The number of applications for​ patents, n, grew dramatically in recent​ years, with growth averaging about 3.43.4​% per year. that​ is, upper n prime left parenthesis t right parenthesisn′(t)equals=0.0340.034upper n left parenthesis t right parenthesisn(t). ​a) find the function that satisfies this equation. assume that tequals=0 corresponds to 19801980​, when approximately 110 comma 000110,000 patent applications were received. ​b) estimate the number of patent applications in 20152015. ​c) estimate the doubling time for upper n left parenthesis t right parenthesisn(t).

Respuesta :

Given that the number of applications for​ patents, n, grew dramatically in recent​ years, with growth averaging about 3.4​% per year.

Part A:

The function that satisfies the equation given that that t = 0 corresponds to 1980​, when approximately 110,000 patent applications were received is given by:

[tex]N(t)=110,000(1+0.034)^t \\ \\ N(t)=110,000(1.034)^t[/tex]

where, N(t) is the number of patent applications received at any particular year, t is the number of years after 1980.



Part B:

In 2015, there are 2015 - 1980 = 35 years after 1980.

The number of patent applications 35 years after 1980 is given by:

[tex]N(t)=110,000(1.034)^{35} \\ \\ =110,000(3.2227)=354,496[/tex]



Part C:

The doubling time for N(t) is the time it takes the number of patents to be 2(110,000) = 220,000

[tex]220,000=110,000(1.034)^t \\ \\ \Rightarrow(1.034)^t= \frac{220,000}{110,000} =2 \\ \\ \Rightarrow t\log{1.034}=\log2 \\ \\ \Rightarrow t= \frac{\log2}{\log1.034} = 20.73[/tex]

Therefore, the doubling time for N(t) is approximately 21 years.
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