Respuesta :
The formula for exponential growth is as follows:
[tex]P=P(0)e^{kt}[/tex]In the problem, the following are the given data:
[tex]\begin{gathered} P(0)=65 \\ P=130 \\ t=32 \end{gathered}[/tex]Substitute the values into the equation and then solve for k. Thus, we have the following.
[tex]\begin{gathered} 130=65e^{k(32)} \\ 130=65e^{32k} \\ \frac{130}{65}=e^{32k} \\ 2=e^{32k} \\ \ln 2=\ln (e^{32k}) \\ \ln 2=32k \\ k=\frac{\ln2}{32} \\ k\approx0.02166 \end{gathered}[/tex]This means that the growth rate is approximately 2.166%.
To identify the temperature at 45 minutes, substitute 45 min to the value of t in the formula. Use 65 for P(0) and 0.02166 for k. Thus, we have the following.
[tex]\begin{gathered} P=P(0)e^{kt} \\ P=65e^{(0.02166)(45)} \\ P=65e^{0.9747} \\ P\approx65(2.65037) \\ P\approx172.274 \end{gathered}[/tex]This means that the object will reach approximately 172.274° in 45 minutes.
To identify the time it will take to reach 230°, substitute 65 for P(0), 230 for P, and 0.02166 for k. Thus, we obtain the following:
[tex]\begin{gathered} P=P(0)e^{kt} \\ 230=65e^{0.02166t} \\ \frac{230}{65}=e^{0.02166t} \\ \ln \frac{230}{65}=\ln (e^{0.02166t}) \\ \ln \frac{230}{65}=0.02166t^{} \\ t=\frac{\ln \frac{230}{65}}{0.02166} \\ t\approx\frac{1.26369}{0.02166} \\ t\approx58.342 \end{gathered}[/tex]Therefore, it will take approximately 58.342 minutes to reach 230°.
