Exercise 10.8-Additional Problems - Chapter 10 Ordinary Differential Equations 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
In a certain chemical reaction the rate of conversion of a substance at time $t$ is proportional to the quantity of the substance still untransformed at that instant. At the end of one hour, 60 grams remain and at the end of 4 hours 21 grams. How many grams of the substance was there initially?
Solution:
Let $\mathrm{A}$ be the substance at time $\mathrm{t}$
$
\frac{d \mathrm{~A}}{d t} \propto \mathrm{A} \Rightarrow \frac{d \mathrm{~A}}{d t}=k \mathrm{~A} \Rightarrow \mathrm{A}=c e^{l t}
$
When $t=1, \mathrm{~A}=60 \Rightarrow c e^k=60$
When $t=4, \mathrm{~A}=21 \Rightarrow c e^{4 k}=21$
(1) $\Rightarrow c^4 e^{4 k}=60^4$
$\frac{(3)}{(2)} \Rightarrow c^3=\frac{60^4}{21} \Rightarrow c=85.15$ (by using log)
Initially i.e., when $\mathrm{t}=0, \mathrm{~A}=\mathrm{c}=85.15 \mathrm{gms}$ (app.)
Hence initially there was $85.15$ gms (approximately) of the substance.
Question 2.
The temperature $\mathrm{T}$ of a cooling object drops at a rate proportional to the difference $\mathrm{T}-\mathrm{S}$, where $\mathrm{S}$ is constant temperature of surrounding medium. If initially $\mathrm{T}=150^{\circ} \mathrm{C}$, find the temperature of the cooling object at any time t.
Solution:
Let $\mathrm{T}$ be the temperature of the cooling object at any time $t$
$\frac{d \mathrm{~T}}{d t} \propto(\mathrm{T}-\mathrm{S}) \Rightarrow \frac{d \mathrm{~T}}{d t}=k(\mathrm{~T}-\mathrm{S}) \Rightarrow \mathrm{T}-\mathrm{S}=c e^{k t}$, where $k$ is negative
$
\Rightarrow \mathrm{T}=\mathrm{S}+c e^{k t}
$

when $\mathrm{t}=0, \mathrm{~T}=150 \Rightarrow 150=\mathrm{S}+\mathrm{c} \Rightarrow \mathrm{c}=150-\mathrm{S}$
$\therefore$ The temperature of the cooling object at any time $\mathrm{T}=\mathrm{S}+(150-\mathrm{S}) \mathrm{e}^{\mathrm{kt}}$
Note:
Since $\mathrm{k}$ is negative, as $\mathrm{t}$ increases $\mathrm{T}$ decreases.
It is a decay problem. Instead of $\mathrm{k}$ one may take $-\mathrm{k}$ where $\mathrm{k}>0$. Then the answer is $\mathrm{T}=\mathrm{S}+(150-\mathrm{S}) \mathrm{e}^{-\mathrm{kt}}$.
Again, as $t$ increases $T$ decreases.
Question 3.
The sum of Rs. 1000 is compounded continuously, the nominal rate of interest being four per cent per annum. In how many years, will the amount be twice the original principal? $\left[\log _e 2=0.6931\right]$

Solution:
Rate of interest $=4 \%$
$
\begin{aligned}
\frac{d x}{d t} & =0.04 x \\
\frac{d x}{x} & =0.04 d t \Rightarrow \log x=0.04 t+c \Rightarrow x=c e^{0.04 t} \\
\text { when } t & =0, x=x_0 \Rightarrow c e^0=x_0 \\
\Rightarrow c & =x_0 \\
\text { when } x & =2 x_0 \\
\Rightarrow 2 x_0 & =x_0 e^{0.04 t} \\
\text { (i.e.,) } 2 & =e^{0.04 t}
\end{aligned}
$
Taking log on both sides,
$
\text { (i.e.,) } \begin{aligned}
\log _e 2 & =0.04 t \\
0.04 t & =0.6931 \\
\therefore t & =\frac{0.6931}{0.04}=17.3275 \approx 17 \text { years. }
\end{aligned}
$
Question 4.
A cup of coffee at temperature $100^{\circ} \mathrm{C}$ is placed in a room whose temperature is $15^{\circ} \mathrm{C}$ and it cools to $60^{\circ} \mathrm{C}$ in 5 minutes. Find its temperature after a further interval of 5 minutes.
Solution:

$
\begin{aligned}
& \frac{d \mathrm{~T}}{d t} \propto \mathrm{T}-15 \Rightarrow \frac{d \mathrm{~T}}{d t}=k(\mathrm{~T}-15) \\
& \int \frac{d \mathrm{~T}}{\mathrm{~T}-15}=k \int d t \\
& \Rightarrow \log (\mathrm{T}-15)=k t+c(\text { or }) \mathrm{T}-15=c e^{k t} \\
& \quad \therefore \mathrm{T}=c e^{k t}+15 \\
& \text { when } \mathrm{T}=100, t=0 \\
& \Rightarrow 100= c e^0+15=c+15 \\
& \therefore c= 100-15=85 \\
& \text { when } t= 5, \mathrm{~T}=60 \\
& \Rightarrow 60= 85 e^{5 k}+15 \\
& \therefore 85 e^{5 k}= 60-15=45 \\
& \therefore e^{5 k}= \frac{45}{85}=\frac{9}{17} \\
& \text { when } t=10, \\
& \mathrm{~T}= 85 e^{10 k}+15=85 e^{(5 k)^2}+15 \\
&= 85\left(\frac{9}{17}\right)^2+15 \\
&= 85\left(\frac{9 \times 9}{17 \times 17}\right)+15 \\
&= 23.823+15=38.823=38.82^{\circ} \mathrm{C} .
\end{aligned}
$
Question 5.
The rate at which the population of a city increases at any time is proportional to the population at that time. If there were 1,30,000 people in the city in 1960 and 1,60,000 in 1990, what population may be anticipated in 2020 ? $\left[\log _{\mathrm{e}}\left(\frac{16}{3}\right)=0.2070 ; \mathrm{e}^{-0.42}=1.52\right]$
Solution:

$
\begin{aligned}
& \text { Given } \frac{d x}{d t} \propto x \Rightarrow \frac{d x}{d t}=k x \\
& \therefore \frac{d x}{x}=k d t \\
& \log x=k t+c \\
& \therefore x=e^{k t+c}=c e^{k t} \\
& \text { At } t=1960, x=130000 \\
& \Rightarrow 1,30,000=c e^{1960 k} \\
& \therefore c=\frac{1,30,000}{e^{1960 k}} \\
& \text { At } t=1990, \quad x=1,60,000 \\
& \Rightarrow 1,60,000=c e^{1990 k} \\
& =\frac{1,30,000}{e^{1960 k}} e^{1990 k} \\
& =1,30,000 e^{30 k} \\
& \therefore e^{30 k}=\frac{1,60,000}{1,30,000}=\frac{16}{3} \\
& \therefore 30 k=\log _e\left(\frac{16}{3}\right)=0.2070 ; k=0.007 \\
&
\end{aligned}
$
$\therefore \mathrm{P}=1,30,000 ; \mathrm{e}^{0.007}$ is the population at any time ' $\mathrm{t}$ '
In the year 2020 , (i.e.,) when $\mathrm{t}=60$
$
\mathrm{x}=130000 \times \mathrm{e}^{0.42}=1,30,000 \times 1.52=1,97,600
$
So, the anticipated population in 2020 will be $1,97,600$