00:01
Hi there.
00:01
In this question, the rate of growth of bacteria is given by the differential equation, d .y over dt equal to 0 .5y and y of 0 is 200 because initially the amount of bacteria is 200.
00:14
In the first part, we have to solve 4 .y.
00:17
Here the given differential initial value problem becomes d .y over dt is equal to 0 .5y with the initial condition, y of 0 is equal to 200.
00:26
So we can rewrite this differential equation as dy over dt.
00:31
Y, d .y divided by y is equal to 0 .5 d t.
00:38
And from here we'll be getting this implies.
00:45
We'll be integrating both the sides and we'll be getting dy by y divided by y is equal to integral 0 .5 dt.
00:53
And from here we'll be getting lny is equal to 0 .5t plus c.
01:01
So taking exponential in both sides we'll be getting e power l n y is equal to e power 0 .5b plus c and we have a result that is e power ln x is equal to x for any x so using this result we'll be getting that y is equal to e power 0 .5 t plus c and this in turn becomes we have another result e power m plus n is equal to e power m multiplied by e power n so here we can use that result and this becomes y is equal to e power 0 .5 t multiplied by e power c and e power c can be considered to be as the constant k so we'll be getting y equal to k e power 0 .5 t.
01:59
So we got this equation and now we need to find the value of k for that we already have y of 0 is equal to 200.
02:06
So we need to put y is equal to 200 and t is equal to 0 in this particular equation.
02:13
So we'll be getting that 200 is equal to k multiplied by e power 0 .5 multiplied by 0 and that is equal to k multiplied by e power 0 and that is equal to k.
02:27
So from here we get that k is equal to 200.
02:33
So we have the solution y of t is equal to 200 e power 0 .5t.
02:41
Now we have to find the average number of bacteria in the population for 0 less than than r equal to t less than equal to 16.
02:50
Average number of the bacteria for 0 less than equal to t less than equal to 16 is equal to integral 0 to 16 dy and which is rewritten as integral 0 to 16 dy over dt d t.
03:03
So the average number is equal to integral integral 0 to 16, 0 .5 multiplied by 200 multiplied by e power 0 .5 d.
03:17
So this is equal to 0 .5 multiplied by 200 integral 0 to 16 e power 0 .5 t d t and that is equal to 0 .5 multiplied by 200 multiplied by e power 0 .5 t d t and that is equal to 0 .5 multiplied by 0 .5 multiplied by e power 0 .5...