Question

between 2006 and 2016, the number of applications for patents, n, grew by about 4.4% per year. that is, n(t)=0.044n(t). a) find the function that satisfies this equation. assume that t = 0 corresponds to 2006, when approximately 444,000 patent applications were received. b) estimate the number of patent applications in 2020. c) estimate the rate of change in the number of patent applications in 2020. a) n(t) =

Explanation:

Step1: Identify the differential - equation type

The differential equation $N^{\prime}(t)=0.044N(t)$ is a first - order separable differential equation of the form $\frac{dN}{dt}=kN$, where $k = 0.044$. The general solution of $\frac{dN}{dt}=kN$ is $N(t)=N_0e^{kt}$, where $N_0$ is the initial value.

Step2: Find the initial value

Given that when $t = 0$ (corresponding to 2006), $N(0)=N_0 = 444000$. So the function $N(t)$ is $N(t)=444000e^{0.044t}$.

Step3: Calculate $t$ for 2020

The year 2020 corresponds to $t=2020 - 2006=14$.

Step4: Estimate the number of patent applications in 2020

Substitute $t = 14$ into $N(t)$: $N(14)=444000e^{0.044\times14}=444000e^{0.616}$. Using a calculator, $N(14)\approx444000\times1.8497\approx821267$.

Step5: Find the derivative of $N(t)$

Since $N(t)=444000e^{0.044t}$, then $N^{\prime}(t)=444000\times0.044e^{0.044t}=19536e^{0.044t}$.

Step6: Estimate the rate of change in 2020

Substitute $t = 14$ into $N^{\prime}(t)$: $N^{\prime}(14)=19536e^{0.616}\approx19536\times1.8497\approx36146.7$.

Answer:

a) $N(t)=444000e^{0.044t}$
b) Approximately $821267$
c) Approximately $36147$