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Continuity And Differentiability

Question
CBSEENMA12036072

The population p(t) at time t of a certain mouse species satisfies the differential equation fraction numerator dp space left parenthesis straight t right parenthesis over denominator dt end fraction space equals space 0.5 space left parenthesis straight t right parenthesis space minus 450. if p (0) = 850, then the  time at which the population becomes zero is

  • 2 log 18

  • log 9

  • 1 half space log space 18
  • log 18

Solution

A.

2 log 18

fraction numerator straight d left parenthesis straight p left parenthesis straight t right parenthesis right parenthesis over denominator dt end fraction space equals space 1 half straight p left parenthesis straight t right parenthesis space minus 450
fraction numerator begin display style straight d left parenthesis straight p left parenthesis straight t right parenthesis right parenthesis end style over denominator begin display style dt end style end fraction space equals space fraction numerator straight p space left parenthesis straight t right parenthesis minus 900 over denominator 2 end fraction
2 integral fraction numerator straight d left parenthesis straight p left parenthesis straight t right parenthesis right parenthesis over denominator straight p left parenthesis straight t right parenthesis minus 900 end fraction equals integral dt
2 space ln space vertical line straight p left parenthesis straight t right parenthesis minus 900 vertical line space equals space straight t plus straight c
straight t equals 0
rightwards double arrow space 2 space ln space 50 space equals space 0 plus straight c
therefore space 2 space ln space vertical line straight p left parenthesis straight t right parenthesis minus 900 vertical line space equals space straight t space plus space 2 space ln space 50
straight p left parenthesis straight t right parenthesis space equals space 0 space
rightwards double arrow 2 space ln space 900 space equals space straight t plus 2 space ln space 50
straight t space equals space 2 space left parenthesis In space 900 minus In space 50 right parenthesis space equals space 2 space ln space open parentheses 900 over 50 close parentheses space
equals space 2 space ln space 18

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