array}{l}
\frac{\partial}{\partial y} (\vec{A} \times \vec{B}) = (x z^2) \hat{i} + (-4 x^4 z) \hat{j} + (2 x^2 y z) \hat{k}; \\
\frac{\partial^2}{\partial x \partial y} (\vec{A} \times \vec{B}) = z^2 \hat{i} - 16 x^3 z \hat{j} + 4 x y z \hat{k};
\end{array} ∂ y ∂ ( A × B ) = ( x z 2 ) i ^ + ( − 4 x 4 z ) j ^ + ( 2 x 2 yz ) k ^ ; ∂ x ∂ y ∂ 2 ( A × B ) = z 2 i ^ − 16 x 3 z j ^ + 4 x yz k ^ ; ( ∂ 2 ∂ x ∂ y ( A ⃗ × B ⃗ ) ) ( 1 , 0 , − 2 ) = 4 i ^ + 32 j ^ + 0 k ^ = 4 i ^ + 32 j ^ . \left(\frac {\partial^ {2}}{\partial x \partial y} (\vec {A} \times \vec {B})\right) _ {(1, 0, - 2)} = 4 \hat {i} + 3 2 \hat {j} + 0 \hat {k} = 4 \hat {i} + 3 2 \hat {j}. ( ∂ x ∂ y ∂ 2 ( A × B ) ) ( 1 , 0 , − 2 ) = 4 i ^ + 32 j ^ + 0 k ^ = 4 i ^ + 32 j ^ .
Answer: 4 i ^ + 32 j ^ 4\hat{i} + 32\hat{j} 4 i ^ + 32 j ^
Question
Solve d2Adt2~4dAdt~5A=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 ; A = e λ t ; A = e ^ {\lambda t}; A = e λ t ; λ 2 e λ t − 4 λ e λ t − 5 e λ t = 0. \lambda^ {2} e ^ {\lambda t} - 4 \lambda e ^ {\lambda t} - 5 e ^ {\lambda t} = 0. λ 2 e λ t − 4 λ e λ t − 5 e λ t = 0.
Dividing by e λ t ≠ 0 e^{\lambda t} \neq 0 e λ t = 0
λ 2 − 4 λ − 5 = 0 → λ = − 1 or λ = 5. \lambda^ {2} - 4 \lambda - 5 = 0 \rightarrow \lambda = - 1 \text{ or } \lambda = 5. λ 2 − 4 λ − 5 = 0 → λ = − 1 or λ = 5. A = C 1 e − t + C 2 e 5 t A = C_{1}e^{-t} + C_{2}e^{5t} A = C 1 e − t + C 2 e 5 t C 1 , C 2 C_1, C_2 C 1 , C 2
Answer: A = C 1 e − t + C 2 e 5 t A = C_1e^{-t} + C_2e^{5t} A = C 1 e − t + C 2 e 5 t
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#339914 on Dec 2023