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# Engineering

A bacterial infection in an internal wound grows as N{}'(t)=N_{o}exp(t),

where the time t is in hours. A dose of antibiotic, taken orally, needs 1 hour to reach the wound.

Once it reaches there, the bacterial population goes down as \frac{dN}{dt}=-5N^{2}.

What will be the plot of \frac{N_{o}}{N} vs. t after 1 hour ? 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

Answers (1)
S solutionqc

From o to 1 hour => N^{t}=N_{o}e^{t}

After 1 hour, it is given : \frac{dN}{dt}=-5N^{2}

\therefore at t = 1 hour , N^{'}=eN_{o}

Now,

\int_{eN_o}^{N}N^{-2}dN=-5\int_{1}^{t}dt

\frac{1}{N}-\frac{1}{eN_o}=5(t-1)

\frac{N_o}{N}=5N_o(t-1)+\frac{1}{e}

\frac{N_o}{N}=5N_ot+(\frac{1}{e}-5N_o)

This equation is similar to the straight line equation (Y=mx+C)

So, the curve will be straight line.

\therefore option (3) is correct.

 

 

 


Option 1)

Option 2)

Option 3)

Option 4)

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