# An influenza virus is spreading according to the function P(t)=50∙(2)^(t/2), where P is the number of people affected after t days?

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An influenza virus is spreading according to the function P(t)=50∙(2)^(t/2), where P is the number of people affected after t days.
a) Determine the number of people that had the virus initially.
b) Determine the number of people that will be infected in one week.
c) Determine how fast the virus will be spreading at the end of one week.
d) Determine the length of time it will take until 1000 people are infected.

Okay, so looks like we are dealing with a situation called an EXPONENTIAL FUNCTION. This is just a way to model very fast growth in populations, or even a fast decrease of something (ie radioactive material). In wild form, these pokemon look like P(t) = (a )x (b)^(t).

I will tell you what math skills you need to solve each problem.

USE ALGEBRA 1) 50 people. Plug in t=0 days into the function. P(0) = 50 x 2^(0/2) => 50 x 2^(0) => 50 x 1 => 50. Notice this is the value sitting outside the " (2)^(t/2) " term. (it looks like a = 50). This is known as the Initial Value for these types of problems.

USE ALGEBRA 2. 50 x 2^(3.5) = 565.69 => 566 people. Like we plugged in before, plug in t = 7 days into the function. P(7) = 50 x 2^(7/2). you get how to do this.

USE CALCULUS 3. ‘How fast’ is code for a rate,ie how fast are things changing, which is a derivative. So remember how to take derivatives of exponential functions? If not, look it up!
It looks like this in general form: P’(t) = 50 x [2^(t/2)] x (1/2) x ln (2). I read that as: (whatever constant ‘a’ stays in front) times (itself; like literally copy the original function behind the a) times (the derivative of the top) times (the natural log of the base).
Now, if you are trying to find HOW FAST something is changing at 1 week, that’s t = 7 days right? so evaluate (plug in) t = 7 in P’(t).

P’(7) = 50 x [2^(7/2)] x (1/2) x ln (2) = 196.05. NOW WHAT ARE THE UNITS?? VERY IMPORTANT. P’(7) = 196.05 people per day.
P(7) is how many total people are sick on the 7th day.
P’(7) is how many people are getting sick every day, on the 7th day.
very different things.

Can u tell me what P"(7) is? That means take the derivative again.
P"(7) is how many people are getting sick per day, per day. It just tells us that the derivative is increasing, or that the rate at which people are getting sick is faster than the people that got sick the day before. meaning that A LOT of people are going to get sick.

USE ALGEBRA 4. almost there. youre done with the actual calculus. Now we just want to know how long it takes (days) until 1000 people are ill, ie what is t when P(t) = 1000? Well, plug in 1000 for P(t) and solve:

P(t) = 50 x 2^(t/2) … rewrite original function
1000 = 50 x 2^(t/2) …set 1000 equal to original function
20 = 2^(t/2) … divide
log (20) =(t/2) x log (2) … use LOGS to get (t/2) down from exponent
log (20) / log (2) = (t/2) … divide
2 x log(20)/log(2) = t => 8.64 days… multiply t by 2 to get it alone