The population P = P(t) at time ‘t’ of a certain species follows the differential equation $\frac{d P}{d t} = 0.5 P - 450 .$ If P(0) = 850, then the time at which population becomes zero is:

Option 1 -

$\frac{1}{2} l o g_{e} 18$

Option 2 -

log_e9

Option 3 -

2log_e18

Option 4 -

log_e18

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

$\frac{d p}{d t} = 0.5 p - 450 a n d P (0) = 850$

$\Rightarrow \frac{d p}{P - 900} = 0.5 d t$

$\int_{850}^{0} \frac{d p}{P - 900} = \int_{0}^{T} 0.5 d t$

$\Rightarrow l n (P - 900) |_{805}^{0} = 0.5 T$

$\frac{T}{2} = l n | \frac{900}{50} | = l n 18$

T = 2 ln 18

<!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1818940951"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mfrac> <mrow> <mi>d</mi> <mi>p</mi> </mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> <mo>.</mo> <mn>5</mn> <mi>p</mi> <mo>−</mo> <mn>4</mn> <mn>5</mn> <mn>0</mn> <mtext> </mtext> <mi>a</mi> <mi>n</mi> <mi>d</mi> <mtext> </mtext> <mtext> </mtext> <mi>P</mi> <mrow> <mo> (</mo> <mrow> <mn>0</mn> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>8</mn> <mn>5</mn> <mn>0</mn> </mrow> </math>           <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1818940952"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mo>⇒</mo> <mfrac> <mrow> <mi>d</mi> <mi>p</mi> </mrow> <mrow> <mi>P</mi> <mo>−</mo> <mn>9</mn> <mn>0</mn> <mn>0</mn> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> <mo>.</mo> <mn>5</mn> <mi>d</mi> <mi>t</mi> </mrow> </math>         <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1818940953"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mstyle displaystyle="true"> <mrow> <munderover> <mo>∫</mo> <mrow> <mn>8</mn> <mn>5</mn> <mn>0</mn> </mrow> <mrow> <mn>0</mn> </mrow> </munderover> <mrow> <mfrac> <mrow> <mi>d</mi> <mi>p</mi> </mrow> <mrow> <mi>P</mi> <mo>−</mo> <mn>9</mn> <mn>0</mn> <mn>0</mn> </mrow> </mfrac> <mo>=</mo> <mstyle displaystyle="true"> <mrow> <munderover> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <mi>T</mi> </mrow> </munderover> <mrow> <mn>0</mn> <mo>.</mo> <mn>5</mn> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mstyle> </mrow> </mrow> </mstyle> </mrow> </math>           <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1818940954"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mo>⇒</mo> <mi>l</mi> <mi>n</mi> <mrow> <mo> (</mo> <mrow> <mi>P</mi> <mo>−</mo> <mn>9</mn> <mn>0</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> <msubsup> <mrow> <mo>|</mo> </mrow> <mrow> <mn>8</mn> <mn>0</mn> <mn>5</mn> </mrow> <mrow> <mn>0</mn> </mrow> </msubsup> <mo>=</mo> <mn>0</mn> <mo>.</mo> <mn>5</mn> <mi>T</mi> </mrow> </math> <math> <mrow> <mfrac> <mrow> <mi>T</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mi>l</mi> <mi>n</mi> <mrow> <mo>|</mo> <mrow> <mfrac> <mrow> <mn>9</mn> <mn>0</mn> <mn>0</mn> </mrow> <mrow> <mn>5</mn> <mn>0</mn> </mrow> </mfrac> </mrow> <mo>|</mo> </mrow> <mo>=</mo> <mi>l</mi> <mi>n</mi> <mn>1</mn> <mn>8</mn> </mrow> </math>           <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1029" DrawAspect="Content" ObjectID="_1818940955"> </o:OLEObject></xml><! [endif]->T = 2 ln 18

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The population P = P(t) at time ‘t’ of a certain species follows the differential equation $\frac{d P}{d t} = 0.5 P - 450 .$ If P(0) = 850, then the time at which population becomes zero is:

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The population P = P(t) at time ‘t’ of a certain species follows the differential equation d P d t = 0 . 5 P − 4 5 0 . If P(0) = 850, then the time at which population becomes zero is:

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The population P = P(t) at time ‘t’ of a certain species follows the differential equation $\frac{d P}{d t} = 0.5 P - 450 .$ If P(0) = 850, then the time at which population becomes zero is: