Solve the problem.
-A cross-section of an irrigation canal is a parabola. If the surface of the water is $40 \mathrm { ft }$ wide and the canal is $26 \mathrm { ft }$ deep at the center, how deep is it $1 \mathrm { ft }$ from the edge?
A) $2.5 \mathrm { ft }$
B) $1.3 \mathrm { ft }$
C) $23.5 \mathrm { ft }$
D) $24.7 \mathrm { ft }$

Question

Quizplus · Accepted Answer

The equation of the parabola is $y=ax^2$. Since the surface of the water is $40 \mathrm { ft }$ wide and the canal is $26 \mathrm { ft }$ deep at the center, we have $y=26$ when $x=20$. Substituting these values into the equation, we get $26=a(20)^2$, so $a=\frac{1}{16}$. Therefore, the equation of the parabola is $y=\frac{1}{16}x^2$. When $x=1$, we have $y=\frac{1}{16}(1)^2=\frac{1}{16}$, so the canal is $2.5 \mathrm { ft }$ deep $1 \mathrm { ft }$ from the edge.

Solve the Problem 40ft40 \mathrm { ft }40ft Wide and the Canal Is

Solve the Problem $40 \mathrm { ft }$ Wide and the Canal Is