question_answer

The angle of intersection of the curves \[{{y}^{2}}=2x/\pi \] and \[y=\sin x\], is                [Roorkee 1998]

A)            \[{{\cot }^{-1}}(-1/\pi )\]          

B)            \[{{\cot }^{-1}}\pi \]

C)            \[{{\cot }^{-1}}(-\pi )\]             

D)            \[{{\cot }^{-1}}(1/\pi )\]

Correct Answer: B

Solution :

           The curves \[{{y}^{2}}=2x/\pi \] and \[y=\sin x\] intersect at \[(0,0)\] and \[(\pi /2,\,1)\]. Let the gradients of the tangents to the curves be \[{{m}_{1}}\] and \[{{m}_{2}}\]respectively. Then \[{{m}_{1}}=\frac{dy}{dx}=\frac{1}{\pi y}\] and \[{{m}_{2}}=\frac{dy}{dx}=\cos x\]            At \[(\pi /2\,,1)\], \[{{m}_{1}}=\frac{1}{\pi }\],                   \[{{m}_{2}}=\cos \frac{\pi }{2}=0\]                    Thus \[\tan \theta =\frac{(1/\pi )-0}{1+(1/\pi )(0)}=\frac{1}{\pi }\]Þ\[\theta ={{\cot }^{-1}}\pi \].