Related Rates Question Concerning Boyle's Law
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Boyle's Law for enclosed gases states that if the volume is kept constant, the pressure P and temperature T are related by the equation P/T=k, where k is a constant. Suppose that the rate of change of the temperature is 3 degrees per hour. What is the rate of change of pressure when T is 250 and P is 500?
I know this is a related rates question involving differentiation, and that dT/dt=3 but I just can't figure out how to tie the values together! Please help, thanks!
calculus derivatives
asked Apr 9 '15 at 5:16
JonJon
1
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add a comment |
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Boyle's Law for enclosed gases states that if the volume is kept constant, the pressure P and temperature T are related by the equation P/T=k, where k is a constant. Suppose that the rate of change of the temperature is 3 degrees per hour. What is the rate of change of pressure when T is 250 and P is 500?
I know this is a related rates question involving differentiation, and that dT/dt=3 but I just can't figure out how to tie the values together! Please help, thanks!
calculus derivatives
asked Apr 9 '15 at 5:16
JonJon
1
$endgroup$
add a comment |
$begingroup$
Boyle's Law for enclosed gases states that if the volume is kept constant, the pressure P and temperature T are related by the equation P/T=k, where k is a constant. Suppose that the rate of change of the temperature is 3 degrees per hour. What is the rate of change of pressure when T is 250 and P is 500?
I know this is a related rates question involving differentiation, and that dT/dt=3 but I just can't figure out how to tie the values together! Please help, thanks!
calculus derivatives
asked Apr 9 '15 at 5:16
JonJon
1
$endgroup$
Boyle's Law for enclosed gases states that if the volume is kept constant, the pressure P and temperature T are related by the equation P/T=k, where k is a constant. Suppose that the rate of change of the temperature is 3 degrees per hour. What is the rate of change of pressure when T is 250 and P is 500?
I know this is a related rates question involving differentiation, and that dT/dt=3 but I just can't figure out how to tie the values together! Please help, thanks!
calculus derivatives
calculus derivatives
asked Apr 9 '15 at 5:16
JonJon
1
asked Apr 9 '15 at 5:16
JonJon
1
asked Apr 9 '15 at 5:16
JonJon
1
asked Apr 9 '15 at 5:16
JonJon
1
asked Apr 9 '15 at 5:16
JonJon
1
1
add a comment |
add a comment |
1 Answer
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I thought Boyle's law involved pressure and volume (not temperature). But anyway, yes, $dfrac{dT}{dt}=3$. So start from the given
$$ dfrac{P}{T}=k$$
So both $P$ and $T$ are functions of time, $t$. $k$ is just a constant. So, differentiating, we need to use the quotient rule:
$$dfrac{ T dfrac{dP}{dt} - P dfrac{dT}{dt}}{T^2} = 0$$.
Along the way you need to understand, why is this equation equated to 0?
You know that $dfrac{dT}{dt} = 3$, so the rest must be easy.
answered Apr 9 '15 at 5:22
cgocgo
809622
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$begingroup$
This is indeed very correct but, may be, it could be set simpler since, rewriting the equation, $P$ is proportional to $T$ and then the result.
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– Claude Leibovici
Apr 9 '15 at 6:07
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I didn't think of that!
$endgroup$
– cgo
Apr 9 '15 at 8:05
$begingroup$
This is the (only) privilege of age ! I can only handle simple things or concepts and, moreover, I am very lazy now ! Cheers :-)
$endgroup$
– Claude Leibovici
Apr 9 '15 at 8:07
add a comment |
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1 Answer
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$begingroup$
I thought Boyle's law involved pressure and volume (not temperature). But anyway, yes, $dfrac{dT}{dt}=3$. So start from the given
$$ dfrac{P}{T}=k$$
So both $P$ and $T$ are functions of time, $t$. $k$ is just a constant. So, differentiating, we need to use the quotient rule:
$$dfrac{ T dfrac{dP}{dt} - P dfrac{dT}{dt}}{T^2} = 0$$.
Along the way you need to understand, why is this equation equated to 0?
You know that $dfrac{dT}{dt} = 3$, so the rest must be easy.
answered Apr 9 '15 at 5:22
cgocgo
809622
$endgroup$
$begingroup$
This is indeed very correct but, may be, it could be set simpler since, rewriting the equation, $P$ is proportional to $T$ and then the result.
$endgroup$
– Claude Leibovici
Apr 9 '15 at 6:07
$begingroup$
I didn't think of that!
$endgroup$
– cgo
Apr 9 '15 at 8:05
$begingroup$
This is the (only) privilege of age ! I can only handle simple things or concepts and, moreover, I am very lazy now ! Cheers :-)
$endgroup$
– Claude Leibovici
Apr 9 '15 at 8:07
add a comment |
$begingroup$
I thought Boyle's law involved pressure and volume (not temperature). But anyway, yes, $dfrac{dT}{dt}=3$. So start from the given
$$ dfrac{P}{T}=k$$
So both $P$ and $T$ are functions of time, $t$. $k$ is just a constant. So, differentiating, we need to use the quotient rule:
$$dfrac{ T dfrac{dP}{dt} - P dfrac{dT}{dt}}{T^2} = 0$$.
Along the way you need to understand, why is this equation equated to 0?
You know that $dfrac{dT}{dt} = 3$, so the rest must be easy.
answered Apr 9 '15 at 5:22
cgocgo
809622
$endgroup$
$begingroup$
This is indeed very correct but, may be, it could be set simpler since, rewriting the equation, $P$ is proportional to $T$ and then the result.
$endgroup$
– Claude Leibovici
Apr 9 '15 at 6:07
$begingroup$
I didn't think of that!
$endgroup$
– cgo
Apr 9 '15 at 8:05
$begingroup$
This is the (only) privilege of age ! I can only handle simple things or concepts and, moreover, I am very lazy now ! Cheers :-)
$endgroup$
– Claude Leibovici
Apr 9 '15 at 8:07
add a comment |
$begingroup$
I thought Boyle's law involved pressure and volume (not temperature). But anyway, yes, $dfrac{dT}{dt}=3$. So start from the given
$$ dfrac{P}{T}=k$$
So both $P$ and $T$ are functions of time, $t$. $k$ is just a constant. So, differentiating, we need to use the quotient rule:
$$dfrac{ T dfrac{dP}{dt} - P dfrac{dT}{dt}}{T^2} = 0$$.
Along the way you need to understand, why is this equation equated to 0?
You know that $dfrac{dT}{dt} = 3$, so the rest must be easy.
answered Apr 9 '15 at 5:22
cgocgo
809622
$endgroup$
I thought Boyle's law involved pressure and volume (not temperature). But anyway, yes, $dfrac{dT}{dt}=3$. So start from the given
$$ dfrac{P}{T}=k$$
So both $P$ and $T$ are functions of time, $t$. $k$ is just a constant. So, differentiating, we need to use the quotient rule:
$$dfrac{ T dfrac{dP}{dt} - P dfrac{dT}{dt}}{T^2} = 0$$.
Along the way you need to understand, why is this equation equated to 0?
You know that $dfrac{dT}{dt} = 3$, so the rest must be easy.
answered Apr 9 '15 at 5:22
cgocgo
809622
answered Apr 9 '15 at 5:22
cgocgo
809622
answered Apr 9 '15 at 5:22
cgocgo
809622
answered Apr 9 '15 at 5:22
cgocgo
809622
809622
$begingroup$
This is indeed very correct but, may be, it could be set simpler since, rewriting the equation, $P$ is proportional to $T$ and then the result.
$endgroup$
– Claude Leibovici
Apr 9 '15 at 6:07
$begingroup$
I didn't think of that!
$endgroup$
– cgo
Apr 9 '15 at 8:05
$begingroup$
This is the (only) privilege of age ! I can only handle simple things or concepts and, moreover, I am very lazy now ! Cheers :-)
$endgroup$
– Claude Leibovici
Apr 9 '15 at 8:07
add a comment |
$begingroup$
This is indeed very correct but, may be, it could be set simpler since, rewriting the equation, $P$ is proportional to $T$ and then the result.
$endgroup$
– Claude Leibovici
Apr 9 '15 at 6:07
$begingroup$
I didn't think of that!
$endgroup$
– cgo
Apr 9 '15 at 8:05
$begingroup$
This is the (only) privilege of age ! I can only handle simple things or concepts and, moreover, I am very lazy now ! Cheers :-)
$endgroup$
– Claude Leibovici
Apr 9 '15 at 8:07
$begingroup$
This is indeed very correct but, may be, it could be set simpler since, rewriting the equation, $P$ is proportional to $T$ and then the result.
$endgroup$
– Claude Leibovici
Apr 9 '15 at 6:07
$begingroup$
This is indeed very correct but, may be, it could be set simpler since, rewriting the equation, $P$ is proportional to $T$ and then the result.
$endgroup$
– Claude Leibovici
Apr 9 '15 at 6:07
$begingroup$
I didn't think of that!
$endgroup$
– cgo
Apr 9 '15 at 8:05
$begingroup$
I didn't think of that!
$endgroup$
– cgo
Apr 9 '15 at 8:05
$begingroup$
This is the (only) privilege of age ! I can only handle simple things or concepts and, moreover, I am very lazy now ! Cheers :-)
$endgroup$
– Claude Leibovici
Apr 9 '15 at 8:07
$begingroup$
This is the (only) privilege of age ! I can only handle simple things or concepts and, moreover, I am very lazy now ! Cheers :-)
$endgroup$
– Claude Leibovici
Apr 9 '15 at 8:07
add a comment |
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