Not sure how to begin this Calc problem. (Weird givens)
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*This is NOT my homework assignment, but that of someone else. I still want to know how to do this. Can someone help me start it off? Much appreciated! *
=============================
$Q(t)$ represents the amount of a certain reactant present at time $t$
The rate of decrease of $Q(t)$ is proportional to $Q^3(t)$. That is, $Q'=-kQ^3$
Given that $Q(0)=frac{1}{8}$, how long will is take for the reactant to be reduced to one half of its original amount?
=============================
Here is my initial mindset:
Is that Q3 supposed to be Q cubed? Or is that just some weird notation?
Q(0) = 1/8, and you want to know when Q(t) = 1/16 (half)
Do you get Q(t) by taking the integral of Q' ?
Then use the initial value of (0, 1/8) to solve for the +C ?
And then set that resulting to 1/16 ?
calculus
asked Sep 29 '14 at 12:12
JackOfAll
2,11843354
add a comment |
up vote
0
down vote
favorite
*This is NOT my homework assignment, but that of someone else. I still want to know how to do this. Can someone help me start it off? Much appreciated! *
=============================
$Q(t)$ represents the amount of a certain reactant present at time $t$
The rate of decrease of $Q(t)$ is proportional to $Q^3(t)$. That is, $Q'=-kQ^3$
Given that $Q(0)=frac{1}{8}$, how long will is take for the reactant to be reduced to one half of its original amount?
=============================
Here is my initial mindset:
Is that Q3 supposed to be Q cubed? Or is that just some weird notation?
Q(0) = 1/8, and you want to know when Q(t) = 1/16 (half)
Do you get Q(t) by taking the integral of Q' ?
Then use the initial value of (0, 1/8) to solve for the +C ?
And then set that resulting to 1/16 ?
calculus
asked Sep 29 '14 at 12:12
JackOfAll
2,11843354
To find the form of $Q(t)$, separate your variables and integrate. (Google 'separation of variables' if you're not familiar with that.)
– Semiclassical
Sep 29 '14 at 12:17
A note on notation... often $Q^3(t)$ would denote $Q(Q(Q(t)))$ rather than $Q(t)^3$ or --- unambiguously --- $(Q(t))^3$. Here it probably denotes the latter.
– JP McCarthy
Sep 29 '14 at 12:20
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
*This is NOT my homework assignment, but that of someone else. I still want to know how to do this. Can someone help me start it off? Much appreciated! *
=============================
$Q(t)$ represents the amount of a certain reactant present at time $t$
The rate of decrease of $Q(t)$ is proportional to $Q^3(t)$. That is, $Q'=-kQ^3$
Given that $Q(0)=frac{1}{8}$, how long will is take for the reactant to be reduced to one half of its original amount?
=============================
Here is my initial mindset:
Is that Q3 supposed to be Q cubed? Or is that just some weird notation?
Q(0) = 1/8, and you want to know when Q(t) = 1/16 (half)
Do you get Q(t) by taking the integral of Q' ?
Then use the initial value of (0, 1/8) to solve for the +C ?
And then set that resulting to 1/16 ?
calculus
asked Sep 29 '14 at 12:12
JackOfAll
2,11843354
*This is NOT my homework assignment, but that of someone else. I still want to know how to do this. Can someone help me start it off? Much appreciated! *
=============================
$Q(t)$ represents the amount of a certain reactant present at time $t$
The rate of decrease of $Q(t)$ is proportional to $Q^3(t)$. That is, $Q'=-kQ^3$
Given that $Q(0)=frac{1}{8}$, how long will is take for the reactant to be reduced to one half of its original amount?
=============================
Here is my initial mindset:
Is that Q3 supposed to be Q cubed? Or is that just some weird notation?
Q(0) = 1/8, and you want to know when Q(t) = 1/16 (half)
Do you get Q(t) by taking the integral of Q' ?
Then use the initial value of (0, 1/8) to solve for the +C ?
And then set that resulting to 1/16 ?
calculus
calculus
asked Sep 29 '14 at 12:12
JackOfAll
2,11843354
asked Sep 29 '14 at 12:12
JackOfAll
2,11843354
edited Sep 29 '14 at 13:14
user940
asked Sep 29 '14 at 12:12
JackOfAll
2,11843354
asked Sep 29 '14 at 12:12
JackOfAll
2,11843354
asked Sep 29 '14 at 12:12
JackOfAll
2,11843354
2,11843354
To find the form of $Q(t)$, separate your variables and integrate. (Google 'separation of variables' if you're not familiar with that.)
– Semiclassical
Sep 29 '14 at 12:17
A note on notation... often $Q^3(t)$ would denote $Q(Q(Q(t)))$ rather than $Q(t)^3$ or --- unambiguously --- $(Q(t))^3$. Here it probably denotes the latter.
– JP McCarthy
Sep 29 '14 at 12:20
add a comment |
To find the form of $Q(t)$, separate your variables and integrate. (Google 'separation of variables' if you're not familiar with that.)
– Semiclassical
Sep 29 '14 at 12:17
A note on notation... often $Q^3(t)$ would denote $Q(Q(Q(t)))$ rather than $Q(t)^3$ or --- unambiguously --- $(Q(t))^3$. Here it probably denotes the latter.
– JP McCarthy
Sep 29 '14 at 12:20
To find the form of $Q(t)$, separate your variables and integrate. (Google 'separation of variables' if you're not familiar with that.)
– Semiclassical
Sep 29 '14 at 12:17
To find the form of $Q(t)$, separate your variables and integrate. (Google 'separation of variables' if you're not familiar with that.)
– Semiclassical
Sep 29 '14 at 12:17
A note on notation... often $Q^3(t)$ would denote $Q(Q(Q(t)))$ rather than $Q(t)^3$ or --- unambiguously --- $(Q(t))^3$. Here it probably denotes the latter.
– JP McCarthy
Sep 29 '14 at 12:20
A note on notation... often $Q^3(t)$ would denote $Q(Q(Q(t)))$ rather than $Q(t)^3$ or --- unambiguously --- $(Q(t))^3$. Here it probably denotes the latter.
– JP McCarthy
Sep 29 '14 at 12:20
add a comment |
1 Answer
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$Q^nleft(tright)$ can be ambiguous, yes. Here it means $left[Qleft(tright)right]^n$.
Integrate $dQ / dt=-kQ^3$:
$$
begin{eqnarray}
frac{dQ}{Q^3} &=& -kdt \
int_{Qleft(0right)}^{Qleft(tright)}frac{dQ}{Q^3} &=& -kint_{0}^{t}dt' \
-frac{1}{2}left[Q^{-2}left(tright) - Q^{-2}left(0right)right] &=& -kt \
Qleft(tright) &=& left[2kt + Q^{-2}left(0right)right]^{-1/2} \
Qleft(tright) &=& left(2kt + 64 right)^{-1/2}
end{eqnarray}
$$
If you just want to answer the question "how long will it take for the reactant to be reduced to one half of its original amount?" stop at the third line above, set $Qleft(t^*right) = Qleft(0right)/2$, and solve for $t^*$:
$$
begin{eqnarray}
t^* &=& frac{1}{2k}left{left[frac{Qleft(0right)}{2}right]^{-2} - Q^{-2}left(0right)right} \
&=& frac{96}{k}
end{eqnarray}
$$
answered Sep 29 '14 at 13:07
Eric Angle
1,63576
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
up vote
1
down vote
$Q^nleft(tright)$ can be ambiguous, yes. Here it means $left[Qleft(tright)right]^n$.
Integrate $dQ / dt=-kQ^3$:
$$
begin{eqnarray}
frac{dQ}{Q^3} &=& -kdt \
int_{Qleft(0right)}^{Qleft(tright)}frac{dQ}{Q^3} &=& -kint_{0}^{t}dt' \
-frac{1}{2}left[Q^{-2}left(tright) - Q^{-2}left(0right)right] &=& -kt \
Qleft(tright) &=& left[2kt + Q^{-2}left(0right)right]^{-1/2} \
Qleft(tright) &=& left(2kt + 64 right)^{-1/2}
end{eqnarray}
$$
If you just want to answer the question "how long will it take for the reactant to be reduced to one half of its original amount?" stop at the third line above, set $Qleft(t^*right) = Qleft(0right)/2$, and solve for $t^*$:
$$
begin{eqnarray}
t^* &=& frac{1}{2k}left{left[frac{Qleft(0right)}{2}right]^{-2} - Q^{-2}left(0right)right} \
&=& frac{96}{k}
end{eqnarray}
$$
answered Sep 29 '14 at 13:07
Eric Angle
1,63576
add a comment |
up vote
1
down vote
$Q^nleft(tright)$ can be ambiguous, yes. Here it means $left[Qleft(tright)right]^n$.
Integrate $dQ / dt=-kQ^3$:
$$
begin{eqnarray}
frac{dQ}{Q^3} &=& -kdt \
int_{Qleft(0right)}^{Qleft(tright)}frac{dQ}{Q^3} &=& -kint_{0}^{t}dt' \
-frac{1}{2}left[Q^{-2}left(tright) - Q^{-2}left(0right)right] &=& -kt \
Qleft(tright) &=& left[2kt + Q^{-2}left(0right)right]^{-1/2} \
Qleft(tright) &=& left(2kt + 64 right)^{-1/2}
end{eqnarray}
$$
If you just want to answer the question "how long will it take for the reactant to be reduced to one half of its original amount?" stop at the third line above, set $Qleft(t^*right) = Qleft(0right)/2$, and solve for $t^*$:
$$
begin{eqnarray}
t^* &=& frac{1}{2k}left{left[frac{Qleft(0right)}{2}right]^{-2} - Q^{-2}left(0right)right} \
&=& frac{96}{k}
end{eqnarray}
$$
answered Sep 29 '14 at 13:07
Eric Angle
1,63576
add a comment |
up vote
1
down vote
up vote
1
down vote
$Q^nleft(tright)$ can be ambiguous, yes. Here it means $left[Qleft(tright)right]^n$.
Integrate $dQ / dt=-kQ^3$:
$$
begin{eqnarray}
frac{dQ}{Q^3} &=& -kdt \
int_{Qleft(0right)}^{Qleft(tright)}frac{dQ}{Q^3} &=& -kint_{0}^{t}dt' \
-frac{1}{2}left[Q^{-2}left(tright) - Q^{-2}left(0right)right] &=& -kt \
Qleft(tright) &=& left[2kt + Q^{-2}left(0right)right]^{-1/2} \
Qleft(tright) &=& left(2kt + 64 right)^{-1/2}
end{eqnarray}
$$
If you just want to answer the question "how long will it take for the reactant to be reduced to one half of its original amount?" stop at the third line above, set $Qleft(t^*right) = Qleft(0right)/2$, and solve for $t^*$:
$$
begin{eqnarray}
t^* &=& frac{1}{2k}left{left[frac{Qleft(0right)}{2}right]^{-2} - Q^{-2}left(0right)right} \
&=& frac{96}{k}
end{eqnarray}
$$
answered Sep 29 '14 at 13:07
Eric Angle
1,63576
$Q^nleft(tright)$ can be ambiguous, yes. Here it means $left[Qleft(tright)right]^n$.
Integrate $dQ / dt=-kQ^3$:
$$
begin{eqnarray}
frac{dQ}{Q^3} &=& -kdt \
int_{Qleft(0right)}^{Qleft(tright)}frac{dQ}{Q^3} &=& -kint_{0}^{t}dt' \
-frac{1}{2}left[Q^{-2}left(tright) - Q^{-2}left(0right)right] &=& -kt \
Qleft(tright) &=& left[2kt + Q^{-2}left(0right)right]^{-1/2} \
Qleft(tright) &=& left(2kt + 64 right)^{-1/2}
end{eqnarray}
$$
If you just want to answer the question "how long will it take for the reactant to be reduced to one half of its original amount?" stop at the third line above, set $Qleft(t^*right) = Qleft(0right)/2$, and solve for $t^*$:
$$
begin{eqnarray}
t^* &=& frac{1}{2k}left{left[frac{Qleft(0right)}{2}right]^{-2} - Q^{-2}left(0right)right} \
&=& frac{96}{k}
end{eqnarray}
$$
answered Sep 29 '14 at 13:07
Eric Angle
1,63576
edited Sep 29 '14 at 13:15
answered Sep 29 '14 at 13:07
Eric Angle
1,63576
answered Sep 29 '14 at 13:07
Eric Angle
1,63576
answered Sep 29 '14 at 13:07
Eric Angle
1,63576
1,63576
add a comment |
add a comment |
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To find the form of $Q(t)$, separate your variables and integrate. (Google 'separation of variables' if you're not familiar with that.)
– Semiclassical
Sep 29 '14 at 12:17
A note on notation... often $Q^3(t)$ would denote $Q(Q(Q(t)))$ rather than $Q(t)^3$ or --- unambiguously --- $(Q(t))^3$. Here it probably denotes the latter.
– JP McCarthy
Sep 29 '14 at 12:20