Answer to Question #211899 in Calculus for Delmundo

Question

1. Suppose a population of virus causing this pandemic known as COVID19

increases according to the law of exponential growth. Let’s say there were 130

viruses on the third day and 380 viruses on the seventh day. Approximately

how many COVID-19 viruses were in the original population?

2. How long will it take for an entrepreneur with an investment worth Php185,000

to double its worth if it is valued at Php260,000 two years after? What is the

exponential equation model for the investment?

3. A certain strain of virus being observed in a laboratory doubles its number

every 12 hours. The virologist counted the virus population to be 350. How

many viruses will there be in 10 hours?

4. The development of vaccines slows down the infection rate of a certain cow

disease exponentially. As of now, 900 cows are infected and in 2 days, only

556 cows are still expected to be infected. How many cows were initially

infected 5 days ago?

Answer 1

Part 1

$P(3)=P_0e^{3k}\\ P(7)=P_0e^{7k}\\ \frac{130}{e^{3k}}=\frac{380}{e^{7k}}\\ \ln \left(130\right)-3k=\ln \left(380\right)-7k\\ k=\frac{\ln \left(\frac{38}{13}\right)}{4} =0.268\\ P_0=\frac{130}{e^{3*0.268}}=58.17958 = 59$

Part 2

$P(t)=P_0e^{tk}\\ 260000=185000e^{2t}\\ \frac{185000e^{2t}}{185000}=\frac{260000}{185000}\\ t=\frac{\ln \left(\frac{52}{37}\right)}{2}=0.17016 years$

Part 3

$P=350*2^{h/12}\\ P=350*2^{10/12}\\ P=623.6=624$

Part 4

$Y(x)=ka^x\\ Y(5)=ka^5=900\\ Y(7)=ka^7=900+556=1956\\ k=270\\ Y(0)= 270$