Using chain rule to find dz/dt leaving answer in terms of t
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Can anyone tell me if I have done this correctly
Using the chain rule Find $ fracdzdt $
if $ z=xy^2, x=e^-3t , y=-sin(2t) $
Leaving answer in terms of $t$
$ fracdzdt =fracpartial zpartial x cdot fracpartial xpartial t + fracpartial zpartial y cdot fracpartial ypartial t$
$ fracdzdt = (y^2 cdot -3e^-3t)+(2xy cdot -2cos(2t))$
$ fracdzdt = (sin^2(2t)cdot-3e^-3t)+(2e^-3tcdot-sin(2t) cdot-2cos(2t))$
Many thanks in advance
differential-equations chain-rule
asked Sep 28 '16 at 1:52
Buddy
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up vote
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favorite
Can anyone tell me if I have done this correctly
Using the chain rule Find $ fracdzdt $
if $ z=xy^2, x=e^-3t , y=-sin(2t) $
Leaving answer in terms of $t$
$ fracdzdt =fracpartial zpartial x cdot fracpartial xpartial t + fracpartial zpartial y cdot fracpartial ypartial t$
$ fracdzdt = (y^2 cdot -3e^-3t)+(2xy cdot -2cos(2t))$
$ fracdzdt = (sin^2(2t)cdot-3e^-3t)+(2e^-3tcdot-sin(2t) cdot-2cos(2t))$
Many thanks in advance
differential-equations chain-rule
asked Sep 28 '16 at 1:52
Buddy
4318
1
Looks fine. One little critique: When writing out the chain rule for $fracdzdt$, you may want to write the derivatives of $x$ and $y$ as total derivatives rather than partial derivatives since each is a function of only $t$.
– user137731
Sep 28 '16 at 1:56
oh. you are absolutley correct thanks for pointing that out. That was where i was getting confused but was just me not paying attention to detail. You are a champ.
– Buddy
Sep 28 '16 at 1:59
Correction - should have read $ fracdzdt =fracpartial zpartial x cdot fracdxdt + fracpartial zpartial y cdot fracdydt$
– Buddy
Sep 28 '16 at 2:00
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Can anyone tell me if I have done this correctly
Using the chain rule Find $ fracdzdt $
if $ z=xy^2, x=e^-3t , y=-sin(2t) $
Leaving answer in terms of $t$
$ fracdzdt =fracpartial zpartial x cdot fracpartial xpartial t + fracpartial zpartial y cdot fracpartial ypartial t$
$ fracdzdt = (y^2 cdot -3e^-3t)+(2xy cdot -2cos(2t))$
$ fracdzdt = (sin^2(2t)cdot-3e^-3t)+(2e^-3tcdot-sin(2t) cdot-2cos(2t))$
Many thanks in advance
differential-equations chain-rule
asked Sep 28 '16 at 1:52
Buddy
4318
Can anyone tell me if I have done this correctly
Using the chain rule Find $ fracdzdt $
if $ z=xy^2, x=e^-3t , y=-sin(2t) $
Leaving answer in terms of $t$
$ fracdzdt =fracpartial zpartial x cdot fracpartial xpartial t + fracpartial zpartial y cdot fracpartial ypartial t$
$ fracdzdt = (y^2 cdot -3e^-3t)+(2xy cdot -2cos(2t))$
$ fracdzdt = (sin^2(2t)cdot-3e^-3t)+(2e^-3tcdot-sin(2t) cdot-2cos(2t))$
Many thanks in advance
differential-equations chain-rule
asked Sep 28 '16 at 1:52
Buddy
4318
asked Sep 28 '16 at 1:52
Buddy
4318
asked Sep 28 '16 at 1:52
Buddy
4318
asked Sep 28 '16 at 1:52
Buddy
4318
4318
1
Looks fine. One little critique: When writing out the chain rule for $fracdzdt$, you may want to write the derivatives of $x$ and $y$ as total derivatives rather than partial derivatives since each is a function of only $t$.
– user137731
Sep 28 '16 at 1:56
oh. you are absolutley correct thanks for pointing that out. That was where i was getting confused but was just me not paying attention to detail. You are a champ.
– Buddy
Sep 28 '16 at 1:59
Correction - should have read $ fracdzdt =fracpartial zpartial x cdot fracdxdt + fracpartial zpartial y cdot fracdydt$
– Buddy
Sep 28 '16 at 2:00
add a comment |Â
1
Looks fine. One little critique: When writing out the chain rule for $fracdzdt$, you may want to write the derivatives of $x$ and $y$ as total derivatives rather than partial derivatives since each is a function of only $t$.
– user137731
Sep 28 '16 at 1:56
oh. you are absolutley correct thanks for pointing that out. That was where i was getting confused but was just me not paying attention to detail. You are a champ.
– Buddy
Sep 28 '16 at 1:59
Correction - should have read $ fracdzdt =fracpartial zpartial x cdot fracdxdt + fracpartial zpartial y cdot fracdydt$
– Buddy
Sep 28 '16 at 2:00
1
1
Looks fine. One little critique: When writing out the chain rule for $fracdzdt$, you may want to write the derivatives of $x$ and $y$ as total derivatives rather than partial derivatives since each is a function of only $t$.
– user137731
Sep 28 '16 at 1:56
Looks fine. One little critique: When writing out the chain rule for $fracdzdt$, you may want to write the derivatives of $x$ and $y$ as total derivatives rather than partial derivatives since each is a function of only $t$.
– user137731
Sep 28 '16 at 1:56
oh. you are absolutley correct thanks for pointing that out. That was where i was getting confused but was just me not paying attention to detail. You are a champ.
– Buddy
Sep 28 '16 at 1:59
oh. you are absolutley correct thanks for pointing that out. That was where i was getting confused but was just me not paying attention to detail. You are a champ.
– Buddy
Sep 28 '16 at 1:59
Correction - should have read $ fracdzdt =fracpartial zpartial x cdot fracdxdt + fracpartial zpartial y cdot fracdydt$
– Buddy
Sep 28 '16 at 2:00
Correction - should have read $ fracdzdt =fracpartial zpartial x cdot fracdxdt + fracpartial zpartial y cdot fracdydt$
– Buddy
Sep 28 '16 at 2:00
add a comment |Â
1 Answer
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We need to find:
$$textz'(t)=fractextdtextz(t)textdt=fractextdtextdtleft(e^-3tcdotleft(-sin(2t)right)^2right)=$$
$$fractextdtextdtleft(e^-3tsin^2(2t)right)=e^-3tsin(2t)left(4cos(2t)-3sin(2t)right)$$
Using:
- The product rule:
$$fractextdtextdtleft(f(t)cdot y(t)right)=f(t)cdotfractextdtextdtleft(y(t)right)+y(t)cdotfractextdtextdtleft(f(t)right)=y(t)cdot f'(t)+f(t)cdot y'(t)$$ - $$fractextdtextdtleft(e^x(t)right)=e^x(t)cdotfractextdtextdtleft(x(t)right)=x'(t)cdot e^x(t)$$
- When $textC$ is a constant:
$$fractextdtextdtleft(textCcdot q(t)right)=textCcdotfractextdtextdtleft(q(t)right)=textCcdot q'(t)$$ - When $textn$ is a constant, using the chain rule:
$$fractextdtextdtleft(w(t)^textnright)=textncdot w(t)^textn-1cdotfractextdtextdtleft(w(t)right)=textncdot w(t)^textn-1cdot w'(t)$$ - When $textm$ is a constant, using the chain rule:
$$fractextdtextdtleft(v(textmt)right)=fractextdtextdtleft(textmtright)cdot v'(textmt)=textmcdot v'(textmt)$$
answered Sep 28 '16 at 19:54
Jan
21.6k31239
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We need to find:
$$textz'(t)=fractextdtextz(t)textdt=fractextdtextdtleft(e^-3tcdotleft(-sin(2t)right)^2right)=$$
$$fractextdtextdtleft(e^-3tsin^2(2t)right)=e^-3tsin(2t)left(4cos(2t)-3sin(2t)right)$$
Using:
- The product rule:
$$fractextdtextdtleft(f(t)cdot y(t)right)=f(t)cdotfractextdtextdtleft(y(t)right)+y(t)cdotfractextdtextdtleft(f(t)right)=y(t)cdot f'(t)+f(t)cdot y'(t)$$ - $$fractextdtextdtleft(e^x(t)right)=e^x(t)cdotfractextdtextdtleft(x(t)right)=x'(t)cdot e^x(t)$$
- When $textC$ is a constant:
$$fractextdtextdtleft(textCcdot q(t)right)=textCcdotfractextdtextdtleft(q(t)right)=textCcdot q'(t)$$ - When $textn$ is a constant, using the chain rule:
$$fractextdtextdtleft(w(t)^textnright)=textncdot w(t)^textn-1cdotfractextdtextdtleft(w(t)right)=textncdot w(t)^textn-1cdot w'(t)$$ - When $textm$ is a constant, using the chain rule:
$$fractextdtextdtleft(v(textmt)right)=fractextdtextdtleft(textmtright)cdot v'(textmt)=textmcdot v'(textmt)$$
answered Sep 28 '16 at 19:54
Jan
21.6k31239
add a comment |Â
up vote
0
down vote
We need to find:
$$textz'(t)=fractextdtextz(t)textdt=fractextdtextdtleft(e^-3tcdotleft(-sin(2t)right)^2right)=$$
$$fractextdtextdtleft(e^-3tsin^2(2t)right)=e^-3tsin(2t)left(4cos(2t)-3sin(2t)right)$$
Using:
- The product rule:
$$fractextdtextdtleft(f(t)cdot y(t)right)=f(t)cdotfractextdtextdtleft(y(t)right)+y(t)cdotfractextdtextdtleft(f(t)right)=y(t)cdot f'(t)+f(t)cdot y'(t)$$ - $$fractextdtextdtleft(e^x(t)right)=e^x(t)cdotfractextdtextdtleft(x(t)right)=x'(t)cdot e^x(t)$$
- When $textC$ is a constant:
$$fractextdtextdtleft(textCcdot q(t)right)=textCcdotfractextdtextdtleft(q(t)right)=textCcdot q'(t)$$ - When $textn$ is a constant, using the chain rule:
$$fractextdtextdtleft(w(t)^textnright)=textncdot w(t)^textn-1cdotfractextdtextdtleft(w(t)right)=textncdot w(t)^textn-1cdot w'(t)$$ - When $textm$ is a constant, using the chain rule:
$$fractextdtextdtleft(v(textmt)right)=fractextdtextdtleft(textmtright)cdot v'(textmt)=textmcdot v'(textmt)$$
answered Sep 28 '16 at 19:54
Jan
21.6k31239
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We need to find:
$$textz'(t)=fractextdtextz(t)textdt=fractextdtextdtleft(e^-3tcdotleft(-sin(2t)right)^2right)=$$
$$fractextdtextdtleft(e^-3tsin^2(2t)right)=e^-3tsin(2t)left(4cos(2t)-3sin(2t)right)$$
Using:
- The product rule:
$$fractextdtextdtleft(f(t)cdot y(t)right)=f(t)cdotfractextdtextdtleft(y(t)right)+y(t)cdotfractextdtextdtleft(f(t)right)=y(t)cdot f'(t)+f(t)cdot y'(t)$$ - $$fractextdtextdtleft(e^x(t)right)=e^x(t)cdotfractextdtextdtleft(x(t)right)=x'(t)cdot e^x(t)$$
- When $textC$ is a constant:
$$fractextdtextdtleft(textCcdot q(t)right)=textCcdotfractextdtextdtleft(q(t)right)=textCcdot q'(t)$$ - When $textn$ is a constant, using the chain rule:
$$fractextdtextdtleft(w(t)^textnright)=textncdot w(t)^textn-1cdotfractextdtextdtleft(w(t)right)=textncdot w(t)^textn-1cdot w'(t)$$ - When $textm$ is a constant, using the chain rule:
$$fractextdtextdtleft(v(textmt)right)=fractextdtextdtleft(textmtright)cdot v'(textmt)=textmcdot v'(textmt)$$
answered Sep 28 '16 at 19:54
Jan
21.6k31239
We need to find:
$$textz'(t)=fractextdtextz(t)textdt=fractextdtextdtleft(e^-3tcdotleft(-sin(2t)right)^2right)=$$
$$fractextdtextdtleft(e^-3tsin^2(2t)right)=e^-3tsin(2t)left(4cos(2t)-3sin(2t)right)$$
Using:
- The product rule:
$$fractextdtextdtleft(f(t)cdot y(t)right)=f(t)cdotfractextdtextdtleft(y(t)right)+y(t)cdotfractextdtextdtleft(f(t)right)=y(t)cdot f'(t)+f(t)cdot y'(t)$$ - $$fractextdtextdtleft(e^x(t)right)=e^x(t)cdotfractextdtextdtleft(x(t)right)=x'(t)cdot e^x(t)$$
- When $textC$ is a constant:
$$fractextdtextdtleft(textCcdot q(t)right)=textCcdotfractextdtextdtleft(q(t)right)=textCcdot q'(t)$$ - When $textn$ is a constant, using the chain rule:
$$fractextdtextdtleft(w(t)^textnright)=textncdot w(t)^textn-1cdotfractextdtextdtleft(w(t)right)=textncdot w(t)^textn-1cdot w'(t)$$ - When $textm$ is a constant, using the chain rule:
$$fractextdtextdtleft(v(textmt)right)=fractextdtextdtleft(textmtright)cdot v'(textmt)=textmcdot v'(textmt)$$
answered Sep 28 '16 at 19:54
Jan
21.6k31239
answered Sep 28 '16 at 19:54
Jan
21.6k31239
answered Sep 28 '16 at 19:54
Jan
21.6k31239
answered Sep 28 '16 at 19:54
Jan
21.6k31239
21.6k31239
add a comment |Â
add a comment |Â
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1
Looks fine. One little critique: When writing out the chain rule for $fracdzdt$, you may want to write the derivatives of $x$ and $y$ as total derivatives rather than partial derivatives since each is a function of only $t$.
– user137731
Sep 28 '16 at 1:56
oh. you are absolutley correct thanks for pointing that out. That was where i was getting confused but was just me not paying attention to detail. You are a champ.
– Buddy
Sep 28 '16 at 1:59
Correction - should have read $ fracdzdt =fracpartial zpartial x cdot fracdxdt + fracpartial zpartial y cdot fracdydt$
– Buddy
Sep 28 '16 at 2:00