Mixing
Proving this vector calculus equation
$begingroup$
Suppose $vec{a}(t)$ is a time dependent vector and $vec{b}$ is a constant vector. I want to show $$frac{d}{dt}left[vec{a} cdot left(frac{d vec{a}}{dt} times vec{b}right)right] = vec{a} cdot left[frac{d^2vec{a}}{dt^2} times vec {b}right].$$
I'm not sure how to go about doing this. The proof need not be rigorous; I'm just confused on which properties to elicit where. Thanks!
multivariable-calculus vector-analysis
edited Jan 19 at 23:57
jordan_glen
1
asked Jan 19 at 23:52
appleapple
445
$endgroup$
add a comment |
$begingroup$
Suppose $vec{a}(t)$ is a time dependent vector and $vec{b}$ is a constant vector. I want to show $$frac{d}{dt}left[vec{a} cdot left(frac{d vec{a}}{dt} times vec{b}right)right] = vec{a} cdot left[frac{d^2vec{a}}{dt^2} times vec {b}right].$$
I'm not sure how to go about doing this. The proof need not be rigorous; I'm just confused on which properties to elicit where. Thanks!
multivariable-calculus vector-analysis
edited Jan 19 at 23:57
jordan_glen
1
asked Jan 19 at 23:52
appleapple
445
$endgroup$
add a comment |
$begingroup$
Suppose $vec{a}(t)$ is a time dependent vector and $vec{b}$ is a constant vector. I want to show $$frac{d}{dt}left[vec{a} cdot left(frac{d vec{a}}{dt} times vec{b}right)right] = vec{a} cdot left[frac{d^2vec{a}}{dt^2} times vec {b}right].$$
I'm not sure how to go about doing this. The proof need not be rigorous; I'm just confused on which properties to elicit where. Thanks!
multivariable-calculus vector-analysis
edited Jan 19 at 23:57
jordan_glen
1
asked Jan 19 at 23:52
appleapple
445
$endgroup$
Suppose $vec{a}(t)$ is a time dependent vector and $vec{b}$ is a constant vector. I want to show $$frac{d}{dt}left[vec{a} cdot left(frac{d vec{a}}{dt} times vec{b}right)right] = vec{a} cdot left[frac{d^2vec{a}}{dt^2} times vec {b}right].$$
I'm not sure how to go about doing this. The proof need not be rigorous; I'm just confused on which properties to elicit where. Thanks!
multivariable-calculus vector-analysis
multivariable-calculus vector-analysis
edited Jan 19 at 23:57
jordan_glen
1
asked Jan 19 at 23:52
appleapple
445
edited Jan 19 at 23:57
jordan_glen
1
asked Jan 19 at 23:52
appleapple
445
edited Jan 19 at 23:57
jordan_glen
1
edited Jan 19 at 23:57
jordan_glen
1
edited Jan 19 at 23:57
jordan_glen
1
1
asked Jan 19 at 23:52
appleapple
445
asked Jan 19 at 23:52
appleapple
445
asked Jan 19 at 23:52
appleapple
445
445
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Apply the derivative
$$frac{d}{dt} left[vec{a} cdot left( frac{dvec{a}}{dt} times vec{b}right) right] = underbrace{frac{dvec{a}}{dt} cdot left( frac{dvec{a}}{dt} times vec{b}right)}_{0} + vec{a} cdot left( frac{d^2vec{a}}{dt^2} times vec{b}right)$$
since $vec{a},' times vec{b}$ is orthogonal to $vec{a},'.$
answered Jan 20 at 0:00
AEngineerAEngineer
1,5441317
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
Apply the derivative
$$frac{d}{dt} left[vec{a} cdot left( frac{dvec{a}}{dt} times vec{b}right) right] = underbrace{frac{dvec{a}}{dt} cdot left( frac{dvec{a}}{dt} times vec{b}right)}_{0} + vec{a} cdot left( frac{d^2vec{a}}{dt^2} times vec{b}right)$$
since $vec{a},' times vec{b}$ is orthogonal to $vec{a},'.$
answered Jan 20 at 0:00
AEngineerAEngineer
1,5441317
$endgroup$
add a comment |
$begingroup$
Apply the derivative
$$frac{d}{dt} left[vec{a} cdot left( frac{dvec{a}}{dt} times vec{b}right) right] = underbrace{frac{dvec{a}}{dt} cdot left( frac{dvec{a}}{dt} times vec{b}right)}_{0} + vec{a} cdot left( frac{d^2vec{a}}{dt^2} times vec{b}right)$$
since $vec{a},' times vec{b}$ is orthogonal to $vec{a},'.$
answered Jan 20 at 0:00
AEngineerAEngineer
1,5441317
$endgroup$
add a comment |
$begingroup$
Apply the derivative
$$frac{d}{dt} left[vec{a} cdot left( frac{dvec{a}}{dt} times vec{b}right) right] = underbrace{frac{dvec{a}}{dt} cdot left( frac{dvec{a}}{dt} times vec{b}right)}_{0} + vec{a} cdot left( frac{d^2vec{a}}{dt^2} times vec{b}right)$$
since $vec{a},' times vec{b}$ is orthogonal to $vec{a},'.$
answered Jan 20 at 0:00
AEngineerAEngineer
1,5441317
$endgroup$
Apply the derivative
$$frac{d}{dt} left[vec{a} cdot left( frac{dvec{a}}{dt} times vec{b}right) right] = underbrace{frac{dvec{a}}{dt} cdot left( frac{dvec{a}}{dt} times vec{b}right)}_{0} + vec{a} cdot left( frac{d^2vec{a}}{dt^2} times vec{b}right)$$
since $vec{a},' times vec{b}$ is orthogonal to $vec{a},'.$
answered Jan 20 at 0:00
AEngineerAEngineer
1,5441317
answered Jan 20 at 0:00
AEngineerAEngineer
1,5441317
answered Jan 20 at 0:00
AEngineerAEngineer
1,5441317
answered Jan 20 at 0:00
AEngineerAEngineer
1,5441317
1,5441317
add a comment |
add a comment |
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