Question

If $x = a(t + \sin t), y = a(1 - \cos t)$, then $\frac{dy}{dx} =$

• A. $\tan \frac{t}{2}$
• B. $-\frac{1}{2}\tan t$
• C. $\frac{1}{2}\tan t$
• D. $-\tan \frac{t}{2}$

Hint:

The parametric equations $x = a(t + \sin t), y = a(1 - \cos t)$ describe a standard cycloid. Its derivative always elegantly simplifies to $\tan(t/2)$. Memorizing this specific result can save you derivation time!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the parametric equations of a curve ($x$ and $y$ in terms of parameter $t$). We need to find the derivative $\frac{dy}{dx}$.

Step 2: Key Formula or Approach:
For parametric differentiation, $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. After dividing, simplify the result using half-angle trigonometric identities: $\sin t = 2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)$ and $1 + \cos t = 2\cos^2\left(\frac{t}{2}\right)$.

Step 3: Detailed Explanation:
Given: $x = a(t + \sin t)$ and $y = a(1 - \cos t)$.
Differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = a(1 + \cos t)$
Differentiate $y$ with respect to $t$:
$\frac{dy}{dt} = a(0 - (-\sin t)) = a\sin t$
Now, find $\frac{dy}{dx}$ by dividing the two derivatives:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a\sin t}{a(1 + \cos t)} = \frac{\sin t}{1 + \cos t}$
Apply half-angle identities to simplify the expression:
$\frac{dy}{dx} = \frac{2\sin \frac{t}{2} \cos \frac{t}{2}}{2\cos^2 \frac{t}{2}}$
Cancel the common terms $2$ and $\cos \frac{t}{2}$:
$\frac{dy}{dx} = \frac{\sin \frac{t}{2}}{\cos \frac{t}{2}} = \tan \frac{t}{2}$

Step 4: Final Answer:
The derivative $\frac{dy}{dx}$ is $\tan \frac{t}{2}$, matching option (A).