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Sagot :
Answer:
a.)Q' = -kQ + r
b.)Q = [tex]\frac{r}{k} [ 1 - e^{-kt} ][/tex]
c.) Limiting long run value of Q = [tex]\frac{r}{k}[/tex]
Step-by-step explanation:
a.)
The rate of change is directly proportional to the quantity, Q
⇒[tex]\frac{dQ}{dt}[/tex] ∝ Q
⇒[tex]\frac{dQ}{dt}[/tex] = -kQ ( because it is decreasing )
Also given, quantity is increasing with a constant rate r
⇒[tex]\frac{dQ}{dt}[/tex] = -kQ + r
⇒Q' = -kQ + r
b.)
As we have
[tex]\frac{dQ}{dt}[/tex] = -kQ + r
⇒[tex]\frac{dQ}{-kQ + r}[/tex] = dt
⇒∫[tex]\frac{dQ}{-kQ + r}[/tex] = ∫dt
⇒-[tex]\frac{1}{k}log(-kQ + r) = t + C[/tex]
⇒log(-kQ + r) = -kt -kC
Taking exponential both side, we get
⇒-kQ + r = [tex]e^{-kt +A}[/tex]
⇒-kQ = -r + [tex]e^{-kt +A}[/tex]
⇒Q = [tex]\frac{r}{k}[/tex] - [tex]\frac{1}{k}[/tex][tex]e^{-kt +A}[/tex]
⇒Q = [tex]\frac{r}{k}[/tex] - [tex]\frac{1}{k}[/tex][tex]e^{-kt}.e^{A}[/tex]
⇒Q = [tex]\frac{r}{k}[/tex] - [tex]\frac{1}{k}[/tex][tex]Be^{-kt}[/tex] .......(1)
Now,
At t = 0, Q = 0
0 = [tex]\frac{r}{k}[/tex] - [tex]\frac{1}{k}[/tex]B
⇒[tex]\frac{1}{k}B = \frac{r}{k}[/tex]
⇒B = r
∴ equation (1) becomes
Q = [tex]\frac{r}{k}[/tex] - [tex]\frac{r}{k}[/tex][tex]e^{-kt}[/tex]
⇒Q = [tex]\frac{r}{k} [ 1 - e^{-kt} ][/tex]
c.)
for limiting long run value of Q
[tex]\lim_{n \to \infty} Q = \lim_{n \to \infty}[/tex] [tex]\frac{r}{k} [ 1 - e^{-kt} ][/tex]
= [tex]\frac{r}{k}[/tex][tex]\lim_{n \to \infty} [ 1 - e^{-kt} ]= \frac{r}{k} [ 1 - e^{\infty} ] = \frac{r}{k}[ 1-0][/tex]
= [tex]\frac{r}{k}[/tex]
⇒Limiting long run value of Q = [tex]\frac{r}{k}[/tex]
Q' = -kQ + r
Q = r/k (1- [tex]\rm e^{-kt}[/tex])
The limiting long-run value of Q = [tex]\rm \frac{r}{k}[/tex]
A drug is administered intravenously at a constant rate of r mg/hour and is excreted at a rate proportional to the quantity present, with a constant of proportionality k > 0.
What is a differential equation?
The differential equation is an equation that contains the derivative of the unknown function.
a) It is given that the rate of change is directly proportional to the quantity Q
[tex]\rm \frac{dQ}{dt}[/tex] ∝ Q
[tex]\rm \frac{dQ}{dt}[/tex] = -kQ
So, Q' = -kQ + r
where quantity Q is increasing with constant rate r and k is unknown constant.
b) Q' = -kQ + r
[tex]\rm \frac{dQ}{dt}[/tex] = -kQ + r
[tex]\rm \frac{dQ}{-kQ+r}=dt\\[/tex]
[tex]\int\limits {\rm \frac{dQ}{-kQ+r} = \int\limits dt[/tex]
log(-kQ + r) = -kt -kC
by taking exponential both side, we get
-kQ + r = [tex]e^{-kt+A}[/tex]
-kQ = -r + [tex]e^{-kt+A}[/tex]
-Q /k= -r/k + [tex]e^{-kt+A}[/tex]/k
Q = -r/k -[tex]\rm \frac{1}{k}Be^{-kt}[/tex]
At t = 0, Q = 0
[tex]\rm \frac{1}{k} B=\frac{r}{k}[/tex]
B = r
by substituting the value in the above equation
Q = r/k (1- [tex]\rm e^{-kt}[/tex])
c) The limiting long-run value of Q
[tex]\rm \lim_{n \to \infty} Q= \lim_{n \to \infty} \frac{r}{k} [1-e^{-kt} ]\\\rm= \frac{r}{k}\lim_{n \to \infty} [1-e^{-kt} ]\\\rm =\frac{r}{k}[1-0]\\\rm=\frac{r}{k}[/tex]
The value Q =[tex]\rm \frac{r}{k}[/tex]
Learn more about differential equations here:
https://brainly.com/question/25731911
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