Question

The angle made by the tangent at \(\theta=\dfrac{\pi}{3}\) on the curve \(x=a(\theta+\sin\theta)\), \(y=a(1-\cos\theta)\) with \(x\)-axis is

• A. \(\dfrac{\pi}{3}\)
• B. \(\dfrac{\pi}{6}\)
• C. \(\dfrac{2\pi}{3}\)
• D. \(\dfrac{5\pi}{6}\)

Hint:

For parametric curves, always use \[ \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \] and then compare the slope with \(\tan\alpha\), where \(\alpha\) is the angle made by the tangent with the \(x\)-axis.

Answer 1

The Correct Option is B

Step 1: Differentiate \(x\) with respect to \(\theta\).
Given, \[ x=a(\theta+\sin\theta) \] Differentiating with respect to \(\theta\), \[ \frac{dx}{d\theta}=a(1+\cos\theta) \]

Step 2: Differentiate \(y\) with respect to \(\theta\).
Given, \[ y=a(1-\cos\theta) \] Differentiating with respect to \(\theta\), \[ \frac{dy}{d\theta}=a\sin\theta \]

Step 3: Find slope of tangent.
The slope of tangent is \[ \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \] Therefore, \[ \frac{dy}{dx}=\frac{a\sin\theta}{a(1+\cos\theta)} \] \[ \frac{dy}{dx}=\frac{\sin\theta}{1+\cos\theta} \] Using the identity, \[ \frac{\sin\theta}{1+\cos\theta}=\tan\frac{\theta}{2} \] So, \[ \frac{dy}{dx}=\tan\frac{\theta}{2} \]

Step 4: Substitute \(\theta=\dfrac{\pi}{3}\).
\[ \frac{dy}{dx}=\tan\left(\frac{\pi}{6}\right) \] \[ \frac{dy}{dx}=\frac{1}{\sqrt{3}} \] If \(\alpha\) is the angle made by the tangent with the \(x\)-axis, then \[ \tan\alpha=\frac{1}{\sqrt{3}} \] Hence, \[ \alpha=\frac{\pi}{6} \]

Step 5: Final conclusion.
Therefore, the angle made by the tangent with the \(x\)-axis is \[ \boxed{\frac{\pi}{6}} \]