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

Answer: (A)

0

Answer 2

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.