From the question it is given that,
The bacteria count is 1,00,000
The number is increased by 10% in 2 hours.
So, let us assume C be the surface area of balloon,
\(( \dfrac{dC}{dt})\) ∝ C
Then,
\( \dfrac{dC}{dt}\) = λC
\( \dfrac{dC}{dt}\) = λ dt
Integrating on both side we get,
∫\( \dfrac{dC}{dt}\) = λ ∫dt
Log C = λt + log k … [equation (i)]
From question, t = 0 when c = 1,00,000,
log (1,00,000) = λ × 0 + log k … [equation (ii)]
log (1,00,000) = log k … [equation (iii)]
And also form question, given t = 2,
c = 1,00,000 + 1,00,000 × (\( \dfrac{10}{100}\))
= 110000
So, from equation (i) we have,
log 110000 = λ × 2 + log K … [equation (iv)]
Now, subtracting equation (ii) from equation (iv), we have,
Log 110000 – log 100000 = 2 λ
Then, log 11 × 10000 – log 10 × 10000 = 2 λ
Log \(\dfrac{ ((11 × 10000)}{(10 × 10000))}\) = 2λ
Log \(\dfrac{11}{10}\) = 2λ
So, λ = \( \dfrac{1}{2}\)log (\( \dfrac{11}{10}\)) … [equation (v)]
Now we need to find the time ‘t’ in which the count reaches 200000.
Then, substituting the values of λ and k from equation (iii) and (v) in equation (i),
Log 200000 = \( \dfrac{1}{2}\)log (\( \dfrac{11}{10}\))t + log 100000
\( \dfrac{1}{2}\)log (\( \dfrac{11}{10}\))t = log 200000 – log 100000
\( \dfrac{1}{2}\)log (\( \dfrac{11}{10}\))t = log\(\dfrac{ 200000}{100000}\)
\( \dfrac{1}{2}\)log(\( \dfrac{11}{10}\))t = log 2
Therefore, the required time t = \(\dfrac{ 2log2}{log}(\dfrac{11}{10}) hours\)
Answered by Aaryan | 1 year agoFind the particular solution satisfying the given condition \( (\dfrac{dy}{dx}) = (x – 1)\dfrac{dy}{dx} = 2x^3y\)
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