Question

1. a snorkeler dives for a shell on a reef. she is at the surface at 1 second and again at 11 seconds. at 2 seconds she is at a depth of 11/3 ft. write an equation that models the divers position with respect to time.

Explanation:

Step1: Assume a quadratic - function form

Since the diver's motion is likely parabolic (going down and coming back up), we assume the equation of the form $y = ax^{2}+bx + c$. The points $(1,0)$ and $(11,0)$ and $(2,\frac{11}{3})$ lie on the curve. Substituting $(1,0)$ into $y = ax^{2}+bx + c$ gives $a\times1^{2}+b\times1 + c=0$, so $a + b + c=0$.

Step2: Substitute the second - point

Substituting $(11,0)$ into $y = ax^{2}+bx + c$ gives $a\times11^{2}+b\times11 + c=0$, so $121a+11b + c=0$.

Step3: Substitute the third - point

Substituting $(2,\frac{11}{3})$ into $y = ax^{2}+bx + c$ gives $a\times2^{2}+b\times2 + c=\frac{11}{3}$, so $4a + 2b + c=\frac{11}{3}$.

Step4: Solve the system of equations

Subtract the first equation $a + b + c=0$ from the second equation $121a+11b + c=0$:
$(121a + 11b + c)-(a + b + c)=0 - 0$, which simplifies to $120a+10b = 0$, or $12a + b=0$, so $b=-12a$.
Subtract the first equation $a + b + c=0$ from the third equation $4a + 2b + c=\frac{11}{3}$:
$(4a + 2b + c)-(a + b + c)=\frac{11}{3}-0$, which simplifies to $3a + b=\frac{11}{3}$.
Substitute $b = - 12a$ into $3a + b=\frac{11}{3}$:
$3a-12a=\frac{11}{3}$, so $-9a=\frac{11}{3}$, and $a=-\frac{11}{27}$.
Since $b=-12a$, then $b=-12\times(-\frac{11}{27})=\frac{44}{9}$.
Substitute $a = -\frac{11}{27}$ and $b=\frac{44}{9}$ into $a + b + c=0$:
$-\frac{11}{27}+\frac{44}{9}+c=0$.
$-\frac{11}{27}+\frac{132}{27}+c=0$, so $\frac{-11 + 132}{27}+c=0$, $\frac{121}{27}+c=0$, and $c=-\frac{121}{27}$.

Answer:

$y=-\frac{11}{27}x^{2}+\frac{44}{9}x-\frac{121}{27}$