Question

Find the ratio in which area enclosed by y = cosx, x-axis, x = 0, x = $\pi$/2, is divided by the curve y = sinx.

Answer 1

1:$\sqrt{2}$

Answer 2

The problem asks us to find the ratio in which the area enclosed by $y = \cos x$, the x-axis, $x = 0$, and $x = \pi/2$ is divided by the curve $y = \sin x$.

1. Calculate the total area: The total area, let's call it $A_{total}$, is the area under the curve $y = \cos x$ from $x = 0$ to $x = \pi/2$. Since $\cos x \ge 0$ in this interval, the area is: $A_{total} = \int_{0}^{\pi/2} \cos x \, dx$ $A_{total} = [\sin x]_{0}^{\pi/2}$ $A_{total} = \sin(\pi/2) - \sin(0)$ $A_{total} = 1 - 0 = 1$ square unit.

2. Understand how the curve $y = \sin x$ divides the area: Let's visualize the curves $y = \sin x$ and $y = \cos x$ in the interval $[0, \pi/2]$.

• At $x=0$, $\cos x = 1$, $\sin x = 0$.
• At $x=\pi/2$, $\cos x = 0$, $\sin x = 1$.
• The curves intersect when $\sin x = \cos x$, which occurs at $x = \pi/4$ in this interval. At this point, $\sin(\pi/4) = \cos(\pi/4) = 1/\sqrt{2}$.

The total area $A_{total}$ is the region $R = \{(x,y) | 0 \le x \le \pi/2, 0 \le y \le \cos x\}$. The curve $y = \sin x$ divides this region $R$ into two parts:

• Part 1 ($A_1$): The area of the region $R$ that lies below or on the curve $y = \sin x$.
• Part 2 ($A_2$): The area of the region $R$ that lies above the curve $y = \sin x$.

Let's determine the boundaries for $A_1$ and $A_2$:

For Part 1 ($A_1$): This area is bounded by $y = \min(\cos x, \sin x)$ and the x-axis.

• For $0 \le x \le \pi/4$, $\sin x \le \cos x$, so $\min(\cos x, \sin x) = \sin x$.
• For $\pi/4 \le x \le \pi/2$, $\cos x \le \sin x$, so $\min(\cos x, \sin x) = \cos x$.

So, $A_1 = \int_{0}^{\pi/4} \sin x \, dx + \int_{\pi/4}^{\pi/2} \cos x \, dx$. $A_1 = [-\cos x]_{0}^{\pi/4} + [\sin x]_{\pi/4}^{\pi/2}$ $A_1 = (-\cos(\pi/4) - (-\cos(0))) + (\sin(\pi/2) - \sin(\pi/4))$ $A_1 = (-1/\sqrt{2} + 1) + (1 - 1/\sqrt{2})$ $A_1 = 2 - 2/\sqrt{2} = 2 - \sqrt{2}$.

For Part 2 ($A_2$): This area is bounded by $y = \cos x$ (upper curve) and $y = \sin x$ (lower curve). This region exists only where $\sin x \le \cos x$, which is for $0 \le x \le \pi/4$. So, $A_2 = \int_{0}^{\pi/4} (\cos x - \sin x) \, dx$. $A_2 = [\sin x + \cos x]_{0}^{\pi/4}$ $A_2 = (\sin(\pi/4) + \cos(\pi/4)) - (\sin(0) + \cos(0))$ $A_2 = (1/\sqrt{2} + 1/\sqrt{2}) - (0 + 1)$ $A_2 = 2/\sqrt{2} - 1 = \sqrt{2} - 1$.

3. Verify the sum of areas: $A_1 + A_2 = (2 - \sqrt{2}) + (\sqrt{2} - 1) = 1$. This matches $A_{total}$, confirming our calculations and interpretation.

4. Find the ratio: The question asks for the ratio in which the area is divided. This usually means the ratio of the two parts. We have $A_1 = 2 - \sqrt{2}$ and $A_2 = \sqrt{2} - 1$. Numerically, $A_1 \approx 2 - 1.414 = 0.586$ and $A_2 \approx 1.414 - 1 = 0.414$. So $A_2$ is the smaller part and $A_1$ is the larger part. The ratio is commonly expressed as smaller to larger.

Ratio = $A_2 : A_1 = (\sqrt{2} - 1) : (2 - \sqrt{2})$ To simplify the ratio, divide both sides by $(\sqrt{2} - 1)$: Ratio = $1 : \frac{2 - \sqrt{2}}{\sqrt{2} - 1}$ $\frac{2 - \sqrt{2}}{\sqrt{2} - 1} = \frac{\sqrt{2}(\sqrt{2} - 1)}{\sqrt{2} - 1} = \sqrt{2}$.

So, the ratio is $1 : \sqrt{2}$.