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

Hint: 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.