Question

If \( x = 2\sqrt{2}\sqrt{\cos 2\theta} \) and \( y = 2\sqrt{2}\sqrt{\sin 2\theta} \), \( 0<\theta<\frac{\pi}{4} \) then the value of \( \frac{dy}{dx} \) at \( \theta = 22\frac{1}{2}^\circ \) is

• A. 1
• B. -1
• C. 0
• D. \( \sqrt{3} \)

Hint:

For parametric equations, sometimes it's easier to find a direct relationship between x and y and then use implicit differentiation. Squaring or using trigonometric identities can often reveal a simple Cartesian equation.

Answer 1

The Correct Option is B

This is a parametric differentiation problem. We will find \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \) and then take their ratio.
First, find \( \frac{dx}{d\theta} \).
\( x = 2\sqrt{2} (\cos 2\theta)^{1/2} \).
Using the chain rule: \( \frac{dx}{d\theta} = 2\sqrt{2} \cdot \frac{1}{2}(\cos 2\theta)^{-1/2} \cdot (-\sin 2\theta) \cdot 2 = \frac{-2\sqrt{2}\sin 2\theta}{\sqrt{\cos 2\theta}} \).
Next, find \( \frac{dy}{d\theta} \).
\( y = 2\sqrt{2} (\sin 2\theta)^{1/2} \).
Using the chain rule: \( \frac{dy}{d\theta} = 2\sqrt{2} \cdot \frac{1}{2}(\sin 2\theta)^{-1/2} \cdot (\cos 2\theta) \cdot 2 = \frac{2\sqrt{2}\cos 2\theta}{\sqrt{\sin 2\theta}} \).
Now, find \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \).
\( \frac{dy}{dx} = \frac{2\sqrt{2}\cos 2\theta / \sqrt{\sin 2\theta}}{-2\sqrt{2}\sin 2\theta / \sqrt{\cos 2\theta}} = - \frac{\cos 2\theta}{\sin 2\theta} \cdot \frac{\sqrt{\cos 2\theta}}{\sqrt{\sin 2\theta}} = - (\cot 2\theta)^{3/2} \).
There must be a mistake. Let's try squaring x and y first.
\( x^2 = 8 \cos 2\theta \) and \( y^2 = 8 \sin 2\theta \).
Differentiate implicitly with respect to \(\theta\):
\( 2x \frac{dx}{d\theta} = -16 \sin 2\theta \implies \frac{dx}{d\theta} = -\frac{8\sin 2\theta}{x} \).
\( 2y \frac{dy}{d\theta} = 16 \cos 2\theta \implies \frac{dy}{d\theta} = \frac{8\cos 2\theta}{y} \).
\( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{8\cos 2\theta / y}{-8\sin 2\theta / x} = -\frac{x \cos 2\theta}{y \sin 2\theta} \).
Substitute back \(x\) and \(y\): \( -\frac{2\sqrt{2}\sqrt{\cos 2\theta} \cos 2\theta}{2\sqrt{2}\sqrt{\sin 2\theta} \sin 2\theta} = - \left( \frac{\cos 2\theta}{\sin 2\theta} \right)^{3/2} = -(\cot 2\theta)^{3/2} \). Still the same result.
Let's find the relationship between \(x\) and \(y\): \(x^2+y^2 = 8(\cos 2\theta + \sin 2\theta)\). This doesn't seem to help.
Let's try differentiating implicitly with respect to \(x\): \( x^2+y^2 = 8(\cos 2\theta+\sin 2\theta) \) is not right. It should be \(x^4+y^4 = 64\). Let's check: \( (8\cos2\theta)^2 + (8\sin2\theta)^2 = 64(\cos^2 2\theta + \sin^2 2\theta) = 64\). Correct.
Differentiate \(x^4+y^4=64\) with respect to \(x\): \( 4x^3 + 4y^3 \frac{dy}{dx} = 0 \).
\( \frac{dy}{dx} = -\frac{x^3}{y^3} \).
We need to evaluate this at \( \theta = 22.5^\circ \).
At \( \theta=22.5^\circ \), \( 2\theta = 45^\circ \).
\( \cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}} \).
At this angle, \( x = 2\sqrt{2}\sqrt{1/\sqrt{2}} \) and \( y = 2\sqrt{2}\sqrt{1/\sqrt{2}} \), so \(x=y\).
Therefore, \( \frac{dy}{dx} = -\frac{x^3}{x^3} = -1 \).