between 2006 and 2016, the number of applications for patents, n, grew by about 4.4% per year. that is…
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)\na) find the function that satisfies this equation. assume that t = 0 corresponds to 2006, when approximately 444,000 patent applications were received\nb) estimate the number of patent applications in 2020\nc) estimate the rate of change in the number of patent applications in 2020.\na) n(t)=444000e^{0.044t}\nb) the number of patent applications in 2020 will be 821267 (round to the nearest whole number as needed.)
Answer
Explanation:
Step1: Solve the differential equation
The differential equation $N'(t)=0.044N(t)$ is a first - order separable differential equation. The general solution of the differential equation $\frac{dN}{dt}=kN$ is $N(t)=N_0e^{kt}$, where $N_0$ is the initial value and $k$ is the growth rate. Here $k = 0.044$ and when $t = 0$ (corresponding to 2006), $N(0)=N_0 = 444000$. So the function is $N(t)=444000e^{0.044t}$.
Step2: Calculate $t$ for 2020
Since $t = 0$ corresponds to 2006, for 2020, $t=2020 - 2006=14$.
Step3: Find the number of patent applications in 2020
Substitute $t = 14$ into $N(t)=444000e^{0.044t}$. So $N(14)=444000e^{0.044\times14}=444000e^{0.616}\approx444000\times1.8272\approx812167$.
Step4: Find the rate of change in 2020
The rate of change is given by $N'(t)$. Since $N'(t)=0.044N(t)$, when $t = 14$, $N'(14)=0.044\times N(14)$. Substitute $N(14)\approx812167$ into it, we get $N'(14)=0.044\times812167\approx35735$.
Answer:
a) $N(t)=444000e^{0.044t}$ b) $812167$ c) $35735$