Solution
- The equation for the potential given is:
[tex]V(t)=320e^{-3.1t}[/tex]Question 1:
- To find when the potential is 150V, we simply substitute the value of V = 150 into the equation and then find the corresponding value of t.
- Thus, we have:
[tex]\begin{gathered} V=150 \\ \\ 150=320e^{-3.1t} \\ \\ \text{ Divide both sides by 320} \\ \\ e^{-3.1t}=\frac{150}{320}=\frac{15}{32} \\ \\ \text{ Take the natural log of both sides} \\ \\ \ln e^{-3.1t}=\ln(\frac{15}{32}) \\ \\ -3.1t=\ln(\frac{15}{32}) \\ \\ \text{ Divide both sides by -3.1} \\ \\ t=-\frac{1}{3.1}\ln(\frac{15}{32}) \\ \\ t=0.2444s\text{ \lparen To 4 decimal places\rparen} \end{gathered}[/tex]Question 2:
- The rate at which the changing occurs is gotten by differentiating the function with respects to time.
- That is,
[tex]\begin{gathered} V(t)=320e^{-3.1t} \\ \\ V^{\prime}(t)=\frac{d}{dt}(320e^{-3.1t}) \\ \\ V^{\prime}(t)=320(-3.1e^{-3.1t}) \\ \\ V^{\prime}(t)=-992e^{-3.1t} \end{gathered}[/tex]- Now that we have the expression for the rate of change of potential with time, we can proceed to find how fast the changing of potential V is happening at t = 0.2444s.
- Thus, we have:
[tex]\begin{gathered} V^{\prime}(t)=-992e^{-3.1t} \\ put\text{ }t=0.2444 \\ \\ V^{\prime}(t)=-992e^{-3.1\times0.2444} \\ \\ \therefore V^{\prime}(t)=-465.02125...\approx-465.0\text{ v/s} \end{gathered}[/tex]Question 3:
- The voltage is changing at -50v/s when we substitute V'(t) = -50 into the equation for V'(t).
- We have that:
[tex]\begin{gathered} V^{\prime}(t)=-992e^{-3.1t} \\ \\ V^{\prime}(t)=-50 \\ \\ -50=-992e^{-3.1t} \\ \\ \text{ Divide both sides by -992} \\ \\ e^{-3.1t}=\frac{-50}{-992}=\frac{25}{496} \\ \\ \text{ Take the natural log of both sides} \\ \ln e^{-3.1t}=\ln(\frac{25}{496}) \\ \\ -3.1t=\ln(\frac{25}{496}) \\ \\ \text{ Divide both sides by -3.1} \\ \\ \therefore t=-\frac{1}{3.1}\ln(\frac{25}{496}) \\ \\ t=0.96377...\approx0.9638seconds\text{ \lparen To 4 decimal places\rparen} \end{gathered}[/tex]Final Answers
Question 1: 0.2444 seconds
Question 2: -465.0v/s
Question 3: 0.9638 seconds
