question_answer

The rate of increase of bacteria in c culture is proportional to the number of bacteria present. If the original number of bacteria doubles in two hours, in how many hours will it be five times?

OR

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point \[(-\,4,\,\,-\,3).\] Find the equation o: the curve given that it passes through \[(-\,2,\,\,1).\]

Answer:

Let the initial count of bacteria\[={{N}_{0}}\]

at any time t the count of bacteria\[=N\]

Then, \[\frac{dN}{dt}\propto N\Rightarrow \frac{dN}{dt}=kN\]

[where, k is a constant]

\[\Rightarrow \]   \[\frac{dN}{N}=kdt\]

[by separating the variable]

On integrating both sides, we get

\[\int{\frac{1}{N}dN}=k\int{dt\Rightarrow \log N=kt+C}\]            ?(i)

We have, \[N={{N}_{0}}\]at \[t=0\]

\[\therefore \]      \[\log {{N}_{0}}=0+C\Rightarrow C=\log {{N}_{0}}\]

On putting \[C=\log {{N}_{0}}\] in Eq. (i), we get

\[\log N=kt+\log {{N}_{0}}\]

\[\Rightarrow \]   \[\log \left( \frac{N}{{{N}_{0}}} \right)=kt\]                               ?(ii)

According to the question,

When\[t=2\,h,\]then \[N=2{{N}_{0}}\]

\[\therefore \]      \[\log \left( \frac{2{{N}_{0}}}{{{N}_{0}}} \right)=2k\]

[putting \[N=2{{N}_{0}}\]and \[t=2\] in Eq. (ii)]

\[\Rightarrow \]   \[k=\frac{1}{2}\log 2\]

On putting \[k=\frac{1}{2}\log 2\]in Eq. (ii), we get

\[\log \left( \frac{N}{{{N}_{0}}} \right)=\left( \frac{1}{2}\log 2 \right)t\]

\[\Rightarrow \]   \[t=\frac{2}{{{\log }_{2}}}\log \left( \frac{N}{{{N}_{0}}} \right)\]

When the count of bacteria becomes 5 times

i.e.\[5{{N}_{0}}\]in \[{{t}_{1}}\] hours.

Then,    \[{{t}_{1}}=\frac{2}{\log 2}\log \left( \frac{5{{N}_{0}}}{{{N}_{0}}} \right)\]

\[\Rightarrow \]   \[{{t}_{1}}=\frac{2}{\log 2}(\log 5)=\frac{2\log 5}{\log 2}h\]

OR

It is given that (x, y) is the point of contact of the curve and its tangent.

The slope of the line segment joining the points\[({{x}_{2}},\,\,{{y}_{2}})=(x,\,\,y)\]and\[({{x}_{1}},\,\,{{y}_{1}})=(-\,4,\,\,-\,3)\].

\[=\frac{y-(-\,3)}{x-(-\,4)}=\frac{y+3}{x+4}\]

\[\left[ \because \text{slope}\,\,\text{of}\,\,\text{tangent}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right]\]

According to the question, (slope of tangent is twice the slope of the line), we must have,

\[\frac{dy}{dx}=2\left( \frac{y+3}{x+4} \right)\]

Now, separating the variables, we get

\[\frac{dy}{y+3}=\left( \frac{2}{x+4} \right)\,\,dx\]

On integrating both sides, we get

\[\int{\frac{dy}{y+3}}=\int{\left( \frac{2}{x+4} \right)\,dx}\]

\[\Rightarrow \]   \[\log |y+3|\,\,=2\log |x+4|+\log |C|\]

\[\Rightarrow \]   \[\log |y+3|\,\,=\log |x+4{{|}^{2}}+\log |C|\]

\[\Rightarrow \]   \[\log \frac{|y+3|}{|x+4{{|}^{2}}}=\log |C|\]

\[\left[ \because \log m-\log n=\log \frac{m}{n} \right]\]

\[\Rightarrow \]   \[\frac{|y+3|}{|x+4{{|}^{2}}}=C\]                         ?(ii)

The curve passes through the point \[(-\,2,\,\,1),\] therefore

\[\frac{|1+3|}{|-\,2+4{{|}^{2}}}=C\Rightarrow C=1\]

On putting \[C=1\]in Eq. (ii), we get

\[\frac{|y+3|}{|x+4{{|}^{2}}}=1\Rightarrow y+3={{(x+4)}^{2}}\]

which is the required equation of curve.