Revision as of 17:39, 26 November 2023 by Admin (Created page with "On July 1, 1999, a person invested 1000 in a fund for which the force of interest at time <math>t</math> is given by <math>\delta_t=\frac{3+2 t}{50}</math>, where <math>t</math> is the number of years since January 1, 1999. Determine the accumulated value of the investment on January 1, 2000. <ul class="mw-excansopts"><li>1036</li><li>1041</li><li>1045</li><li>1046</li><li>1051</li></ul> '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests...")
ABy Admin
Nov 26'23
Exercise
On July 1, 1999, a person invested 1000 in a fund for which the force of interest at time [math]t[/math] is given by [math]\delta_t=\frac{3+2 t}{50}[/math], where [math]t[/math] is the number of years since January 1, 1999. Determine the accumulated value of the investment on January 1, 2000.
- 1036
- 1041
- 1045
- 1046
- 1051
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.
ABy Admin
Nov 26'23
Solution: D
[[math]]
\begin{aligned}
\textrm{AccAmt} & =1000 * \exp \left(\int_{.5}^1 \frac{3+2 t}{50}\right)=1000 * \exp \left(\left.\frac{3 t+t^2}{50}\right|_{.5} ^1\right) \\
& =1000 * \exp ((4-7 / 4) / 50)=1046.028
\end{aligned}
[[/math]]
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.