The Force$\ { F }$ (in newtons) of a hydraulic cylinder in a press is proportional to the square of $\sec x$ where $x$ is the distance (in meters) that the cylinder is extended in its cycle. The domain of $F$ is $\left[ 0 , \frac { \pi } { 3 } \right]$ and $F ( 0 ) = 150$. Find $F$ as a function of $x$.
A) $F ( x ) = 150 \sec ^ { 2 } x$
B) $F ( x ) = 150 + \sec ^ { 2 } x$
C) $F ( x ) = 300 \sec ^ { 2 } x$
D) $F ( x ) = \sec ^ { 2 } x + 1$
E) $F ( x ) = 75 \sec ^ { 2 } x$

Question

Quizplus · Accepted Answer

Given that $F(x) = k \sec^2 x$ and $F(0) = 150$, substituting $x = 0$ gives $F(0) = k \sec^2(0) = k(1) = 150$, so $k = 150$. Therefore, $F(x) = 150 \sec^2 x$.

The Force F\ { F } F (In Newtons) of a Hydraulic Cylinder in a Press Is

The Force $\ { F }$ (In Newtons) of a Hydraulic Cylinder in a Press Is