Mixing
Vector functions and differentiation
$begingroup$
I am confused with the solution to the following question. I'm not sure how to follow the steps from the 2nd line onwards and how it relates to the chain rule. Can someone please break this down for me? Why has ${bf|{a}|}^2 = bf{a}cdot{a}$ been used?
Exercise:
Let a be a vector function. Calculate $frac{d(f({bf{|a|}})}{dt}$.
Solution:
begin{align}
frac{d(f({|bf{a}|})}{dt} &= frac{d}{dt}({|bf{a}|}) f'({|bf{a}|}) \ {|bf{a}|}^2 &= bf{a}cdot{a} quad \ 2{|bf{a}|} frac{d}{dt} ({|bf{a}|}) &= 2{bf{a}cdot{dot{a}}} \ frac{d}{dt}(f({|bf{a}|}) &= frac{{bf{a} cdot{a}}}{{|bf{a}|}} f'({|bf{a}|})
end{align}
calculus derivatives vectors proof-explanation
edited Jan 10 at 17:19
the_fox
2,79221537
$endgroup$
add a comment |
$begingroup$
I am confused with the solution to the following question. I'm not sure how to follow the steps from the 2nd line onwards and how it relates to the chain rule. Can someone please break this down for me? Why has ${bf|{a}|}^2 = bf{a}cdot{a}$ been used?
Exercise:
Let a be a vector function. Calculate $frac{d(f({bf{|a|}})}{dt}$.
Solution:
begin{align}
frac{d(f({|bf{a}|})}{dt} &= frac{d}{dt}({|bf{a}|}) f'({|bf{a}|}) \ {|bf{a}|}^2 &= bf{a}cdot{a} quad \ 2{|bf{a}|} frac{d}{dt} ({|bf{a}|}) &= 2{bf{a}cdot{dot{a}}} \ frac{d}{dt}(f({|bf{a}|}) &= frac{{bf{a} cdot{a}}}{{|bf{a}|}} f'({|bf{a}|})
end{align}
calculus derivatives vectors proof-explanation
edited Jan 10 at 17:19
the_fox
2,79221537
$endgroup$
add a comment |
$begingroup$
I am confused with the solution to the following question. I'm not sure how to follow the steps from the 2nd line onwards and how it relates to the chain rule. Can someone please break this down for me? Why has ${bf|{a}|}^2 = bf{a}cdot{a}$ been used?
Exercise:
Let a be a vector function. Calculate $frac{d(f({bf{|a|}})}{dt}$.
Solution:
begin{align}
frac{d(f({|bf{a}|})}{dt} &= frac{d}{dt}({|bf{a}|}) f'({|bf{a}|}) \ {|bf{a}|}^2 &= bf{a}cdot{a} quad \ 2{|bf{a}|} frac{d}{dt} ({|bf{a}|}) &= 2{bf{a}cdot{dot{a}}} \ frac{d}{dt}(f({|bf{a}|}) &= frac{{bf{a} cdot{a}}}{{|bf{a}|}} f'({|bf{a}|})
end{align}
calculus derivatives vectors proof-explanation
edited Jan 10 at 17:19
the_fox
2,79221537
$endgroup$
I am confused with the solution to the following question. I'm not sure how to follow the steps from the 2nd line onwards and how it relates to the chain rule. Can someone please break this down for me? Why has ${bf|{a}|}^2 = bf{a}cdot{a}$ been used?
Exercise:
Let a be a vector function. Calculate $frac{d(f({bf{|a|}})}{dt}$.
Solution:
begin{align}
frac{d(f({|bf{a}|})}{dt} &= frac{d}{dt}({|bf{a}|}) f'({|bf{a}|}) \ {|bf{a}|}^2 &= bf{a}cdot{a} quad \ 2{|bf{a}|} frac{d}{dt} ({|bf{a}|}) &= 2{bf{a}cdot{dot{a}}} \ frac{d}{dt}(f({|bf{a}|}) &= frac{{bf{a} cdot{a}}}{{|bf{a}|}} f'({|bf{a}|})
end{align}
calculus derivatives vectors proof-explanation
calculus derivatives vectors proof-explanation
edited Jan 10 at 17:19
the_fox
2,79221537
edited Jan 10 at 17:19
the_fox
2,79221537
edited Jan 10 at 17:19
the_fox
2,79221537
edited Jan 10 at 17:19
the_fox
2,79221537
edited Jan 10 at 17:19
the_fox
2,79221537
2,79221537
asked Jan 10 at 16:55
user503154
add a comment |
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3 Answers
3
active
oldest
votes
$begingroup$
Maybe if you expand in components it will help
$$
|{bf a}|^2 = {bf a}cdot {bf a} = sum_k a_k a_k
$$
So that
$$
frac{{rm d}}{{rm d}t}|{bf a}|^2 = frac{{rm d}}{{rm d}t}sum_k a_k a_k = sum_k a_k frac{{rm d}a_k}{{rm d}t} + frac{{rm d}a_k}{{rm d}t}a_k = 2sum_k a_k frac{{rm d}a_k}{{rm d}t} = 2sum_k a_k dot{a}_k = 2 {bf a}cdot dot{bf a}
$$
answered Jan 10 at 17:03
caveraccaverac
14.6k31130
$endgroup$
add a comment |
$begingroup$
The reason why the identity is used is to avoid having to deal directly with the norm function. Instead the product rule is used and the calculation is simpler, as we are taking the derivative of known functions.
We could apply the same idea to compute the derivative of $sqrt{x}$ from scratch by only knowing the derivative of $x$:
Using that (for positive $x$):
$$(sqrt{x})^2=x$$ we apply the derivative on both sides and use the product rule
$$2sqrt{x}cdot frac{d}{dx}(sqrt{x})^=frac{d}{dx}((sqrt{x})^2)=frac{d}{dx}x=1$$
and thus
$$frac{d}{dx}sqrt{x}=frac1{2sqrt{x}}$$
answered Jan 10 at 17:15
b00n heTb00n heT
10.3k12235
$endgroup$
add a comment |
$begingroup$
$|mathbf a|^2=mathbf acdotmathbf a$ has been used because you want to express the final answer in terms of $displaystylemathbf{dot a}=frac{dmathbf a}{dt}$.
Chain rule gives$$frac{df(|mathbf a|)}{dt}=frac{df(|mathbf a|)}{d|mathbf a|}frac{d|mathbf a|}{dt}=f'(|mathbf a|)frac{d|mathbf a|}{dt}$$
$|mathbf a|$ is also a function of $mathbf a$. Use the chain rule to differentiate $|mathbf a|^2=mathbf acdotmathbf a$ with respect to $t$ to get,$$2|mathbf a|frac{d|mathbf a|}{dt}=mathbf acdotfrac{dmathbf a}{dt}+frac{dmathbf a}{dt}cdotmathbf a=2mathbf acdotdot{mathbf a}implies frac{d|mathbf a|}{dt}=frac{mathbf acdotdot{mathbf a}}{|mathbf a|}$$Alternatively, you can also differentiate $|mathbf a|=sqrt{mathbf acdotmathbf a}$ with respect to $t$ to get$$frac{d|mathbf a|}{dt}=frac12cdotfrac{mathbf acdotfrac{dmathbf a}{dt}+frac{dmathbf a}{dt}cdotmathbf a}{sqrt{mathbf acdotmathbf a}}=frac{mathbf acdotdot{mathbf a}}{|mathbf a|}$$
This gives the answer as$$frac{df(|mathbf a|)}{dt}=f'(|mathbf a|)frac{mathbf acdotdot{mathbf a}}{|mathbf a|}$$
answered Jan 10 at 17:13
Shubham JohriShubham Johri
5,102717
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Maybe if you expand in components it will help
$$
|{bf a}|^2 = {bf a}cdot {bf a} = sum_k a_k a_k
$$
So that
$$
frac{{rm d}}{{rm d}t}|{bf a}|^2 = frac{{rm d}}{{rm d}t}sum_k a_k a_k = sum_k a_k frac{{rm d}a_k}{{rm d}t} + frac{{rm d}a_k}{{rm d}t}a_k = 2sum_k a_k frac{{rm d}a_k}{{rm d}t} = 2sum_k a_k dot{a}_k = 2 {bf a}cdot dot{bf a}
$$
answered Jan 10 at 17:03
caveraccaverac
14.6k31130
$endgroup$
add a comment |
$begingroup$
Maybe if you expand in components it will help
$$
|{bf a}|^2 = {bf a}cdot {bf a} = sum_k a_k a_k
$$
So that
$$
frac{{rm d}}{{rm d}t}|{bf a}|^2 = frac{{rm d}}{{rm d}t}sum_k a_k a_k = sum_k a_k frac{{rm d}a_k}{{rm d}t} + frac{{rm d}a_k}{{rm d}t}a_k = 2sum_k a_k frac{{rm d}a_k}{{rm d}t} = 2sum_k a_k dot{a}_k = 2 {bf a}cdot dot{bf a}
$$
answered Jan 10 at 17:03
caveraccaverac
14.6k31130
$endgroup$
add a comment |
$begingroup$
Maybe if you expand in components it will help
$$
|{bf a}|^2 = {bf a}cdot {bf a} = sum_k a_k a_k
$$
So that
$$
frac{{rm d}}{{rm d}t}|{bf a}|^2 = frac{{rm d}}{{rm d}t}sum_k a_k a_k = sum_k a_k frac{{rm d}a_k}{{rm d}t} + frac{{rm d}a_k}{{rm d}t}a_k = 2sum_k a_k frac{{rm d}a_k}{{rm d}t} = 2sum_k a_k dot{a}_k = 2 {bf a}cdot dot{bf a}
$$
answered Jan 10 at 17:03
caveraccaverac
14.6k31130
$endgroup$
Maybe if you expand in components it will help
$$
|{bf a}|^2 = {bf a}cdot {bf a} = sum_k a_k a_k
$$
So that
$$
frac{{rm d}}{{rm d}t}|{bf a}|^2 = frac{{rm d}}{{rm d}t}sum_k a_k a_k = sum_k a_k frac{{rm d}a_k}{{rm d}t} + frac{{rm d}a_k}{{rm d}t}a_k = 2sum_k a_k frac{{rm d}a_k}{{rm d}t} = 2sum_k a_k dot{a}_k = 2 {bf a}cdot dot{bf a}
$$
answered Jan 10 at 17:03
caveraccaverac
14.6k31130
answered Jan 10 at 17:03
caveraccaverac
14.6k31130
answered Jan 10 at 17:03
caveraccaverac
14.6k31130
answered Jan 10 at 17:03
caveraccaverac
14.6k31130
14.6k31130
add a comment |
add a comment |
$begingroup$
The reason why the identity is used is to avoid having to deal directly with the norm function. Instead the product rule is used and the calculation is simpler, as we are taking the derivative of known functions.
We could apply the same idea to compute the derivative of $sqrt{x}$ from scratch by only knowing the derivative of $x$:
Using that (for positive $x$):
$$(sqrt{x})^2=x$$ we apply the derivative on both sides and use the product rule
$$2sqrt{x}cdot frac{d}{dx}(sqrt{x})^=frac{d}{dx}((sqrt{x})^2)=frac{d}{dx}x=1$$
and thus
$$frac{d}{dx}sqrt{x}=frac1{2sqrt{x}}$$
answered Jan 10 at 17:15
b00n heTb00n heT
10.3k12235
$endgroup$
add a comment |
$begingroup$
The reason why the identity is used is to avoid having to deal directly with the norm function. Instead the product rule is used and the calculation is simpler, as we are taking the derivative of known functions.
We could apply the same idea to compute the derivative of $sqrt{x}$ from scratch by only knowing the derivative of $x$:
Using that (for positive $x$):
$$(sqrt{x})^2=x$$ we apply the derivative on both sides and use the product rule
$$2sqrt{x}cdot frac{d}{dx}(sqrt{x})^=frac{d}{dx}((sqrt{x})^2)=frac{d}{dx}x=1$$
and thus
$$frac{d}{dx}sqrt{x}=frac1{2sqrt{x}}$$
answered Jan 10 at 17:15
b00n heTb00n heT
10.3k12235
$endgroup$
add a comment |
$begingroup$
The reason why the identity is used is to avoid having to deal directly with the norm function. Instead the product rule is used and the calculation is simpler, as we are taking the derivative of known functions.
We could apply the same idea to compute the derivative of $sqrt{x}$ from scratch by only knowing the derivative of $x$:
Using that (for positive $x$):
$$(sqrt{x})^2=x$$ we apply the derivative on both sides and use the product rule
$$2sqrt{x}cdot frac{d}{dx}(sqrt{x})^=frac{d}{dx}((sqrt{x})^2)=frac{d}{dx}x=1$$
and thus
$$frac{d}{dx}sqrt{x}=frac1{2sqrt{x}}$$
answered Jan 10 at 17:15
b00n heTb00n heT
10.3k12235
$endgroup$
The reason why the identity is used is to avoid having to deal directly with the norm function. Instead the product rule is used and the calculation is simpler, as we are taking the derivative of known functions.
We could apply the same idea to compute the derivative of $sqrt{x}$ from scratch by only knowing the derivative of $x$:
Using that (for positive $x$):
$$(sqrt{x})^2=x$$ we apply the derivative on both sides and use the product rule
$$2sqrt{x}cdot frac{d}{dx}(sqrt{x})^=frac{d}{dx}((sqrt{x})^2)=frac{d}{dx}x=1$$
and thus
$$frac{d}{dx}sqrt{x}=frac1{2sqrt{x}}$$
answered Jan 10 at 17:15
b00n heTb00n heT
10.3k12235
answered Jan 10 at 17:15
b00n heTb00n heT
10.3k12235
answered Jan 10 at 17:15
b00n heTb00n heT
10.3k12235
answered Jan 10 at 17:15
b00n heTb00n heT
10.3k12235
10.3k12235
add a comment |
add a comment |
$begingroup$
$|mathbf a|^2=mathbf acdotmathbf a$ has been used because you want to express the final answer in terms of $displaystylemathbf{dot a}=frac{dmathbf a}{dt}$.
Chain rule gives$$frac{df(|mathbf a|)}{dt}=frac{df(|mathbf a|)}{d|mathbf a|}frac{d|mathbf a|}{dt}=f'(|mathbf a|)frac{d|mathbf a|}{dt}$$
$|mathbf a|$ is also a function of $mathbf a$. Use the chain rule to differentiate $|mathbf a|^2=mathbf acdotmathbf a$ with respect to $t$ to get,$$2|mathbf a|frac{d|mathbf a|}{dt}=mathbf acdotfrac{dmathbf a}{dt}+frac{dmathbf a}{dt}cdotmathbf a=2mathbf acdotdot{mathbf a}implies frac{d|mathbf a|}{dt}=frac{mathbf acdotdot{mathbf a}}{|mathbf a|}$$Alternatively, you can also differentiate $|mathbf a|=sqrt{mathbf acdotmathbf a}$ with respect to $t$ to get$$frac{d|mathbf a|}{dt}=frac12cdotfrac{mathbf acdotfrac{dmathbf a}{dt}+frac{dmathbf a}{dt}cdotmathbf a}{sqrt{mathbf acdotmathbf a}}=frac{mathbf acdotdot{mathbf a}}{|mathbf a|}$$
This gives the answer as$$frac{df(|mathbf a|)}{dt}=f'(|mathbf a|)frac{mathbf acdotdot{mathbf a}}{|mathbf a|}$$
answered Jan 10 at 17:13
Shubham JohriShubham Johri
5,102717
$endgroup$
add a comment |
$begingroup$
$|mathbf a|^2=mathbf acdotmathbf a$ has been used because you want to express the final answer in terms of $displaystylemathbf{dot a}=frac{dmathbf a}{dt}$.
Chain rule gives$$frac{df(|mathbf a|)}{dt}=frac{df(|mathbf a|)}{d|mathbf a|}frac{d|mathbf a|}{dt}=f'(|mathbf a|)frac{d|mathbf a|}{dt}$$
$|mathbf a|$ is also a function of $mathbf a$. Use the chain rule to differentiate $|mathbf a|^2=mathbf acdotmathbf a$ with respect to $t$ to get,$$2|mathbf a|frac{d|mathbf a|}{dt}=mathbf acdotfrac{dmathbf a}{dt}+frac{dmathbf a}{dt}cdotmathbf a=2mathbf acdotdot{mathbf a}implies frac{d|mathbf a|}{dt}=frac{mathbf acdotdot{mathbf a}}{|mathbf a|}$$Alternatively, you can also differentiate $|mathbf a|=sqrt{mathbf acdotmathbf a}$ with respect to $t$ to get$$frac{d|mathbf a|}{dt}=frac12cdotfrac{mathbf acdotfrac{dmathbf a}{dt}+frac{dmathbf a}{dt}cdotmathbf a}{sqrt{mathbf acdotmathbf a}}=frac{mathbf acdotdot{mathbf a}}{|mathbf a|}$$
This gives the answer as$$frac{df(|mathbf a|)}{dt}=f'(|mathbf a|)frac{mathbf acdotdot{mathbf a}}{|mathbf a|}$$
answered Jan 10 at 17:13
Shubham JohriShubham Johri
5,102717
$endgroup$
add a comment |
$begingroup$
$|mathbf a|^2=mathbf acdotmathbf a$ has been used because you want to express the final answer in terms of $displaystylemathbf{dot a}=frac{dmathbf a}{dt}$.
Chain rule gives$$frac{df(|mathbf a|)}{dt}=frac{df(|mathbf a|)}{d|mathbf a|}frac{d|mathbf a|}{dt}=f'(|mathbf a|)frac{d|mathbf a|}{dt}$$
$|mathbf a|$ is also a function of $mathbf a$. Use the chain rule to differentiate $|mathbf a|^2=mathbf acdotmathbf a$ with respect to $t$ to get,$$2|mathbf a|frac{d|mathbf a|}{dt}=mathbf acdotfrac{dmathbf a}{dt}+frac{dmathbf a}{dt}cdotmathbf a=2mathbf acdotdot{mathbf a}implies frac{d|mathbf a|}{dt}=frac{mathbf acdotdot{mathbf a}}{|mathbf a|}$$Alternatively, you can also differentiate $|mathbf a|=sqrt{mathbf acdotmathbf a}$ with respect to $t$ to get$$frac{d|mathbf a|}{dt}=frac12cdotfrac{mathbf acdotfrac{dmathbf a}{dt}+frac{dmathbf a}{dt}cdotmathbf a}{sqrt{mathbf acdotmathbf a}}=frac{mathbf acdotdot{mathbf a}}{|mathbf a|}$$
This gives the answer as$$frac{df(|mathbf a|)}{dt}=f'(|mathbf a|)frac{mathbf acdotdot{mathbf a}}{|mathbf a|}$$
answered Jan 10 at 17:13
Shubham JohriShubham Johri
5,102717
$endgroup$
$|mathbf a|^2=mathbf acdotmathbf a$ has been used because you want to express the final answer in terms of $displaystylemathbf{dot a}=frac{dmathbf a}{dt}$.
Chain rule gives$$frac{df(|mathbf a|)}{dt}=frac{df(|mathbf a|)}{d|mathbf a|}frac{d|mathbf a|}{dt}=f'(|mathbf a|)frac{d|mathbf a|}{dt}$$
$|mathbf a|$ is also a function of $mathbf a$. Use the chain rule to differentiate $|mathbf a|^2=mathbf acdotmathbf a$ with respect to $t$ to get,$$2|mathbf a|frac{d|mathbf a|}{dt}=mathbf acdotfrac{dmathbf a}{dt}+frac{dmathbf a}{dt}cdotmathbf a=2mathbf acdotdot{mathbf a}implies frac{d|mathbf a|}{dt}=frac{mathbf acdotdot{mathbf a}}{|mathbf a|}$$Alternatively, you can also differentiate $|mathbf a|=sqrt{mathbf acdotmathbf a}$ with respect to $t$ to get$$frac{d|mathbf a|}{dt}=frac12cdotfrac{mathbf acdotfrac{dmathbf a}{dt}+frac{dmathbf a}{dt}cdotmathbf a}{sqrt{mathbf acdotmathbf a}}=frac{mathbf acdotdot{mathbf a}}{|mathbf a|}$$
This gives the answer as$$frac{df(|mathbf a|)}{dt}=f'(|mathbf a|)frac{mathbf acdotdot{mathbf a}}{|mathbf a|}$$
answered Jan 10 at 17:13
Shubham JohriShubham Johri
5,102717
edited Jan 10 at 17:25
answered Jan 10 at 17:13
Shubham JohriShubham Johri
5,102717
answered Jan 10 at 17:13
Shubham JohriShubham Johri
5,102717
answered Jan 10 at 17:13
Shubham JohriShubham Johri
5,102717
5,102717
add a comment |
add a comment |
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