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Problem in Tom M Apostol's Calculus Vol-II book regarding Partial Derivatives.
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I've found some difficulties in solving the following problem from Apostol, Calculus Vol-II: p-292
If $k$ is a positive constant and
$$g(x,t)=fracx2sqrtkt$$
let
$$f(x,t)= int_0^g(x,t)e^-u^2du$$
Show that
$$fracpartial fpartial x= e^-g^2fracpartial gpartial x$$
and
$$fracpartial fpartial t= e^-g^2fracpartial gpartial t$$
In this problem I want to use only the definitions of partial derivatives and basic theorems on that (as demanded by the exercise).
My Solution
I try to find the partial derivative of $f$ with respect to $x$. So we have
beginalign
D_1f(x,t) &= lim_hto 0fracf(x+h, t)-f(x,t)h \
&=lim_hto 0frac1h left[int_0^g(x+h,t)e^-u^2du-int_0^g(x,t)e^-u^2du right] \
&=lim_hto 0frac1hint_g(x,t)^g(x+h,t)e^-u^2du quad text(since, for fixed $x$ and $ t$, $g(x,t)$ is increasing)
endalign
Here, I can't figure out this limit. Also, for the other one, we have
$$D_2f(x,t)=lim_hto 0frac1hint_g(x,t+h)^g(x,t)e^-u^2du$$
Please help.
multivariable-calculus partial-derivative
edited Jul 16 at 2:00
Mattos
2,65221121
asked Jul 16 at 1:19
Indrajit Ghosh
602415
 |Â
show 1 more comment
up vote
1
down vote
favorite
I've found some difficulties in solving the following problem from Apostol, Calculus Vol-II: p-292
If $k$ is a positive constant and
$$g(x,t)=fracx2sqrtkt$$
let
$$f(x,t)= int_0^g(x,t)e^-u^2du$$
Show that
$$fracpartial fpartial x= e^-g^2fracpartial gpartial x$$
and
$$fracpartial fpartial t= e^-g^2fracpartial gpartial t$$
In this problem I want to use only the definitions of partial derivatives and basic theorems on that (as demanded by the exercise).
My Solution
I try to find the partial derivative of $f$ with respect to $x$. So we have
beginalign
D_1f(x,t) &= lim_hto 0fracf(x+h, t)-f(x,t)h \
&=lim_hto 0frac1h left[int_0^g(x+h,t)e^-u^2du-int_0^g(x,t)e^-u^2du right] \
&=lim_hto 0frac1hint_g(x,t)^g(x+h,t)e^-u^2du quad text(since, for fixed $x$ and $ t$, $g(x,t)$ is increasing)
endalign
Here, I can't figure out this limit. Also, for the other one, we have
$$D_2f(x,t)=lim_hto 0frac1hint_g(x,t+h)^g(x,t)e^-u^2du$$
Please help.
multivariable-calculus partial-derivative
edited Jul 16 at 2:00
Mattos
2,65221121
asked Jul 16 at 1:19
Indrajit Ghosh
602415
The intuition is that since h is small, the integrand is constant in the range of integration, so you can take it out of the integral (with a value of $u$ somewhere in the interval between $g(x,t)$ and $g(x+h, t)$. The integral can then be evaluated to $g(x+h, t) - g(x, t)$ and when you divide this by h and let h go to 0, you get the partial derivative of g wrt x.
– NickD
Jul 16 at 1:45
I see..........
– Indrajit Ghosh
Jul 16 at 2:14
Can it be done rigorously..?
– Indrajit Ghosh
Jul 16 at 2:14
@IndrajitGhosh Why can't you use l'Hopitals rule to evaluate the limit in $D_1 f$? It just becomes an exercise in applying the FTOC.
– Mattos
Jul 16 at 2:20
I'm talking about applying l'Hopital rule to $$lim_h to 0 frac1h int_g(x,t)^g(x+h,t) e^-u^2 du$$
– Mattos
Jul 16 at 2:28
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I've found some difficulties in solving the following problem from Apostol, Calculus Vol-II: p-292
If $k$ is a positive constant and
$$g(x,t)=fracx2sqrtkt$$
let
$$f(x,t)= int_0^g(x,t)e^-u^2du$$
Show that
$$fracpartial fpartial x= e^-g^2fracpartial gpartial x$$
and
$$fracpartial fpartial t= e^-g^2fracpartial gpartial t$$
In this problem I want to use only the definitions of partial derivatives and basic theorems on that (as demanded by the exercise).
My Solution
I try to find the partial derivative of $f$ with respect to $x$. So we have
beginalign
D_1f(x,t) &= lim_hto 0fracf(x+h, t)-f(x,t)h \
&=lim_hto 0frac1h left[int_0^g(x+h,t)e^-u^2du-int_0^g(x,t)e^-u^2du right] \
&=lim_hto 0frac1hint_g(x,t)^g(x+h,t)e^-u^2du quad text(since, for fixed $x$ and $ t$, $g(x,t)$ is increasing)
endalign
Here, I can't figure out this limit. Also, for the other one, we have
$$D_2f(x,t)=lim_hto 0frac1hint_g(x,t+h)^g(x,t)e^-u^2du$$
Please help.
multivariable-calculus partial-derivative
edited Jul 16 at 2:00
Mattos
2,65221121
asked Jul 16 at 1:19
Indrajit Ghosh
602415
I've found some difficulties in solving the following problem from Apostol, Calculus Vol-II: p-292
If $k$ is a positive constant and
$$g(x,t)=fracx2sqrtkt$$
let
$$f(x,t)= int_0^g(x,t)e^-u^2du$$
Show that
$$fracpartial fpartial x= e^-g^2fracpartial gpartial x$$
and
$$fracpartial fpartial t= e^-g^2fracpartial gpartial t$$
In this problem I want to use only the definitions of partial derivatives and basic theorems on that (as demanded by the exercise).
My Solution
I try to find the partial derivative of $f$ with respect to $x$. So we have
beginalign
D_1f(x,t) &= lim_hto 0fracf(x+h, t)-f(x,t)h \
&=lim_hto 0frac1h left[int_0^g(x+h,t)e^-u^2du-int_0^g(x,t)e^-u^2du right] \
&=lim_hto 0frac1hint_g(x,t)^g(x+h,t)e^-u^2du quad text(since, for fixed $x$ and $ t$, $g(x,t)$ is increasing)
endalign
Here, I can't figure out this limit. Also, for the other one, we have
$$D_2f(x,t)=lim_hto 0frac1hint_g(x,t+h)^g(x,t)e^-u^2du$$
Please help.
multivariable-calculus partial-derivative
edited Jul 16 at 2:00
Mattos
2,65221121
asked Jul 16 at 1:19
Indrajit Ghosh
602415
edited Jul 16 at 2:00
Mattos
2,65221121
edited Jul 16 at 2:00
Mattos
2,65221121
edited Jul 16 at 2:00
Mattos
2,65221121
2,65221121
asked Jul 16 at 1:19
Indrajit Ghosh
602415
asked Jul 16 at 1:19
Indrajit Ghosh
602415
asked Jul 16 at 1:19
Indrajit Ghosh
602415
602415
The intuition is that since h is small, the integrand is constant in the range of integration, so you can take it out of the integral (with a value of $u$ somewhere in the interval between $g(x,t)$ and $g(x+h, t)$. The integral can then be evaluated to $g(x+h, t) - g(x, t)$ and when you divide this by h and let h go to 0, you get the partial derivative of g wrt x.
– NickD
Jul 16 at 1:45
I see..........
– Indrajit Ghosh
Jul 16 at 2:14
Can it be done rigorously..?
– Indrajit Ghosh
Jul 16 at 2:14
@IndrajitGhosh Why can't you use l'Hopitals rule to evaluate the limit in $D_1 f$? It just becomes an exercise in applying the FTOC.
– Mattos
Jul 16 at 2:20
I'm talking about applying l'Hopital rule to $$lim_h to 0 frac1h int_g(x,t)^g(x+h,t) e^-u^2 du$$
– Mattos
Jul 16 at 2:28
 |Â
show 1 more comment
The intuition is that since h is small, the integrand is constant in the range of integration, so you can take it out of the integral (with a value of $u$ somewhere in the interval between $g(x,t)$ and $g(x+h, t)$. The integral can then be evaluated to $g(x+h, t) - g(x, t)$ and when you divide this by h and let h go to 0, you get the partial derivative of g wrt x.
– NickD
Jul 16 at 1:45
I see..........
– Indrajit Ghosh
Jul 16 at 2:14
Can it be done rigorously..?
– Indrajit Ghosh
Jul 16 at 2:14
@IndrajitGhosh Why can't you use l'Hopitals rule to evaluate the limit in $D_1 f$? It just becomes an exercise in applying the FTOC.
– Mattos
Jul 16 at 2:20
I'm talking about applying l'Hopital rule to $$lim_h to 0 frac1h int_g(x,t)^g(x+h,t) e^-u^2 du$$
– Mattos
Jul 16 at 2:28
The intuition is that since h is small, the integrand is constant in the range of integration, so you can take it out of the integral (with a value of $u$ somewhere in the interval between $g(x,t)$ and $g(x+h, t)$. The integral can then be evaluated to $g(x+h, t) - g(x, t)$ and when you divide this by h and let h go to 0, you get the partial derivative of g wrt x.
– NickD
Jul 16 at 1:45
The intuition is that since h is small, the integrand is constant in the range of integration, so you can take it out of the integral (with a value of $u$ somewhere in the interval between $g(x,t)$ and $g(x+h, t)$. The integral can then be evaluated to $g(x+h, t) - g(x, t)$ and when you divide this by h and let h go to 0, you get the partial derivative of g wrt x.
– NickD
Jul 16 at 1:45
I see..........
– Indrajit Ghosh
Jul 16 at 2:14
I see..........
– Indrajit Ghosh
Jul 16 at 2:14
Can it be done rigorously..?
– Indrajit Ghosh
Jul 16 at 2:14
Can it be done rigorously..?
– Indrajit Ghosh
Jul 16 at 2:14
@IndrajitGhosh Why can't you use l'Hopitals rule to evaluate the limit in $D_1 f$? It just becomes an exercise in applying the FTOC.
– Mattos
Jul 16 at 2:20
@IndrajitGhosh Why can't you use l'Hopitals rule to evaluate the limit in $D_1 f$? It just becomes an exercise in applying the FTOC.
– Mattos
Jul 16 at 2:20
I'm talking about applying l'Hopital rule to $$lim_h to 0 frac1h int_g(x,t)^g(x+h,t) e^-u^2 du$$
– Mattos
Jul 16 at 2:28
I'm talking about applying l'Hopital rule to $$lim_h to 0 frac1h int_g(x,t)^g(x+h,t) e^-u^2 du$$
– Mattos
Jul 16 at 2:28
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Similar to Chris Custer's answer, consider the general case of
$$f(x,t)=int_0^g(x,t) h(u) , du$$ Then, using the fundamental theorem of calculus,
$$fracpartial fpartial x= h(g(x,t)), fracpartial g(x,t)partial x$$
$$fracpartial fpartial t= h(g(x,t)), fracpartial g(x,t)partial t$$ and we can continue for higher derivatives using the chain rule.
answered Jul 16 at 6:22
Claude Leibovici
112k1055126
add a comment |Â
up vote
0
down vote
The two partials are computed by using the fundamental theorem of calculus (FTC), and the chain rule...
Since the upper limit of integration is a function of $x$ and $t$, you need the chain rule... this means multiplying by $fracpartial gpartial x$ and $fracpartial gpartial t$ respectively...
FTC just tells us to replace the variable ($u$) in the integrand with the upper limit of integration ($g$)...
The result follows immediately.
answered Jul 16 at 3:34
Chris Custer
5,4582622
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Similar to Chris Custer's answer, consider the general case of
$$f(x,t)=int_0^g(x,t) h(u) , du$$ Then, using the fundamental theorem of calculus,
$$fracpartial fpartial x= h(g(x,t)), fracpartial g(x,t)partial x$$
$$fracpartial fpartial t= h(g(x,t)), fracpartial g(x,t)partial t$$ and we can continue for higher derivatives using the chain rule.
answered Jul 16 at 6:22
Claude Leibovici
112k1055126
add a comment |Â
up vote
2
down vote
accepted
Similar to Chris Custer's answer, consider the general case of
$$f(x,t)=int_0^g(x,t) h(u) , du$$ Then, using the fundamental theorem of calculus,
$$fracpartial fpartial x= h(g(x,t)), fracpartial g(x,t)partial x$$
$$fracpartial fpartial t= h(g(x,t)), fracpartial g(x,t)partial t$$ and we can continue for higher derivatives using the chain rule.
answered Jul 16 at 6:22
Claude Leibovici
112k1055126
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Similar to Chris Custer's answer, consider the general case of
$$f(x,t)=int_0^g(x,t) h(u) , du$$ Then, using the fundamental theorem of calculus,
$$fracpartial fpartial x= h(g(x,t)), fracpartial g(x,t)partial x$$
$$fracpartial fpartial t= h(g(x,t)), fracpartial g(x,t)partial t$$ and we can continue for higher derivatives using the chain rule.
answered Jul 16 at 6:22
Claude Leibovici
112k1055126
Similar to Chris Custer's answer, consider the general case of
$$f(x,t)=int_0^g(x,t) h(u) , du$$ Then, using the fundamental theorem of calculus,
$$fracpartial fpartial x= h(g(x,t)), fracpartial g(x,t)partial x$$
$$fracpartial fpartial t= h(g(x,t)), fracpartial g(x,t)partial t$$ and we can continue for higher derivatives using the chain rule.
answered Jul 16 at 6:22
Claude Leibovici
112k1055126
answered Jul 16 at 6:22
Claude Leibovici
112k1055126
answered Jul 16 at 6:22
Claude Leibovici
112k1055126
answered Jul 16 at 6:22
Claude Leibovici
112k1055126
112k1055126
add a comment |Â
add a comment |Â
up vote
0
down vote
The two partials are computed by using the fundamental theorem of calculus (FTC), and the chain rule...
Since the upper limit of integration is a function of $x$ and $t$, you need the chain rule... this means multiplying by $fracpartial gpartial x$ and $fracpartial gpartial t$ respectively...
FTC just tells us to replace the variable ($u$) in the integrand with the upper limit of integration ($g$)...
The result follows immediately.
answered Jul 16 at 3:34
Chris Custer
5,4582622
add a comment |Â
up vote
0
down vote
The two partials are computed by using the fundamental theorem of calculus (FTC), and the chain rule...
Since the upper limit of integration is a function of $x$ and $t$, you need the chain rule... this means multiplying by $fracpartial gpartial x$ and $fracpartial gpartial t$ respectively...
FTC just tells us to replace the variable ($u$) in the integrand with the upper limit of integration ($g$)...
The result follows immediately.
answered Jul 16 at 3:34
Chris Custer
5,4582622
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The two partials are computed by using the fundamental theorem of calculus (FTC), and the chain rule...
Since the upper limit of integration is a function of $x$ and $t$, you need the chain rule... this means multiplying by $fracpartial gpartial x$ and $fracpartial gpartial t$ respectively...
FTC just tells us to replace the variable ($u$) in the integrand with the upper limit of integration ($g$)...
The result follows immediately.
answered Jul 16 at 3:34
Chris Custer
5,4582622
The two partials are computed by using the fundamental theorem of calculus (FTC), and the chain rule...
Since the upper limit of integration is a function of $x$ and $t$, you need the chain rule... this means multiplying by $fracpartial gpartial x$ and $fracpartial gpartial t$ respectively...
FTC just tells us to replace the variable ($u$) in the integrand with the upper limit of integration ($g$)...
The result follows immediately.
answered Jul 16 at 3:34
Chris Custer
5,4582622
answered Jul 16 at 3:34
Chris Custer
5,4582622
answered Jul 16 at 3:34
Chris Custer
5,4582622
answered Jul 16 at 3:34
Chris Custer
5,4582622
5,4582622
add a comment |Â
add a comment |Â
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The intuition is that since h is small, the integrand is constant in the range of integration, so you can take it out of the integral (with a value of $u$ somewhere in the interval between $g(x,t)$ and $g(x+h, t)$. The integral can then be evaluated to $g(x+h, t) - g(x, t)$ and when you divide this by h and let h go to 0, you get the partial derivative of g wrt x.
– NickD
Jul 16 at 1:45
I see..........
– Indrajit Ghosh
Jul 16 at 2:14
Can it be done rigorously..?
– Indrajit Ghosh
Jul 16 at 2:14
@IndrajitGhosh Why can't you use l'Hopitals rule to evaluate the limit in $D_1 f$? It just becomes an exercise in applying the FTOC.
– Mattos
Jul 16 at 2:20
I'm talking about applying l'Hopital rule to $$lim_h to 0 frac1h int_g(x,t)^g(x+h,t) e^-u^2 du$$
– Mattos
Jul 16 at 2:28