Derivative with Respect to a Variable Times Constant? And Then Factoring-Out That Constant as a Fraction?

Derivative with Respect to a Variable Times Constant? And Then Factoring-Out That Constant as a Fraction?

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

I have the PDE $dfracpartialTpartialt = alpha dfrac1r dfracpartialpartialrleft( r dfracpartialTpartialr right)$

I have done change of variables and now have $r^* = dfracra$, $dfracpartialTpartialt = dfrac(T_i - T_w)t_0 dfracpartialupartialt^*$, and $dfracpartialTpartialr = dfrac(T_i - T_w)a dfracpartialupartialr^*$.

I am told that the original PDE can be rewritten as $dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^* left( r^* dfracpartialupartialr^* right)$.

Doing the necessary substitutions, if my calculations are correct, we get

$$dfrac(T_i - T_w)t_0 dfracpartialupartialt^* = dfracalphar^* a dfracpartialpartialrleft( r^* a left[ dfrac(T_i - T_w)a dfrac partialu partialr^* right] right)$$

And, if it is correct to do so (?), we substitute $r = r^* a$ into $dfrac partial partialr$ to get

$$dfrac(T_i - T_w)t_0 dfracpartialupartialt^* = dfracalphar^* a dfracpartialpartial(r^* a)left( r^* a left[ dfrac(T_i - T_w)a dfrac partialu partialr^* right] right)$$

I have never seen such a thing, so I have no idea if this is even mathematically correct, but assuming it is, in order to get the rewritten PDE, we would have to factor out the $a$ to get

$$dfrac(T_i - T_w)t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^*left( r^* a left[ dfrac(T_i - T_w)a dfrac partialu partialr^* right] right)$$

And doing the necessary cancellations, we get

$$dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^*left( r^* dfrac partialu partialr^* right)$$

But, as I said, I have never seen someone differentiating with respect to a variable multiplied by a constant, and then factoring out the constant as the fraction $dfrac1a$ ($a$ in this case). Can someone please explain whether or not this is valid and why?

asked Jul 24 at 16:14

The Pointer

2,4702829

add a commentÂ |Â

up vote
1
down vote

favorite

I have the PDE $dfracpartialTpartialt = alpha dfrac1r dfracpartialpartialrleft( r dfracpartialTpartialr right)$

I have done change of variables and now have $r^* = dfracra$, $dfracpartialTpartialt = dfrac(T_i - T_w)t_0 dfracpartialupartialt^*$, and $dfracpartialTpartialr = dfrac(T_i - T_w)a dfracpartialupartialr^*$.

I am told that the original PDE can be rewritten as $dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^* left( r^* dfracpartialupartialr^* right)$.

Doing the necessary substitutions, if my calculations are correct, we get

$$dfrac(T_i - T_w)t_0 dfracpartialupartialt^* = dfracalphar^* a dfracpartialpartialrleft( r^* a left[ dfrac(T_i - T_w)a dfrac partialu partialr^* right] right)$$

And, if it is correct to do so (?), we substitute $r = r^* a$ into $dfrac partial partialr$ to get

$$dfrac(T_i - T_w)t_0 dfracpartialupartialt^* = dfracalphar^* a dfracpartialpartial(r^* a)left( r^* a left[ dfrac(T_i - T_w)a dfrac partialu partialr^* right] right)$$

I have never seen such a thing, so I have no idea if this is even mathematically correct, but assuming it is, in order to get the rewritten PDE, we would have to factor out the $a$ to get

$$dfrac(T_i - T_w)t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^*left( r^* a left[ dfrac(T_i - T_w)a dfrac partialu partialr^* right] right)$$

And doing the necessary cancellations, we get

$$dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^*left( r^* dfrac partialu partialr^* right)$$

But, as I said, I have never seen someone differentiating with respect to a variable multiplied by a constant, and then factoring out the constant as the fraction $dfrac1a$ ($a$ in this case). Can someone please explain whether or not this is valid and why?

asked Jul 24 at 16:14

The Pointer

2,4702829

add a commentÂ |Â

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I have the PDE $dfracpartialTpartialt = alpha dfrac1r dfracpartialpartialrleft( r dfracpartialTpartialr right)$

I have done change of variables and now have $r^* = dfracra$, $dfracpartialTpartialt = dfrac(T_i - T_w)t_0 dfracpartialupartialt^*$, and $dfracpartialTpartialr = dfrac(T_i - T_w)a dfracpartialupartialr^*$.

I am told that the original PDE can be rewritten as $dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^* left( r^* dfracpartialupartialr^* right)$.

Doing the necessary substitutions, if my calculations are correct, we get

$$dfrac(T_i - T_w)t_0 dfracpartialupartialt^* = dfracalphar^* a dfracpartialpartialrleft( r^* a left[ dfrac(T_i - T_w)a dfrac partialu partialr^* right] right)$$

And, if it is correct to do so (?), we substitute $r = r^* a$ into $dfrac partial partialr$ to get

$$dfrac(T_i - T_w)t_0 dfracpartialupartialt^* = dfracalphar^* a dfracpartialpartial(r^* a)left( r^* a left[ dfrac(T_i - T_w)a dfrac partialu partialr^* right] right)$$

I have never seen such a thing, so I have no idea if this is even mathematically correct, but assuming it is, in order to get the rewritten PDE, we would have to factor out the $a$ to get

$$dfrac(T_i - T_w)t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^*left( r^* a left[ dfrac(T_i - T_w)a dfrac partialu partialr^* right] right)$$

And doing the necessary cancellations, we get

$$dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^*left( r^* dfrac partialu partialr^* right)$$

But, as I said, I have never seen someone differentiating with respect to a variable multiplied by a constant, and then factoring out the constant as the fraction $dfrac1a$ ($a$ in this case). Can someone please explain whether or not this is valid and why?

asked Jul 24 at 16:14

The Pointer

2,4702829

I have the PDE $dfracpartialTpartialt = alpha dfrac1r dfracpartialpartialrleft( r dfracpartialTpartialr right)$

I have done change of variables and now have $r^* = dfracra$, $dfracpartialTpartialt = dfrac(T_i - T_w)t_0 dfracpartialupartialt^*$, and $dfracpartialTpartialr = dfrac(T_i - T_w)a dfracpartialupartialr^*$.

I am told that the original PDE can be rewritten as $dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^* left( r^* dfracpartialupartialr^* right)$.

Doing the necessary substitutions, if my calculations are correct, we get

$$dfrac(T_i - T_w)t_0 dfracpartialupartialt^* = dfracalphar^* a dfracpartialpartialrleft( r^* a left[ dfrac(T_i - T_w)a dfrac partialu partialr^* right] right)$$

And, if it is correct to do so (?), we substitute $r = r^* a$ into $dfrac partial partialr$ to get

$$dfrac(T_i - T_w)t_0 dfracpartialupartialt^* = dfracalphar^* a dfracpartialpartial(r^* a)left( r^* a left[ dfrac(T_i - T_w)a dfrac partialu partialr^* right] right)$$

I have never seen such a thing, so I have no idea if this is even mathematically correct, but assuming it is, in order to get the rewritten PDE, we would have to factor out the $a$ to get

$$dfrac(T_i - T_w)t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^*left( r^* a left[ dfrac(T_i - T_w)a dfrac partialu partialr^* right] right)$$

And doing the necessary cancellations, we get

$$dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^*left( r^* dfrac partialu partialr^* right)$$

But, as I said, I have never seen someone differentiating with respect to a variable multiplied by a constant, and then factoring out the constant as the fraction $dfrac1a$ ($a$ in this case). Can someone please explain whether or not this is valid and why?

asked Jul 24 at 16:14

The Pointer

2,4702829

asked Jul 24 at 16:14

The Pointer

2,4702829

asked Jul 24 at 16:14

The Pointer

2,4702829

asked Jul 24 at 16:14

The Pointer

2,4702829

2,4702829

add a commentÂ |Â

add a commentÂ |Â

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

It is perfectly possible, to show you why use the chain rule. Here's (perhaps) a simpler version, call

begineqnarray
T &=& (T_i - T_w)u \
r &=& a r^* \
t &=& t_0 t^*
endeqnarray

So that

$$
fracpartial partial r = fracrm dr^*rm drfracpartial partial r^* = fracrm d(r/a)rm drfracpartialpartial r^* = frac1afracpartialpartial r^*
$$

Similarly

$$
fracpartialpartial t = frac1t_0fracpartialpartial t^*
$$

And your equation becomes

begineqnarray
requirecancel
frac1t_0fracpartialpartial t^*[colorbluecancel(T_i - T_w)u] &=& fracalphaa^2frac1r^*fracpartial partial r^*left( fraccolorredcancela r^*colorredcancela fracpartial partial r^*[colorbluecancel(T_i - T_w)u]right) \
implies~~frac1t_0 fracpartial upartial t^* &=& fracalphaa^2 frac1r^*fracpartialpartial r^*left(r^* fracpartial upartial r^* right)
endeqnarray

answered Jul 24 at 17:01

caverac

11k2927

Thanks for the response. How did you get $fracpartial partial r = fracrm dr^*rm drfracpartial partial r^*$ from $fracpartial partial r$? After all, it is not differentiating anything? Seems odd.
â€“Â The Pointer
Jul 24 at 17:04

1

@ThePointer Sure, imagine a generic function $f$ $$fracpartial fpartial r = fracrm dr^*rm drfracpartial fpartial r^* = fracrm d(r/a)rm drfracpartial fpartial r^* = frac1afracpartial fpartial r^*$$
â€“Â caverac
Jul 24 at 17:17

Oh, I see. Thank you for the clarification. So it seems that I did the calculations incorrectly. Thank you very much for this! Very enlightening.
â€“Â The Pointer
Jul 24 at 17:25

1

@ThePointer I wouldn't say your procedure is incorrect, doing $$ fracpartial fpartial (a r^*) = frac1afracpartial fpartial r^* $$ is perfectly fine, it is just a shortcut for the chain rule
â€“Â caverac
Jul 24 at 18:11

1

Ok, thanks for the clarification.
â€“Â The Pointer
Jul 24 at 18:18

add a commentÂ |Â

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

Â

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861512%2fderivative-with-respect-to-a-variable-times-constant-and-then-factoring-out-tha%23new-answer', 'question_page');

);

Post as a guest

Name

Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

accepted

It is perfectly possible, to show you why use the chain rule. Here's (perhaps) a simpler version, call

begineqnarray
T &=& (T_i - T_w)u \
r &=& a r^* \
t &=& t_0 t^*
endeqnarray

So that

$$
fracpartial partial r = fracrm dr^*rm drfracpartial partial r^* = fracrm d(r/a)rm drfracpartialpartial r^* = frac1afracpartialpartial r^*
$$

Similarly

$$
fracpartialpartial t = frac1t_0fracpartialpartial t^*
$$

And your equation becomes

begineqnarray
requirecancel
frac1t_0fracpartialpartial t^*[colorbluecancel(T_i - T_w)u] &=& fracalphaa^2frac1r^*fracpartial partial r^*left( fraccolorredcancela r^*colorredcancela fracpartial partial r^*[colorbluecancel(T_i - T_w)u]right) \
implies~~frac1t_0 fracpartial upartial t^* &=& fracalphaa^2 frac1r^*fracpartialpartial r^*left(r^* fracpartial upartial r^* right)
endeqnarray

answered Jul 24 at 17:01

caverac

11k2927

Thanks for the response. How did you get $fracpartial partial r = fracrm dr^*rm drfracpartial partial r^*$ from $fracpartial partial r$? After all, it is not differentiating anything? Seems odd.
â€“Â The Pointer
Jul 24 at 17:04

1

@ThePointer Sure, imagine a generic function $f$ $$fracpartial fpartial r = fracrm dr^*rm drfracpartial fpartial r^* = fracrm d(r/a)rm drfracpartial fpartial r^* = frac1afracpartial fpartial r^*$$
â€“Â caverac
Jul 24 at 17:17

Oh, I see. Thank you for the clarification. So it seems that I did the calculations incorrectly. Thank you very much for this! Very enlightening.
â€“Â The Pointer
Jul 24 at 17:25

1

@ThePointer I wouldn't say your procedure is incorrect, doing $$ fracpartial fpartial (a r^*) = frac1afracpartial fpartial r^* $$ is perfectly fine, it is just a shortcut for the chain rule
â€“Â caverac
Jul 24 at 18:11

1

Ok, thanks for the clarification.
â€“Â The Pointer
Jul 24 at 18:18

add a commentÂ |Â

up vote
1
down vote

accepted

It is perfectly possible, to show you why use the chain rule. Here's (perhaps) a simpler version, call

begineqnarray
T &=& (T_i - T_w)u \
r &=& a r^* \
t &=& t_0 t^*
endeqnarray

So that

$$
fracpartial partial r = fracrm dr^*rm drfracpartial partial r^* = fracrm d(r/a)rm drfracpartialpartial r^* = frac1afracpartialpartial r^*
$$

Similarly

$$
fracpartialpartial t = frac1t_0fracpartialpartial t^*
$$

And your equation becomes

begineqnarray
requirecancel
frac1t_0fracpartialpartial t^*[colorbluecancel(T_i - T_w)u] &=& fracalphaa^2frac1r^*fracpartial partial r^*left( fraccolorredcancela r^*colorredcancela fracpartial partial r^*[colorbluecancel(T_i - T_w)u]right) \
implies~~frac1t_0 fracpartial upartial t^* &=& fracalphaa^2 frac1r^*fracpartialpartial r^*left(r^* fracpartial upartial r^* right)
endeqnarray

answered Jul 24 at 17:01

caverac

11k2927

Thanks for the response. How did you get $fracpartial partial r = fracrm dr^*rm drfracpartial partial r^*$ from $fracpartial partial r$? After all, it is not differentiating anything? Seems odd.
â€“Â The Pointer
Jul 24 at 17:04

1

@ThePointer Sure, imagine a generic function $f$ $$fracpartial fpartial r = fracrm dr^*rm drfracpartial fpartial r^* = fracrm d(r/a)rm drfracpartial fpartial r^* = frac1afracpartial fpartial r^*$$
â€“Â caverac
Jul 24 at 17:17

Oh, I see. Thank you for the clarification. So it seems that I did the calculations incorrectly. Thank you very much for this! Very enlightening.
â€“Â The Pointer
Jul 24 at 17:25

1

@ThePointer I wouldn't say your procedure is incorrect, doing $$ fracpartial fpartial (a r^*) = frac1afracpartial fpartial r^* $$ is perfectly fine, it is just a shortcut for the chain rule
â€“Â caverac
Jul 24 at 18:11

1

Ok, thanks for the clarification.
â€“Â The Pointer
Jul 24 at 18:18

add a commentÂ |Â

up vote
1
down vote

accepted

up vote
1
down vote

accepted

It is perfectly possible, to show you why use the chain rule. Here's (perhaps) a simpler version, call

begineqnarray
T &=& (T_i - T_w)u \
r &=& a r^* \
t &=& t_0 t^*
endeqnarray

So that

$$
fracpartial partial r = fracrm dr^*rm drfracpartial partial r^* = fracrm d(r/a)rm drfracpartialpartial r^* = frac1afracpartialpartial r^*
$$

Similarly

$$
fracpartialpartial t = frac1t_0fracpartialpartial t^*
$$

And your equation becomes

begineqnarray
requirecancel
frac1t_0fracpartialpartial t^*[colorbluecancel(T_i - T_w)u] &=& fracalphaa^2frac1r^*fracpartial partial r^*left( fraccolorredcancela r^*colorredcancela fracpartial partial r^*[colorbluecancel(T_i - T_w)u]right) \
implies~~frac1t_0 fracpartial upartial t^* &=& fracalphaa^2 frac1r^*fracpartialpartial r^*left(r^* fracpartial upartial r^* right)
endeqnarray

answered Jul 24 at 17:01

caverac

11k2927

It is perfectly possible, to show you why use the chain rule. Here's (perhaps) a simpler version, call

begineqnarray
T &=& (T_i - T_w)u \
r &=& a r^* \
t &=& t_0 t^*
endeqnarray

So that

$$
fracpartial partial r = fracrm dr^*rm drfracpartial partial r^* = fracrm d(r/a)rm drfracpartialpartial r^* = frac1afracpartialpartial r^*
$$

Similarly

$$
fracpartialpartial t = frac1t_0fracpartialpartial t^*
$$

And your equation becomes

begineqnarray
requirecancel
frac1t_0fracpartialpartial t^*[colorbluecancel(T_i - T_w)u] &=& fracalphaa^2frac1r^*fracpartial partial r^*left( fraccolorredcancela r^*colorredcancela fracpartial partial r^*[colorbluecancel(T_i - T_w)u]right) \
implies~~frac1t_0 fracpartial upartial t^* &=& fracalphaa^2 frac1r^*fracpartialpartial r^*left(r^* fracpartial upartial r^* right)
endeqnarray

answered Jul 24 at 17:01

caverac

11k2927

answered Jul 24 at 17:01

caverac

11k2927

answered Jul 24 at 17:01

caverac

11k2927

answered Jul 24 at 17:01

caverac

11k2927

11k2927

Thanks for the response. How did you get $fracpartial partial r = fracrm dr^*rm drfracpartial partial r^*$ from $fracpartial partial r$? After all, it is not differentiating anything? Seems odd.
â€“Â The Pointer
Jul 24 at 17:04

1

@ThePointer Sure, imagine a generic function $f$ $$fracpartial fpartial r = fracrm dr^*rm drfracpartial fpartial r^* = fracrm d(r/a)rm drfracpartial fpartial r^* = frac1afracpartial fpartial r^*$$
â€“Â caverac
Jul 24 at 17:17

Oh, I see. Thank you for the clarification. So it seems that I did the calculations incorrectly. Thank you very much for this! Very enlightening.
â€“Â The Pointer
Jul 24 at 17:25

1

@ThePointer I wouldn't say your procedure is incorrect, doing $$ fracpartial fpartial (a r^*) = frac1afracpartial fpartial r^* $$ is perfectly fine, it is just a shortcut for the chain rule
â€“Â caverac
Jul 24 at 18:11

1

Ok, thanks for the clarification.
â€“Â The Pointer
Jul 24 at 18:18

add a commentÂ |Â

Thanks for the response. How did you get $fracpartial partial r = fracrm dr^*rm drfracpartial partial r^*$ from $fracpartial partial r$? After all, it is not differentiating anything? Seems odd.
â€“Â The Pointer
Jul 24 at 17:04

1

@ThePointer Sure, imagine a generic function $f$ $$fracpartial fpartial r = fracrm dr^*rm drfracpartial fpartial r^* = fracrm d(r/a)rm drfracpartial fpartial r^* = frac1afracpartial fpartial r^*$$
â€“Â caverac
Jul 24 at 17:17

Oh, I see. Thank you for the clarification. So it seems that I did the calculations incorrectly. Thank you very much for this! Very enlightening.
â€“Â The Pointer
Jul 24 at 17:25

1

@ThePointer I wouldn't say your procedure is incorrect, doing $$ fracpartial fpartial (a r^*) = frac1afracpartial fpartial r^* $$ is perfectly fine, it is just a shortcut for the chain rule
â€“Â caverac
Jul 24 at 18:11

1

Ok, thanks for the clarification.
â€“Â The Pointer
Jul 24 at 18:18

Thanks for the response. How did you get $fracpartial partial r = fracrm dr^*rm drfracpartial partial r^*$ from $fracpartial partial r$? After all, it is not differentiating anything? Seems odd.
â€“Â The Pointer
Jul 24 at 17:04

Thanks for the response. How did you get $fracpartial partial r = fracrm dr^*rm drfracpartial partial r^*$ from $fracpartial partial r$? After all, it is not differentiating anything? Seems odd.
â€“Â The Pointer
Jul 24 at 17:04

1

1

@ThePointer Sure, imagine a generic function $f$ $$fracpartial fpartial r = fracrm dr^*rm drfracpartial fpartial r^* = fracrm d(r/a)rm drfracpartial fpartial r^* = frac1afracpartial fpartial r^*$$
â€“Â caverac
Jul 24 at 17:17

@ThePointer Sure, imagine a generic function $f$ $$fracpartial fpartial r = fracrm dr^*rm drfracpartial fpartial r^* = fracrm d(r/a)rm drfracpartial fpartial r^* = frac1afracpartial fpartial r^*$$
â€“Â caverac
Jul 24 at 17:17

Oh, I see. Thank you for the clarification. So it seems that I did the calculations incorrectly. Thank you very much for this! Very enlightening.
â€“Â The Pointer
Jul 24 at 17:25

Oh, I see. Thank you for the clarification. So it seems that I did the calculations incorrectly. Thank you very much for this! Very enlightening.
â€“Â The Pointer
Jul 24 at 17:25

1

1

@ThePointer I wouldn't say your procedure is incorrect, doing $$ fracpartial fpartial (a r^*) = frac1afracpartial fpartial r^* $$ is perfectly fine, it is just a shortcut for the chain rule
â€“Â caverac
Jul 24 at 18:11

@ThePointer I wouldn't say your procedure is incorrect, doing $$ fracpartial fpartial (a r^*) = frac1afracpartial fpartial r^* $$ is perfectly fine, it is just a shortcut for the chain rule
â€“Â caverac
Jul 24 at 18:11

1

1

Ok, thanks for the clarification.
â€“Â The Pointer
Jul 24 at 18:18

Ok, thanks for the clarification.
â€“Â The Pointer
Jul 24 at 18:18

add a commentÂ |Â

Â

draft saved

draft discarded

Â

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861512%2fderivative-with-respect-to-a-variable-times-constant-and-then-factoring-out-tha%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Name

Email

Name

Email

Name

Email

Name

Email

Comments