Question

Consider the family F1 of curve lying in the region
{(x,y) εR²: y>0 and 0<x<π }
and given by 
\(y=\frac{c(1-cosx)}{sinx},\) Where c is a positive real number.
Let F₂ be the family of orthogonal trajectories to f₁. consider the curve C belonging to the family F2 passing through the point (\(\frac{π}{3}\),1). if a is a real number such that (\(\frac{π}{4}\), a) lies on C, then the value of a⁴ is equal _____to (Rounded off to two decimal places).

Answer 1

Correct Answer: 2

Based on the problem statement provided in the image, here is the step-by-step solution.

 

1. Simplify the Equation for Family $F_1$

 

The given family of curves $F_1$ is defined by:

 

$$y = \frac{c(1 - \cos x)}{\sin x}$$

Using trigonometric half-angle identities:

$1 - \cos x = 2\sin^2(x/2)$

$\sin x = 2\sin(x/2)\cos(x/2)$

Substitute these into the equation:

 

$$y = \frac{c \cdot 2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)}$$

$$y = c \tan(x/2)$$

 

2. Find the Differential Equation for $F_1$

 

To find the differential equation, we differentiate $y$ with respect to $x$:

 

$$\frac{dy}{dx} = c \cdot \frac{1}{2} \sec^2(x/2)$$

We need to eliminate the parameter $c$. From the simplified equation, we have $c = y \cot(x/2)$. Substituting this back into the derivative:

 

$$\frac{dy}{dx} = \left( y \cot(x/2) \right) \cdot \frac{1}{2} \sec^2(x/2)$$

$$\frac{dy}{dx} = \frac{y}{2} \cdot \frac{\cos(x/2)}{\sin(x/2)} \cdot \frac{1}{\cos^2(x/2)}$$

$$\frac{dy}{dx} = \frac{y}{2\sin(x/2)\cos(x/2)}$$

Using the identity $\sin x = 2\sin(x/2)\cos(x/2)$, we get the differential equation for $F_1$:

 

$$\left(\frac{dy}{dx}\right)_{F_1} = \frac{y}{\sin x}$$

 

3. Find the Differential Equation for Family $F_2$ (Orthogonal Trajectories)

 

For orthogonal trajectories, the product of the slopes is $-1$. Therefore, we replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$:

 

$$-\frac{dx}{dy} = \frac{y}{\sin x}$$

Rearranging variables to separate them:

 

$$\sin x \, dx = -y \, dy$$

$$\sin x \, dx + y \, dy = 0$$

 

4. Solve for the Curve $C$ in Family $F_2$

 

Integrate both sides:

 

$$\int \sin x \, dx + \int y \, dy = 0$$

$$-\cos x + \frac{y^2}{2} = K$$

So, the general equation for family $F_2$ is:

 

$$\frac{y^2}{2} - \cos x = K$$

Find the constant $K$:

We are given that the curve $C$ passes through the point $\left(\frac{\pi}{3}, 1\right)$.

Substitute $x = \frac{\pi}{3}$ and $y = 1$:

 

$$\frac{1^2}{2} - \cos\left(\frac{\pi}{3}\right) = K$$

$$\frac{1}{2} - \frac{1}{2} = K$$

$$K = 0$$

Thus, the equation of curve $C$ is:

 

$$\frac{y^2}{2} - \cos x = 0 \implies y^2 = 2\cos x$$

 

5. Find the Value of $a$ and Calculate $a^4$

 

We are given that the point $\left(\frac{\pi}{4}, a\right)$ lies on curve $C$.

Substitute $x = \frac{\pi}{4}$ and $y = a$ into the equation $y^2 = 2\cos x$:

 

$$a^2 = 2\cos\left(\frac{\pi}{4}\right)$$

$$a^2 = 2 \cdot \frac{1}{\sqrt{2}}$$

$$a^2 = \sqrt{2}$$

To find $a^4$, we square $a^2$:

 

$$a^4 = (\sqrt{2})^2$$

$$a^4 = 2$$

The question asks for the value rounded to two decimal places.

 

$$a^4 = 2.00$$

Final Answer:

The value of $a^4$ is equal to 2.00.