You are given that [math]a(t)=K t^2+L t+M[/math], for [math]0 \leq t \leq 2[/math], and that [math]a(0)=100, a(1)=110[/math], and [math]a(2)=136[/math].

Determine the force of interest at time [math]t=1 / 2[/math].

• .030
• .049
• .061
• .095
• .097

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.

Solution: E

[math]100=a(0)=M .110=a(1)=K+L+M[/math] so [math]K+L=10.136=a(2)=4 K+2 L+M[/math] so [math]18=2 K+L[/math] so [math]K=8[/math] and [math]L=2[/math]. Thus [math]a(t)=8 t^2+2 t+100[/math]. So [math]\delta(t)=\frac{a^{\prime}(t)}{a(t)}=\frac{16 t+2}{8 t^2+2 t+100}[/math] so [math]\delta(.5)=\frac{10}{103}=.097[/math].

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.