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.
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