Question

between 2006 and 2016, the number of applications for patents, n, grew by about 4.1% per year. that is, n(t)=0.041n(t). a) find the function that satisfies this equation. assume that t = 0 corresponds to 2006, when approximately 453,000 patent applications were received. b) estimate the number of patent applications in 2021. c) estimate the rate of change in the number of patent applications in 2021. a) n(t)=453000e^{0.041t} b) the number of patent applications in 2021 will be (round to the nearest whole number as needed.)

Explanation:

Step1: Determine the value of t for 2021

Since \(t = 0\) corresponds to 2006, for 2021, \(t=2021 - 2006=15\).

Step2: Substitute t into the function

We have the function \(N(t)=453000e^{0.041t}\). Substitute \(t = 15\) into it:
\[N(15)=453000e^{0.041\times15}\]
\[N(15)=453000e^{0.615}\]
Since \(e^{0.615}\approx1.8499\), then \(N(15)=453000\times1.8499\approx837005\).

Step3: Find the derivative of the function

The derivative of \(N(t)=453000e^{0.041t}\) is \(N'(t)=0.041\times453000e^{0.041t}\) according to the chain - rule \((e^{ax})'=ae^{ax}\).

Step4: Substitute t = 15 into the derivative

Substitute \(t = 15\) into \(N'(t)\):
\[N'(15)=0.041\times453000e^{0.041\times15}\]
\[N'(15)=0.041\times453000\times e^{0.615}\]
\[N'(15)=0.041\times453000\times1.8499\]
\[N'(15)\approx34317\]

Answer:

b) 837005
c) 34317