question_answer

Find the equation of tangents to the curve\[y=\cos (x+y),\] \[-\,2\pi \le x\le 2\pi ,\] that   are parallel to the line \[x+2y=0.\]

OR

Find the radius of the smallest circle with centre on Y-axis and passing through the point (7, 3).

Answer:

Given equation of curve is

\[y=\cos (x+y)\]

On differentiating both sides w.r.t. x, we get

\[\frac{dy}{dx}=-\sin (x+y)\left( 1+\frac{dy}{dx} \right)\]

\[\Rightarrow \]   \[\frac{dy}{dx}=-\sin (x+y)-\sin (x+y)\frac{dy}{dx}\]

\[\Rightarrow \]   \[\frac{dy}{dx}+\sin (x+y)\frac{dy}{dx}=-\sin (x+y)\]

\[\Rightarrow \]   \[\frac{dy}{dx}=\frac{-\sin (x+y)}{1+\sin (x+y)}\]

\[\therefore \] Slope of tangent at \[(x,\,\,y)=\frac{-\sin (x+y)}{1+\sin (x+y)}\]

Since, the tangents to the given curve are parallel to the line \[x+2y=0,\]whose slope is \[\frac{1}{2}.\]

\[\therefore \]      \[\frac{-\sin (x+y)}{1+\sin (x+y)}=\frac{-\,1}{2}\]

\[\Rightarrow \]   \[2\sin (x+y)=1+\sin (x+y)\]

\[\Rightarrow \]   \[\sin (x+y)=1\]                          ?(i)

\[\Rightarrow \]   \[\sin (x+y)=\sin \frac{\pi }{2}\]

\[\Rightarrow \]   \[x+y=n\pi +{{(-\,1)}^{n}}\frac{\pi }{2},\] \[n\in Z\]      ?(ii)

\[\because \] \[\cos (x+y)=y\] and \[sin(x+y)=1\]

\[\therefore \] \[co{{s}^{2}}(x+y)+si{{n}^{2}}(x+y)={{y}^{2}}+1\]

\[\Rightarrow \]   \[1={{y}^{2}}+1\] \[\Rightarrow \] y = 0

From Eq. (ii), we have

\[x=n\pi +{{(-\,1)}^{n}}\frac{\pi }{2}\]

\[\Rightarrow \]  \[x=\pm \frac{\pi }{2},\,\,\pm \frac{3\pi }{2}\]           \[[\because -\,2\pi \le x\le 2\pi ]\]

But \[x=\frac{-\,\pi }{2}\] and \[\frac{3\pi }{2}\] does not satisfy Eq. (i).

\[\therefore \] The points are \[\left( \frac{\pi }{2},\,\,0 \right)\] and \[\left( \frac{-\,3\pi }{2},\,\,0 \right).\]

Hence, the required equations of tangents are

\[y-0=\frac{-\,1}{2}\left( x+\frac{3\pi }{2} \right)\]

Or         \[2x+4y+3\pi =0\]

and       \[y-0=\frac{-\,1}{2}\left( x-\frac{\pi }{2} \right)\]

Or         \[2x+4y-\pi =0\]

OR

Let the centre of the circle on Y-axis be (0, k). Then, radius of circle = distance between (0, k) and (7, 3).

\[\Rightarrow \]   \[r=\sqrt{{{(7-0)}^{2}}+{{(3-k)}^{2}}}\]                       ?(i)

\[\Rightarrow \]   \[{{r}^{2}}={{7}^{2}}+{{(3-k)}^{2}}\][on squaring both sides]

\[\Rightarrow \]   \[r={{7}^{2}}+{{(3-k)}^{2}},\] Where \[{{r}^{2}}=R\]

On differentiating both sides w.r.t. k, we get

\[\frac{dR}{dk}=0+2\,(3-k)(-\,1)\]

\[\Rightarrow \]   \[\frac{dR}{dk}=-\,2\,(3-k)\]                                  ?(ii)

For maximum or minimum, put \[\frac{dR}{dk}=0\]

\[\Rightarrow \]   \[-\,2(3-k)=0\]

\[\Rightarrow \]   \[3-k=0\] \[\Rightarrow \] k = 3

Again, differentiating both sides of Eq. (ii) w.r.t. k, we get \[\frac{{{d}^{2}}R}{d{{k}^{2}}}=-\,2(-\,1)=2\]

At k = 3,  \[\frac{{{d}^{2}}R}{d{{k}^{2}}}=2>0\]

So, R Is minimum at k = 3.

On putting k = 3 in Eq. (i), we get,

\[r=\sqrt{{{(7-0)}^{2}}+{{(3-3)}^{2}}}=\sqrt{{{7}^{2}}+0}=7\]

Hence, radius of the smallest circle is 7 units.