Question

Find the equation of the curve passing through $\left( {1,\dfrac{\pi }{4}} \right)$ and having slope $\dfrac{{\sin 2y}}{{x + \tan y}}$ at $(x,y)$ .

Answer 1

To find the equation of a curve when a slope is given, we first need to find the integration factor, which is $\tan \dfrac{\pi }{4} = 1$ where P is derived from the slope $\dfrac{{dy}}{{dx}}$ . The slope has to be expressed in the form, $\dfrac{{dx}}{{dy}} + Px = Q$ from this we will get our values of $P\& Q$ . After finding the integrating factor we will find the equation of the curve by $xI.F = \int {Qdy}$ .

Complete step by step answer:
It is given that the curve has a slope $\dfrac{{\sin 2y}}{{x + \tan y}}$ at $(x,y)$ .
Therefore, $\dfrac{{dy}}{{dx}} = \dfrac{{\sin 2y}}{{x + \tan y}}$
Let’s reciprocate it for our convenience, $\dfrac{{dx}}{{dy}} = \dfrac{{x + \tan y}}{{\sin 2y}}$
$\Rightarrow \dfrac{{dx}}{{dy}} = \dfrac{x}{{\sin 2y}} + \dfrac{{\tan y}}{{\sin 2y}}$
On further simplification we get,
$\Rightarrow \dfrac{{dx}}{{dy}} - \dfrac{x}{{\sin 2y}} = \dfrac{{\tan y}}{{\sin 2y}}$
We know that $Sin2y = 2\cos y\sin y$ by applying this we get
$\Rightarrow \dfrac{{dx}}{{dy}} - \dfrac{x}{{\sin 2y}} = \dfrac{{\tan y}}{{2\cos y\sin y}}$
From the trigonometry identities, $\dfrac{1}{{\sin \theta }} = \csc \theta$
$\Rightarrow \dfrac{{dx}}{{dy}} - x\csc 2y = \dfrac{{\tan y}}{{2\cos y\sin y}}$
Again, from the trigonometry identity we have $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
$\Rightarrow \dfrac{{dx}}{{dy}} - x\csc 2y = \dfrac{{\sin y}}{{\cos y(2\cos y\sin y)}}$
Simplifying the above expression, we get
$\Rightarrow \dfrac{{dx}}{{dy}} - x\csc 2y = \dfrac{1}{{2{{\cos }^2}y}}$
From the trigonometry identity we have, $\dfrac{1}{{\cos \theta }} = \sec \theta$
$\Rightarrow \dfrac{{dx}}{{dy}} - x\csc 2y = \dfrac{1}{2}{\sec ^2}y$
Now we can see that the above equation is of the form $\dfrac{{dx}}{{dy}} + Px = Q$ .
From this, we get $P = - \csc 2y$ and $Q = \dfrac{1}{2}{\sec ^2}y$ .
Let’s find the integration factor $I.F = {e^{\int {Pdy} }}$ .
$I.F = {e^{ - \int {\csc 2y} dy}}$
On integrating the part, we get
$I.F = {e^{ - \dfrac{1}{2}( - 1)\ln (\csc 2y + \cot 2y)}}$
On simplifying this we get,
$I.F = {e^{\dfrac{1}{2}\ln (\csc 2y + \cot 2y)}}$
Let’s simplify it further,
$I.F = {e^{\ln {{(\csc 2y + \cot 2y)}^{1/2}}}}$
We know that ${e^{\ln x}} = x$ , so we get
$I.F = {(\csc 2y + \cot 2y)^{1/2}}$
Now we have the integration factor. Let us find the equation of the curve.
$x\sqrt {\csc 2y + \cot 2y} = \int {{{\sec }^2}ydy}$
On integrating the above equation, we get
$x\sqrt {\csc 2y + \cot 2y} = \dfrac{1}{2}\tan y + c$ , where c is the integration factor.
This is the required equation of the curve. Also, it is given that this curve is passing through the point $\left( {1,\dfrac{\pi }{4}} \right)$ . Let’s substitute this point in the equation of the curve.
$(x,y) = \left( {1,\dfrac{\pi }{4}} \right)$ in $x\sqrt {\csc 2y + \cot 2y} = \dfrac{1}{2}\tan y + c$ we get,
$1\sqrt {\csc 2\left( {\dfrac{\pi }{4}} \right) + \cot 2\left( {\dfrac{\pi }{4}} \right)} = \dfrac{1}{2}\tan \left( {\dfrac{\pi }{4}} \right) + c$
Since, $\csc \dfrac{\pi }{2} = 1,\cot \dfrac{\pi }{2} = 0$ and $\tan \dfrac{\pi }{4} = 1$ we get $c = 0$ .
$\therefore x\sqrt {\csc 2y + \cot 2y} = \dfrac{1}{2}\tan y$ is the required equation of the curve passing through the point $\left( {1,\dfrac{\pi }{4}} \right)$ .

Note: In finding the equation of the curve when the point is given, we have to substitute that point in the curve equation. The slope of a curve is nothing but the small displacement in the graph (i.e., Rate of change $\dfrac{{dy}}{{dx}}$ ). Thus, we need to first find the elements P and Q from the slope then find the integrating factor to get the equation of the curve.