Question

A curve passes through the point $\left( {1,\dfrac{\pi }{6}} \right)$, let the slope of the curve at each point (x, y) be $\dfrac{y}{x} + \sec \left( {\dfrac{y}{x}} \right)$, x > 0. Then the equation of the curve is:
$\left( a \right)\sin \left( {\dfrac{y}{x}} \right) = \log x + \dfrac{1}{2}$
$\left( b \right)\csc \left( {\dfrac{y}{x}} \right) = \log x + 2$
$\left( c \right)\sec \left( {\dfrac{{2y}}{x}} \right) = \log x + 2$
$\left( d \right)\cos \left( {\dfrac{{2y}}{x}} \right) = \log x + \dfrac{1}{2}$

Answer 1

In this particular question uses the concept that the slope of the curve is represented as $\dfrac{{dy}}{{dx}}$ so equate them and substitute y = vx, in the equation so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given slope of the curve,
$\dfrac{y}{x} + \sec \left( {\dfrac{y}{x}} \right)$
Now as we know that the slope of the curve is represented as $\dfrac{{dy}}{{dx}}$.
So equate both of then we have,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{x} + \sec \left( {\dfrac{y}{x}} \right)$.................. (1)
Now let, y = vx............... (2)
Differentiate this equation w.r.t x we have,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}vx$
Now as we know that $\dfrac{d}{{dx}}mn = m\dfrac{d}{{dx}}n + n\dfrac{d}{{dx}}m$ so we have,
$\Rightarrow \dfrac{{dy}}{{dx}} = v\dfrac{d}{{dx}}x + x\dfrac{d}{{dx}}v = v + x\dfrac{{dv}}{{dx}}$.................. (3)
Now substitute the values from equation (2) and (3) in equation (1) we have,
$\Rightarrow v +x \dfrac{{dv}}{{dx}} = v + \sec v$
$\Rightarrow x\dfrac{{dv}}{{dx}} = \sec v$
$\Rightarrow \dfrac{{dv}}{{\sec v}} = \dfrac{{dx}}{x}$
Now integrate both sides we have,
$\Rightarrow \int {\dfrac{{dv}}{{\sec v}}} = \int {\dfrac{{dx}}{x}}$
Now as we know that cos x = (1/sec x) so we have,
$\Rightarrow \int {\cos vdv} = \int {\dfrac{{dx}}{x}}$
Now as we know that $\int {\cos xdx = \sin x,\int {\dfrac{1}{x}dx} = \log x} + c$, where c is some integration constant.
$\Rightarrow \sin v = \log x + c$
Now substitute the value of v we have,
$\Rightarrow \sin \left( {\dfrac{y}{x}} \right) = \log x + c$............. (4)
Now it is given that the curve is passing from the point $\left( {1,\dfrac{\pi }{6}} \right)$, so this point satisfies equation (4) so we have,
$\Rightarrow \sin \left( {\dfrac{\pi }{6}} \right) = \log 1 + c$
$\Rightarrow \dfrac{1}{2} = 0 + c$, $\left[ {\because \sin \dfrac{\pi }{6} = \dfrac{1}{2},\log 1 = 0} \right]$
$\Rightarrow c = \dfrac{1}{2}$
Now substitute this value in equation (4) we have,
$\Rightarrow \sin \left( {\dfrac{y}{x}} \right) = \log x + \dfrac{1}{2}$
So this is the required curve.
Hence option (a) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic differentiation and integrating properties such as $\dfrac{d}{{dx}}mn = m\dfrac{d}{{dx}}n + n\dfrac{d}{{dx}}m$, $\int {\cos xdx = \sin x,\int {\dfrac{1}{x}dx} = \log x} + c$, where c is some integration constant.