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

We are given the following equations:

\[ x = e^\theta \sin \theta \] 

\[ y = e^{\theta} \cos \theta \]

Next, we differentiate both sides of the first equation with respect to \(\theta\):

\[ \frac{d}{d\theta}(x) = \frac{d}{d\theta}(e^{\theta} \sin \theta) \]

Using the product rule, we get:

\[ \frac{dx}{d\theta} = e^\theta \sin \theta + e^\theta \cos \theta \]

Now, we differentiate both sides of the second equation with respect to \(\theta\):

\[ \frac{d}{d\theta}(y) = \frac{d}{d\theta}(e^\theta \cos \theta) \]

Using the product rule, we get:

\[ \frac{dy}{d\theta} = e^\theta \cos \theta - e^\theta \sin \theta \]

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 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 \(\tan \theta = 1\), we 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.

Answer 2

We are given the parametric equations: \[ x = e^{\theta} \sin \theta, \quad y = e^{\theta} \cos \theta \] To find \(\frac{dy}{dx}\), use: \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \] Differentiate both: \[ \frac{dx}{d\theta} = \frac{d}{d\theta}(e^\theta \sin \theta) = e^\theta \sin \theta + e^\theta \cos \theta \] \[ \frac{dy}{d\theta} = \frac{d}{d\theta}(e^\theta \cos \theta) = e^\theta \cos \theta - e^\theta \sin \theta \] So, \[ \frac{dy}{dx} = \frac{e^\theta(\cos \theta - \sin \theta)}{e^\theta(\sin \theta + \cos \theta)} = \frac{\cos \theta - \sin \theta}{\sin \theta + \cos \theta} \] Now find the value of \( \theta \) that gives the point \( (1,1) \): From \( x = e^\theta \sin \theta = 1 \) and \( y = e^\theta \cos \theta = 1 \) Divide both: \[ \frac{x}{y} = \frac{\sin \theta}{\cos \theta} = \tan \theta = 1 \Rightarrow \theta = \frac{\pi}{4} \] Substitute \( \theta = \frac{\pi}{4} \) into the expression: \[ \frac{dy}{dx} = \frac{\cos \theta - \sin \theta}{\sin \theta + \cos \theta} = \frac{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} = \frac{0}{\frac{2}{\sqrt{2}}} = 0 \] Hence, \(\left. \frac{dy}{dx} \right|_{(1,1)} = 0\) 

Answer 3

We are given x = e^θ sin θ and y = e^θ cos θ. We need to find dy/dx at the point (1, 1).

First, let's find dx/dθ and dy/dθ:

\(\frac{dx}{d\theta} = \frac{d}{d\theta}(e^\theta \sin \theta) = e^\theta \sin \theta + e^\theta \cos \theta = e^\theta(\sin \theta + \cos \theta)\)

\(\frac{dy}{d\theta} = \frac{d}{d\theta}(e^\theta \cos \theta) = e^\theta \cos \theta - e^\theta \sin \theta = e^\theta(\cos \theta - \sin \theta)\)

Now, we can find dy/dx using the chain rule:

\(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{e^\theta(\cos \theta - \sin \theta)}{e^\theta(\sin \theta + \cos \theta)} = \frac{\cos \theta - \sin \theta}{\sin \theta + \cos \theta}\)

Next, we need to find the value of θ for which x = 1 and y = 1:

e^θ sin θ = 1

e^θ cos θ = 1

Dividing the two equations:

\(\frac{e^\theta \sin \theta}{e^\theta \cos \theta} = \frac{1}{1}\)

tan θ = 1

θ = \(\frac{\pi}{4}\) (since x and y are both positive)

Now, substitute θ = π/4 into the expression for dy/dx:

\(\frac{dy}{dx} \Big|_{\theta = \frac{\pi}{4}} = \frac{\cos(\frac{\pi}{4}) - \sin(\frac{\pi}{4})}{\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4})} = \frac{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} = \frac{0}{\frac{2}{\sqrt{2}}} = 0\)

Therefore, \(\left. \frac{dy}{dx} \right|_{(1,1)} = 0\)

Answer: 0