Question

If \(x = e^\theta \sin \theta, \quad y = e^\theta \cos \theta\), where \(θ\) is a parameter , then \(\left. \frac{dy}{dx} \right|_{(1,1)}\) is equal to

• A. 0
• B. \(-\frac{1}{2}\)
• C. \(\frac{1}{2}\)
• D. \(-\frac{1}{4}\)

Answer 1

The Correct Option is A

Given:
\(x = e^\theta \sin \theta\)
\(y = e^{\theta} \cos \theta\)
Differentiating both sides of the first equation with respect to θ:
\(\frac{d}{d\theta}(x) = \frac{d}{d\theta}(e^{\theta} \sin \theta)\)
Using the product rule, we have:
\(\frac{dx}{d\theta} = e^\theta \sin \theta + e^\theta \cos \theta\)
Differentiating both sides of the second equation with respect to θ:
\(\frac{d}{d\theta}(y) = \frac{d}{d\theta}(e^\theta \cos \theta)\)
Using the product rule, we have:
\(\frac{dy}{d\theta} = e^\theta \cos \theta - e^\theta \sin \theta\)
Now, to find \(\frac{dy}{dx}\), we divide \(\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\)
\(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\)
Substituting the expressions for \(\frac{dy}{d\theta} \) and \(\frac{dx}{d\theta} \), we get:
\(\frac{dy}{dx} = \frac{e^\theta \cos \theta - e^\theta \sin \theta}{e^\theta \sin \theta + e^\theta \cos \theta}\)
Next, we can substitute the values of x and y at (1,1) into the equation to evaluate \(\frac{dy}{dx}:\)
\(x = e^\theta \sin \theta = 1\)
\(y = e^\theta \cos \theta = 1\)
Dividing the second equation by the first equation, we get:
\(\tan \theta = \frac{y}{x} = \frac{1}{1} = 1\)
Simplifying the expression for \(\frac{dy}{dx}\) using trigonometric identities:
\(\frac{dy}{dx} = \frac{\cos \theta - \sin \theta}{\sin \theta + \cos \theta}\)
Since \(tan θ = 1\), we can substitute \(\frac{\cos \theta - \sin \theta}{\sin \theta + \cos \theta} = \frac{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} = \frac{0}{\sqrt{2}} = 0\)
Therefore, at (1,1), \(\frac{dy}{dx}\) is equal to 0, which corresponds to option (A) 0.