00:01
In this portion, we are given a function p of t is equal to 10 times natural law of 0 .19 plus 1 over 0 .190 plus 1.
00:16
This represents the number of people infected t days after the spread of the pandemic.
00:24
We need to obtain the number of days after which the number of days, after which the number of days, after which the patient infected will start getting decreased.
00:36
Now they will start decreasing after the function p of t, that is number of people infected, has reached a maximum value.
00:47
After that, it will start decreasing.
00:50
Now let us first obtain the number of days after which the number of patients infected would be maximum.
00:59
For that, we'll need to differentiate the given function.
01:02
Refrantiate the given function with this pectivity, that is d over dt of 3 of t is equal to d over dt of 10 times natural of 0 .190 plus 1 over 0 .19 plus 1.
01:22
Solving this we get p dash of t is equal to 10 times of d over dt of natural love of 0 .190 plus 1.
01:34
Over 0 .190 plus 1.
01:39
Further, using the quotient rule, the derivative is given by denominator times the derivative of numerator, the derivative log x is 1 over x, therefore 1 over 0 .291, and then the derivative of 0 .190 is 0 .19, minus numerator derivative of denominator over denominator square further.
02:24
So all in this we get, this will cancel out with this and we are left with 10 times of 0 .19 times 1 minus natural log of 0 .19 plus 1 plus 1 over 0 .19 plus 1 over 0 .19 .000.
02:43
20 plus 1 whole square...