# Trigonometric Functions Question (Area underneath a curve). Help please?

## An arched window has a base length of 4m and a height of 2m. The arch is to be either an arc of a parabola or a half-period of a sine curve. (a) If the arch is the arch of a parabola, the equation of the curve is $f \left(x\right) = a x \left(4 - x\right)$. Show that the value of $a$ is $\frac{1}{2}$. (b) If the arch is sinusoidal, the equation is the form $g \left(x\right) = A \sin \left(\frac{\pi x}{4}\right)$. Find the value of $A$. (c) Calculate the area for each window design and hence, determine which one has less area. Thanks!

Oct 3, 2017

Area of f(x) : $5.33 {m}^{2}$
Area of g(x) : $5.09 {m}^{2}$

The sinusoidal design has lesser area.

#### Explanation:

The height of the parabolic design is at the vertex, with the coordinate $\left(2 , 2\right)$:
$2 = a \cdot 2 \cdot \left(4 - 2\right)$
$a = \frac{1}{2}$
The height of the sinusoidal design is the amplitude, so $A = 2$
Now the area under the graph: (I assume you know about integral)
The parabolic design:
Area = ${\int}_{0}^{4} \frac{1}{2} \cdot x \cdot \left(4 - x\right) d \setminus x = 5.33 {m}^{2}$
The sinsoidal design:
Area = ${\int}_{0}^{4} 2 \cdot \sin \left(\pi \cdot \frac{x}{4}\right) d \setminus x = 5.09 {m}^{2}$

Therefore, the parabolic design is bigger.