Question

If $x=a\sin 2\theta (1+\cos 2\theta ),y=b\cos 2\theta (1-\cos 2\theta )$, then $dy/dx=$?

• A. $(b\tan \theta )/a$
• B. $(a\tan \theta )/b$
• C. $a/(b\tan \theta )$
• D. $b/a\tan \theta$

Answer 1

The correct option is A

$(b\tan \theta )/a$

Step 1: Simplify $x=a\sin 2\theta (1+\cos 2\theta )$.

We know that $\sin 2\theta =2\sin \theta \cos \theta$

$\begin{array}{rcl}& \Rightarrow & \cos 2\theta =2\cos 2\theta –1\\ \Rightarrow x& =& 2a\sin \theta \cos \theta (1+2\cos 2\theta -1)\\ & =& 2a\sin \theta \cos \theta \times 2\cos 2\theta \\ & =& 4a\sin \theta \cos 3\theta \end{array}$

Step 2: Differentiate $x$ w.r.t. $\theta$

$\begin{array}{rcl}dx/d\theta & =& 4a(\cos \theta \times \cos 3\theta +\sin \theta \times 3\cos 2\theta \times -\sin \theta )\\ & =& 4a(\cos 4\theta –3\sin 2\theta \cos 2\theta )\\ & =& 4a\cos 2\theta (\cos 2\theta –3\sin 2\theta )\end{array}$

Step 3: Simplify $y=b\cos 2\theta (1-\cos 2\theta )$.

We know that $\cos 2\theta =1-2{\sin }^2\theta$

$\begin{array}{rcl}& \Rightarrow & y=b(1-2\sin 2\theta )(1-1+2\sin 2\theta )\\ & =& b(1-2\sin 2\theta )\times 2\sin 2\theta \\ & =& 2b(\sin 2\theta –2\sin 4\theta )\end{array}$

Step 4: Differentiate $y$ w.r.t. $\theta$

$\begin{array}{rcl}dy/d\theta & =& 2b(2\sin \theta \cos \theta –2\times 4\sin 3\theta \cos \theta )\\ & =& 2b(2\sin \theta \cos \theta –4\times 2\sin \theta \cos \theta \sin 2\theta )\\ & =& 2b(\sin 2\theta –4\times \sin 2\theta \sin 2\theta )\\ & =& 2b\sin 2\theta (1–4\sin 2\theta )\end{array}$

Step 5: Find $dy/dx$.

$\begin{array}{rcl}dy/dx& =& (dy/d\theta )/(dx/d\theta )\\ & =& 2b\sin 2\theta (1–4\sin 2\theta )/4a\cos 2\theta (\cos 2\theta –3\sin 2\theta )\\ & =& b\times 2\sin \theta \cos \theta (1–4\sin 2\theta )/2a\cos 2\theta \left(\right(1-\sin 2\theta )-3\sin 2\theta )\\ & =& b\sin \theta \cos \theta (1–4\sin 2\theta )/a\cos 2\theta (1-4\sin 2\theta )\\ & =& b\sin \theta /a\cos \theta \\ & =& (b\tan \theta /a)\end{array}$