Mixing
Differentiation the form $frac{dy}{dt}$ with respect to y.
$begingroup$
It's been awhile since I took differential equations, so I am unsure if my manipulation is correct. Isn't it true that if we set $h(t,y)=y,$ then $h_{ty}=h_{yt}$? This would imply $$dfrac{partial}{partial y}dfrac{partial y}{partial t}=dfrac{partial}{partial t}dfrac{partial y}{partial y}=dfrac{partial}{partial t}0=0.$$
calculus pde partial-derivative
asked Jan 13 at 23:03
MelodyMelody
80012
$endgroup$
add a comment |
$begingroup$
It's been awhile since I took differential equations, so I am unsure if my manipulation is correct. Isn't it true that if we set $h(t,y)=y,$ then $h_{ty}=h_{yt}$? This would imply $$dfrac{partial}{partial y}dfrac{partial y}{partial t}=dfrac{partial}{partial t}dfrac{partial y}{partial y}=dfrac{partial}{partial t}0=0.$$
calculus pde partial-derivative
asked Jan 13 at 23:03
MelodyMelody
80012
$endgroup$
1
$begingroup$
I'm not sure that I understand your notation. Is y a function? Is it a function of t? Is $h_{ty}$ the derivative of h with respect to t and y? Is $h_{yt}$ the derivative of h with respect to y and t? With your notation, what would the definition of $h_y$ be?
$endgroup$
– NicNic8
Jan 13 at 23:06
$begingroup$
The variable $y$ is function of $t$, and $t$ alone. The notation $h_{ty}$ is $dfrac{partial }{partial t}dfrac{partial }{partial y}h$. Similarly for $h_{yt},$ we just swap the order of the differential operators.
$endgroup$
– Melody
Jan 13 at 23:18
1
$begingroup$
What is given in the problem? Is $h(t,y)=y$ and $h_{ty}=h_{yt}$ both given?
$endgroup$
– NicNic8
Jan 13 at 23:51
$begingroup$
@NicNic8 Setting $h(t,y)=y$ was for convenience. We're not given that $h_{ty}=h_{yt},$ I just thought that the mixed partials were always equivalent. It's been years since I learned this stuff. We're given that $f=dy/dt$ and $f$ is smooth in $y$ as well as $t$. We're also given that $y$ is smooth. No other details are given.
$endgroup$
– Melody
Jan 14 at 0:50
1
$begingroup$
My attempt would be to go via the inverse function, $frac{dy}{dt} = frac{1}{frac{dt}{dy}}$, and to differentiate that with respect to $y$.
$endgroup$
– Christoph
Jan 14 at 2:39
add a comment |
$begingroup$
It's been awhile since I took differential equations, so I am unsure if my manipulation is correct. Isn't it true that if we set $h(t,y)=y,$ then $h_{ty}=h_{yt}$? This would imply $$dfrac{partial}{partial y}dfrac{partial y}{partial t}=dfrac{partial}{partial t}dfrac{partial y}{partial y}=dfrac{partial}{partial t}0=0.$$
calculus pde partial-derivative
asked Jan 13 at 23:03
MelodyMelody
80012
$endgroup$
It's been awhile since I took differential equations, so I am unsure if my manipulation is correct. Isn't it true that if we set $h(t,y)=y,$ then $h_{ty}=h_{yt}$? This would imply $$dfrac{partial}{partial y}dfrac{partial y}{partial t}=dfrac{partial}{partial t}dfrac{partial y}{partial y}=dfrac{partial}{partial t}0=0.$$
calculus pde partial-derivative
calculus pde partial-derivative
asked Jan 13 at 23:03
MelodyMelody
80012
asked Jan 13 at 23:03
MelodyMelody
80012
asked Jan 13 at 23:03
MelodyMelody
80012
asked Jan 13 at 23:03
MelodyMelody
80012
asked Jan 13 at 23:03
MelodyMelody
80012
80012
1
$begingroup$
I'm not sure that I understand your notation. Is y a function? Is it a function of t? Is $h_{ty}$ the derivative of h with respect to t and y? Is $h_{yt}$ the derivative of h with respect to y and t? With your notation, what would the definition of $h_y$ be?
$endgroup$
– NicNic8
Jan 13 at 23:06
$begingroup$
The variable $y$ is function of $t$, and $t$ alone. The notation $h_{ty}$ is $dfrac{partial }{partial t}dfrac{partial }{partial y}h$. Similarly for $h_{yt},$ we just swap the order of the differential operators.
$endgroup$
– Melody
Jan 13 at 23:18
1
$begingroup$
What is given in the problem? Is $h(t,y)=y$ and $h_{ty}=h_{yt}$ both given?
$endgroup$
– NicNic8
Jan 13 at 23:51
$begingroup$
@NicNic8 Setting $h(t,y)=y$ was for convenience. We're not given that $h_{ty}=h_{yt},$ I just thought that the mixed partials were always equivalent. It's been years since I learned this stuff. We're given that $f=dy/dt$ and $f$ is smooth in $y$ as well as $t$. We're also given that $y$ is smooth. No other details are given.
$endgroup$
– Melody
Jan 14 at 0:50
1
$begingroup$
My attempt would be to go via the inverse function, $frac{dy}{dt} = frac{1}{frac{dt}{dy}}$, and to differentiate that with respect to $y$.
$endgroup$
– Christoph
Jan 14 at 2:39
add a comment |
1
$begingroup$
I'm not sure that I understand your notation. Is y a function? Is it a function of t? Is $h_{ty}$ the derivative of h with respect to t and y? Is $h_{yt}$ the derivative of h with respect to y and t? With your notation, what would the definition of $h_y$ be?
$endgroup$
– NicNic8
Jan 13 at 23:06
$begingroup$
The variable $y$ is function of $t$, and $t$ alone. The notation $h_{ty}$ is $dfrac{partial }{partial t}dfrac{partial }{partial y}h$. Similarly for $h_{yt},$ we just swap the order of the differential operators.
$endgroup$
– Melody
Jan 13 at 23:18
1
$begingroup$
What is given in the problem? Is $h(t,y)=y$ and $h_{ty}=h_{yt}$ both given?
$endgroup$
– NicNic8
Jan 13 at 23:51
$begingroup$
@NicNic8 Setting $h(t,y)=y$ was for convenience. We're not given that $h_{ty}=h_{yt},$ I just thought that the mixed partials were always equivalent. It's been years since I learned this stuff. We're given that $f=dy/dt$ and $f$ is smooth in $y$ as well as $t$. We're also given that $y$ is smooth. No other details are given.
$endgroup$
– Melody
Jan 14 at 0:50
1
$begingroup$
My attempt would be to go via the inverse function, $frac{dy}{dt} = frac{1}{frac{dt}{dy}}$, and to differentiate that with respect to $y$.
$endgroup$
– Christoph
Jan 14 at 2:39
1
1
$begingroup$
I'm not sure that I understand your notation. Is y a function? Is it a function of t? Is $h_{ty}$ the derivative of h with respect to t and y? Is $h_{yt}$ the derivative of h with respect to y and t? With your notation, what would the definition of $h_y$ be?
$endgroup$
– NicNic8
Jan 13 at 23:06
$begingroup$
I'm not sure that I understand your notation. Is y a function? Is it a function of t? Is $h_{ty}$ the derivative of h with respect to t and y? Is $h_{yt}$ the derivative of h with respect to y and t? With your notation, what would the definition of $h_y$ be?
$endgroup$
– NicNic8
Jan 13 at 23:06
$begingroup$
The variable $y$ is function of $t$, and $t$ alone. The notation $h_{ty}$ is $dfrac{partial }{partial t}dfrac{partial }{partial y}h$. Similarly for $h_{yt},$ we just swap the order of the differential operators.
$endgroup$
– Melody
Jan 13 at 23:18
$begingroup$
The variable $y$ is function of $t$, and $t$ alone. The notation $h_{ty}$ is $dfrac{partial }{partial t}dfrac{partial }{partial y}h$. Similarly for $h_{yt},$ we just swap the order of the differential operators.
$endgroup$
– Melody
Jan 13 at 23:18
1
1
$begingroup$
What is given in the problem? Is $h(t,y)=y$ and $h_{ty}=h_{yt}$ both given?
$endgroup$
– NicNic8
Jan 13 at 23:51
$begingroup$
What is given in the problem? Is $h(t,y)=y$ and $h_{ty}=h_{yt}$ both given?
$endgroup$
– NicNic8
Jan 13 at 23:51
$begingroup$
@NicNic8 Setting $h(t,y)=y$ was for convenience. We're not given that $h_{ty}=h_{yt},$ I just thought that the mixed partials were always equivalent. It's been years since I learned this stuff. We're given that $f=dy/dt$ and $f$ is smooth in $y$ as well as $t$. We're also given that $y$ is smooth. No other details are given.
$endgroup$
– Melody
Jan 14 at 0:50
$begingroup$
@NicNic8 Setting $h(t,y)=y$ was for convenience. We're not given that $h_{ty}=h_{yt},$ I just thought that the mixed partials were always equivalent. It's been years since I learned this stuff. We're given that $f=dy/dt$ and $f$ is smooth in $y$ as well as $t$. We're also given that $y$ is smooth. No other details are given.
$endgroup$
– Melody
Jan 14 at 0:50
1
1
$begingroup$
My attempt would be to go via the inverse function, $frac{dy}{dt} = frac{1}{frac{dt}{dy}}$, and to differentiate that with respect to $y$.
$endgroup$
– Christoph
Jan 14 at 2:39
$begingroup$
My attempt would be to go via the inverse function, $frac{dy}{dt} = frac{1}{frac{dt}{dy}}$, and to differentiate that with respect to $y$.
$endgroup$
– Christoph
Jan 14 at 2:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Without following your calculus, a straightforward method is :
$$frac{d}{dy}left(frac{dy}{dt}right)=frac{d}{dy}left(frac{1}{frac{dt}{dy}}right)=-frac{frac{d^2t}{dy^2}}{left(frac{dt}{dy}right)^2}$$
because to differentiate wrt $y$ we have to define which function of $y$ (not of $t$) must be considered. Thus it is not $frac{dy}{dt}$ because this is a function of $t$. It is $left(frac{dt}{dy}right)^{-1}$ which is a function of $y$ as required.
answered Jan 14 at 6:28
JJacquelinJJacquelin
43.9k21853
$endgroup$
add a comment |
$begingroup$
Setting h(t,y) to y appears bogus.
When y is a function of t, one usually writes y = y(t).
Then $y_y = 1$, not the zero you claim, though $y_{yt}$ = 0.
$y_t = y'$ since y is a function of t.
$y_{ty} = y'_y$ doesn't happen because y' is not a function of y.
answered Jan 14 at 0:00
William ElliotWilliam Elliot
8,1212720
$endgroup$
1
$begingroup$
Why is it bogus to write $h(t,y)=y$? I see the mistake of setting $y_y=0.$ If $y'$ is not a function of $y$, then how do you differentiate it with respect to y? My professor claims we can do this for homework.
$endgroup$
– Melody
Jan 14 at 0:44
$begingroup$
@Melody. h = y + 0t.
$endgroup$
– William Elliot
Jan 14 at 5:51
$begingroup$
@Melody. d/dy dy/dt = d$^2$y/dt$^2$ × dt/dy.
$endgroup$
– William Elliot
Jan 14 at 6:02
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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$begingroup$
Without following your calculus, a straightforward method is :
$$frac{d}{dy}left(frac{dy}{dt}right)=frac{d}{dy}left(frac{1}{frac{dt}{dy}}right)=-frac{frac{d^2t}{dy^2}}{left(frac{dt}{dy}right)^2}$$
because to differentiate wrt $y$ we have to define which function of $y$ (not of $t$) must be considered. Thus it is not $frac{dy}{dt}$ because this is a function of $t$. It is $left(frac{dt}{dy}right)^{-1}$ which is a function of $y$ as required.
answered Jan 14 at 6:28
JJacquelinJJacquelin
43.9k21853
$endgroup$
add a comment |
$begingroup$
Without following your calculus, a straightforward method is :
$$frac{d}{dy}left(frac{dy}{dt}right)=frac{d}{dy}left(frac{1}{frac{dt}{dy}}right)=-frac{frac{d^2t}{dy^2}}{left(frac{dt}{dy}right)^2}$$
because to differentiate wrt $y$ we have to define which function of $y$ (not of $t$) must be considered. Thus it is not $frac{dy}{dt}$ because this is a function of $t$. It is $left(frac{dt}{dy}right)^{-1}$ which is a function of $y$ as required.
answered Jan 14 at 6:28
JJacquelinJJacquelin
43.9k21853
$endgroup$
add a comment |
$begingroup$
Without following your calculus, a straightforward method is :
$$frac{d}{dy}left(frac{dy}{dt}right)=frac{d}{dy}left(frac{1}{frac{dt}{dy}}right)=-frac{frac{d^2t}{dy^2}}{left(frac{dt}{dy}right)^2}$$
because to differentiate wrt $y$ we have to define which function of $y$ (not of $t$) must be considered. Thus it is not $frac{dy}{dt}$ because this is a function of $t$. It is $left(frac{dt}{dy}right)^{-1}$ which is a function of $y$ as required.
answered Jan 14 at 6:28
JJacquelinJJacquelin
43.9k21853
$endgroup$
Without following your calculus, a straightforward method is :
$$frac{d}{dy}left(frac{dy}{dt}right)=frac{d}{dy}left(frac{1}{frac{dt}{dy}}right)=-frac{frac{d^2t}{dy^2}}{left(frac{dt}{dy}right)^2}$$
because to differentiate wrt $y$ we have to define which function of $y$ (not of $t$) must be considered. Thus it is not $frac{dy}{dt}$ because this is a function of $t$. It is $left(frac{dt}{dy}right)^{-1}$ which is a function of $y$ as required.
answered Jan 14 at 6:28
JJacquelinJJacquelin
43.9k21853
answered Jan 14 at 6:28
JJacquelinJJacquelin
43.9k21853
answered Jan 14 at 6:28
JJacquelinJJacquelin
43.9k21853
answered Jan 14 at 6:28
JJacquelinJJacquelin
43.9k21853
43.9k21853
add a comment |
add a comment |
$begingroup$
Setting h(t,y) to y appears bogus.
When y is a function of t, one usually writes y = y(t).
Then $y_y = 1$, not the zero you claim, though $y_{yt}$ = 0.
$y_t = y'$ since y is a function of t.
$y_{ty} = y'_y$ doesn't happen because y' is not a function of y.
answered Jan 14 at 0:00
William ElliotWilliam Elliot
8,1212720
$endgroup$
1
$begingroup$
Why is it bogus to write $h(t,y)=y$? I see the mistake of setting $y_y=0.$ If $y'$ is not a function of $y$, then how do you differentiate it with respect to y? My professor claims we can do this for homework.
$endgroup$
– Melody
Jan 14 at 0:44
$begingroup$
@Melody. h = y + 0t.
$endgroup$
– William Elliot
Jan 14 at 5:51
$begingroup$
@Melody. d/dy dy/dt = d$^2$y/dt$^2$ × dt/dy.
$endgroup$
– William Elliot
Jan 14 at 6:02
add a comment |
$begingroup$
Setting h(t,y) to y appears bogus.
When y is a function of t, one usually writes y = y(t).
Then $y_y = 1$, not the zero you claim, though $y_{yt}$ = 0.
$y_t = y'$ since y is a function of t.
$y_{ty} = y'_y$ doesn't happen because y' is not a function of y.
answered Jan 14 at 0:00
William ElliotWilliam Elliot
8,1212720
$endgroup$
1
$begingroup$
Why is it bogus to write $h(t,y)=y$? I see the mistake of setting $y_y=0.$ If $y'$ is not a function of $y$, then how do you differentiate it with respect to y? My professor claims we can do this for homework.
$endgroup$
– Melody
Jan 14 at 0:44
$begingroup$
@Melody. h = y + 0t.
$endgroup$
– William Elliot
Jan 14 at 5:51
$begingroup$
@Melody. d/dy dy/dt = d$^2$y/dt$^2$ × dt/dy.
$endgroup$
– William Elliot
Jan 14 at 6:02
add a comment |
$begingroup$
Setting h(t,y) to y appears bogus.
When y is a function of t, one usually writes y = y(t).
Then $y_y = 1$, not the zero you claim, though $y_{yt}$ = 0.
$y_t = y'$ since y is a function of t.
$y_{ty} = y'_y$ doesn't happen because y' is not a function of y.
answered Jan 14 at 0:00
William ElliotWilliam Elliot
8,1212720
$endgroup$
Setting h(t,y) to y appears bogus.
When y is a function of t, one usually writes y = y(t).
Then $y_y = 1$, not the zero you claim, though $y_{yt}$ = 0.
$y_t = y'$ since y is a function of t.
$y_{ty} = y'_y$ doesn't happen because y' is not a function of y.
answered Jan 14 at 0:00
William ElliotWilliam Elliot
8,1212720
edited Jan 14 at 0:05
answered Jan 14 at 0:00
William ElliotWilliam Elliot
8,1212720
answered Jan 14 at 0:00
William ElliotWilliam Elliot
8,1212720
answered Jan 14 at 0:00
William ElliotWilliam Elliot
8,1212720
8,1212720
1
$begingroup$
Why is it bogus to write $h(t,y)=y$? I see the mistake of setting $y_y=0.$ If $y'$ is not a function of $y$, then how do you differentiate it with respect to y? My professor claims we can do this for homework.
$endgroup$
– Melody
Jan 14 at 0:44
$begingroup$
@Melody. h = y + 0t.
$endgroup$
– William Elliot
Jan 14 at 5:51
$begingroup$
@Melody. d/dy dy/dt = d$^2$y/dt$^2$ × dt/dy.
$endgroup$
– William Elliot
Jan 14 at 6:02
add a comment |
1
$begingroup$
Why is it bogus to write $h(t,y)=y$? I see the mistake of setting $y_y=0.$ If $y'$ is not a function of $y$, then how do you differentiate it with respect to y? My professor claims we can do this for homework.
$endgroup$
– Melody
Jan 14 at 0:44
$begingroup$
@Melody. h = y + 0t.
$endgroup$
– William Elliot
Jan 14 at 5:51
$begingroup$
@Melody. d/dy dy/dt = d$^2$y/dt$^2$ × dt/dy.
$endgroup$
– William Elliot
Jan 14 at 6:02
1
1
$begingroup$
Why is it bogus to write $h(t,y)=y$? I see the mistake of setting $y_y=0.$ If $y'$ is not a function of $y$, then how do you differentiate it with respect to y? My professor claims we can do this for homework.
$endgroup$
– Melody
Jan 14 at 0:44
$begingroup$
Why is it bogus to write $h(t,y)=y$? I see the mistake of setting $y_y=0.$ If $y'$ is not a function of $y$, then how do you differentiate it with respect to y? My professor claims we can do this for homework.
$endgroup$
– Melody
Jan 14 at 0:44
$begingroup$
@Melody. h = y + 0t.
$endgroup$
– William Elliot
Jan 14 at 5:51
$begingroup$
@Melody. h = y + 0t.
$endgroup$
– William Elliot
Jan 14 at 5:51
$begingroup$
@Melody. d/dy dy/dt = d$^2$y/dt$^2$ × dt/dy.
$endgroup$
– William Elliot
Jan 14 at 6:02
$begingroup$
@Melody. d/dy dy/dt = d$^2$y/dt$^2$ × dt/dy.
$endgroup$
– William Elliot
Jan 14 at 6:02
add a comment |
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$begingroup$
I'm not sure that I understand your notation. Is y a function? Is it a function of t? Is $h_{ty}$ the derivative of h with respect to t and y? Is $h_{yt}$ the derivative of h with respect to y and t? With your notation, what would the definition of $h_y$ be?
$endgroup$
– NicNic8
Jan 13 at 23:06
$begingroup$
The variable $y$ is function of $t$, and $t$ alone. The notation $h_{ty}$ is $dfrac{partial }{partial t}dfrac{partial }{partial y}h$. Similarly for $h_{yt},$ we just swap the order of the differential operators.
$endgroup$
– Melody
Jan 13 at 23:18
1
$begingroup$
What is given in the problem? Is $h(t,y)=y$ and $h_{ty}=h_{yt}$ both given?
$endgroup$
– NicNic8
Jan 13 at 23:51
$begingroup$
@NicNic8 Setting $h(t,y)=y$ was for convenience. We're not given that $h_{ty}=h_{yt},$ I just thought that the mixed partials were always equivalent. It's been years since I learned this stuff. We're given that $f=dy/dt$ and $f$ is smooth in $y$ as well as $t$. We're also given that $y$ is smooth. No other details are given.
$endgroup$
– Melody
Jan 14 at 0:50
1
$begingroup$
My attempt would be to go via the inverse function, $frac{dy}{dt} = frac{1}{frac{dt}{dy}}$, and to differentiate that with respect to $y$.
$endgroup$
– Christoph
Jan 14 at 2:39