Mixing
Proportionality for A-Level Maths
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During the decay of a radioactive substance, the rate at which mass is lost is proportional to the mass present at the instant. Use $m$ for the mass of the substance in grams and $t$ for the time in seconds. Initially there is 24 g of the substance and the mass is decreasing at a rate of 1.2$g$$s^{-1}$. Write a differential equation for the rate of change of mass.
My workings go as follows:
$frac{dm}{dt}$ $∝$ $m$
$m$ = 24 when $frac{dm}{dt}$ = -1.2
therefore, $frac{dm}{dt}$ = $km$
However, as $frac{dm}{dt}$ is negative, and $m$ is positive, $k$ must be negative for the initial condition to be true.
Therefore:
$frac{dm}{dt}$ = -$km$
-1.2 = -24$k$
$k$ = $frac{1}{20}$
Therefore:
$frac{dm}{dt}$ = - $frac{m}{20}$
or $frac{dm}{dt}$ = - 0.05$m$
Is my method correct? Because the textbook claimed the answer was $frac{dm}{dt}$ = -0.005$m$
mathematical-modeling
asked Jan 14 at 17:51
Sam ConnellSam Connell
62
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add a comment |
$begingroup$
During the decay of a radioactive substance, the rate at which mass is lost is proportional to the mass present at the instant. Use $m$ for the mass of the substance in grams and $t$ for the time in seconds. Initially there is 24 g of the substance and the mass is decreasing at a rate of 1.2$g$$s^{-1}$. Write a differential equation for the rate of change of mass.
My workings go as follows:
$frac{dm}{dt}$ $∝$ $m$
$m$ = 24 when $frac{dm}{dt}$ = -1.2
therefore, $frac{dm}{dt}$ = $km$
However, as $frac{dm}{dt}$ is negative, and $m$ is positive, $k$ must be negative for the initial condition to be true.
Therefore:
$frac{dm}{dt}$ = -$km$
-1.2 = -24$k$
$k$ = $frac{1}{20}$
Therefore:
$frac{dm}{dt}$ = - $frac{m}{20}$
or $frac{dm}{dt}$ = - 0.05$m$
Is my method correct? Because the textbook claimed the answer was $frac{dm}{dt}$ = -0.005$m$
mathematical-modeling
asked Jan 14 at 17:51
Sam ConnellSam Connell
62
$endgroup$
add a comment |
$begingroup$
During the decay of a radioactive substance, the rate at which mass is lost is proportional to the mass present at the instant. Use $m$ for the mass of the substance in grams and $t$ for the time in seconds. Initially there is 24 g of the substance and the mass is decreasing at a rate of 1.2$g$$s^{-1}$. Write a differential equation for the rate of change of mass.
My workings go as follows:
$frac{dm}{dt}$ $∝$ $m$
$m$ = 24 when $frac{dm}{dt}$ = -1.2
therefore, $frac{dm}{dt}$ = $km$
However, as $frac{dm}{dt}$ is negative, and $m$ is positive, $k$ must be negative for the initial condition to be true.
Therefore:
$frac{dm}{dt}$ = -$km$
-1.2 = -24$k$
$k$ = $frac{1}{20}$
Therefore:
$frac{dm}{dt}$ = - $frac{m}{20}$
or $frac{dm}{dt}$ = - 0.05$m$
Is my method correct? Because the textbook claimed the answer was $frac{dm}{dt}$ = -0.005$m$
mathematical-modeling
asked Jan 14 at 17:51
Sam ConnellSam Connell
62
$endgroup$
During the decay of a radioactive substance, the rate at which mass is lost is proportional to the mass present at the instant. Use $m$ for the mass of the substance in grams and $t$ for the time in seconds. Initially there is 24 g of the substance and the mass is decreasing at a rate of 1.2$g$$s^{-1}$. Write a differential equation for the rate of change of mass.
My workings go as follows:
$frac{dm}{dt}$ $∝$ $m$
$m$ = 24 when $frac{dm}{dt}$ = -1.2
therefore, $frac{dm}{dt}$ = $km$
However, as $frac{dm}{dt}$ is negative, and $m$ is positive, $k$ must be negative for the initial condition to be true.
Therefore:
$frac{dm}{dt}$ = -$km$
-1.2 = -24$k$
$k$ = $frac{1}{20}$
Therefore:
$frac{dm}{dt}$ = - $frac{m}{20}$
or $frac{dm}{dt}$ = - 0.05$m$
Is my method correct? Because the textbook claimed the answer was $frac{dm}{dt}$ = -0.005$m$
mathematical-modeling
mathematical-modeling
asked Jan 14 at 17:51
Sam ConnellSam Connell
62
asked Jan 14 at 17:51
Sam ConnellSam Connell
62
asked Jan 14 at 17:51
Sam ConnellSam Connell
62
asked Jan 14 at 17:51
Sam ConnellSam Connell
62
asked Jan 14 at 17:51
Sam ConnellSam Connell
62
62
add a comment |
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1 Answer
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Your method looks essentially correct. Maybe the book has a typographical error.
answered Jan 14 at 18:18
J. W. TannerJ. W. Tanner
1,9351114
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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$begingroup$
Your method looks essentially correct. Maybe the book has a typographical error.
answered Jan 14 at 18:18
J. W. TannerJ. W. Tanner
1,9351114
$endgroup$
add a comment |
$begingroup$
Your method looks essentially correct. Maybe the book has a typographical error.
answered Jan 14 at 18:18
J. W. TannerJ. W. Tanner
1,9351114
$endgroup$
add a comment |
$begingroup$
Your method looks essentially correct. Maybe the book has a typographical error.
answered Jan 14 at 18:18
J. W. TannerJ. W. Tanner
1,9351114
$endgroup$
Your method looks essentially correct. Maybe the book has a typographical error.
answered Jan 14 at 18:18
J. W. TannerJ. W. Tanner
1,9351114
answered Jan 14 at 18:18
J. W. TannerJ. W. Tanner
1,9351114
answered Jan 14 at 18:18
J. W. TannerJ. W. Tanner
1,9351114
answered Jan 14 at 18:18
J. W. TannerJ. W. Tanner
1,9351114
1,9351114
add a comment |
add a comment |
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