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Question: If \( x = \cos ec\,\theta - \sin \,\theta \) and \( y = \cos e{c^n}\theta - {\sin ^n}\theta \) , T...

If x=cos⁡ec θ−sin⁡ θx = \cos ec\,\theta - \sin \,\theta and y=cos⁡ecnθ−sin⁡nθy = \cos e{c^n}\theta - {\sin ^n}\theta ,
Then show that (x2+y)(dydx)2−n2(y2+4)=0\left( {{x^2} + y} \right){\left( {\dfrac{{dy}}{{dx}}} \right)^2} - {n^2}\left( {{y^2} + 4} \right) = 0 .

Explanation

Solution

Hint : (dydx)2{\left( {\dfrac{{dy}}{{dx}}} \right)^2} is not equal to d2ydx2\dfrac{{{d^2}y}}{{d{x^2}}} as (dydx)2{\left( {\dfrac{{dy}}{{dx}}} \right)^2} is the square of dydx\dfrac{{dy}}{{dx}} and d2ydx2\dfrac{{{d^2}y}}{{d{x^2}}} is the double differentiation of yy with respect to xx . sin⁡nθ{\sin ^n}\theta on differentiation do not cos⁡nθ{\cos ^n}\theta . as it's totally different thing. It's (sin⁡ θ)n{\left( {\sin \,\theta } \right)^n} which will be differentiated by the rule ddxxn=nxn−1\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}} method. The parametric form is a method in which you differentiate two values with same value and later divide them, for example
if x=sin⁡θx = \sin \theta and y=cos⁡ θy = \cos \,\theta , Find dydx\dfrac{{dy}}{{dx}}
Thus, dxdθ=cos⁡θ\dfrac{{dx}}{{d\theta }} = \cos \theta and dydθ=−sin⁡θ\dfrac{{dy}}{{d\theta }} = - \sin \theta
Thus, dydx=dydθdxdθ=−sin⁡θcos⁡θ=tan⁡θ\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}} = \dfrac{{ - \sin \theta }}{{\cos \theta }} = \tan \theta

Complete step-by-step answer :
Given: (x2+y)(dydx)2−n2(y2+4)=0\left( {{x^2} + y} \right){\left( {\dfrac{{dy}}{{dx}}} \right)^2} - {n^2}\left( {{y^2} + 4} \right) = 0
First differentiate xx and yy with respect to θ\theta to get the values,
x=cos⁡ec θ−sin⁡θ ⇒dxdθ=−cot⁡θ.cos⁡ec θ−cos⁡θ        =−cos⁡θsin⁡θ×cos⁡ec θ−cos⁡θ        =−cos⁡θ(cos⁡ec θsin⁡θ+1) dxdθ=−cos⁡θsin⁡θ(cos⁡ec θ+sin⁡θ)      −−−−−−(i)  x = \cos ec\,\theta - \sin \theta \\\ \Rightarrow \dfrac{{dx}}{{d\theta }} = - \cot \theta .\cos ec\,\theta - \cos \theta \\\ \,\,\,\,\,\,\, = \dfrac{{ - \cos \theta }}{{\sin \theta }} \times \cos ec\,\theta - \cos \theta \\\ \,\,\,\,\,\,\, = - \cos \theta \left( {\dfrac{{\cos ec\,\theta }}{{\sin \theta }} + 1} \right) \\\ \dfrac{{dx}}{{d\theta }} = - \dfrac{{\cos \theta }}{{\sin \theta }}\left( {\cos ec\,\theta + \sin \theta } \right)\,\,\,\,\,\, - - - - - - \left( i \right) \\\
Now, for yy ,

y=cos⁡ec nθ−sin⁡nθ ⇒dydθ=ddθ(cos⁡ec nθ−sin⁡nθ)        =ncos⁡ec n−1θ(−cot⁡θ.cos⁡ec θ)−nsin⁡n−1θ.cos⁡θ        =−ncot⁡θ(cos⁡ecn θ+sin⁡n−1θ.sin⁡θ) dydθ=−ncot⁡θ(cos⁡ecn θ+sin⁡nθ)      −−−−−−(ii)  y = \cos ec{\,^n}\theta - {\sin ^n}\theta \\\ \Rightarrow \dfrac{{dy}}{{d\theta }} = \dfrac{d}{{d\theta }}\left( {\cos ec{\,^n}\theta - {{\sin }^n}\theta } \right) \\\ \,\,\,\,\,\,\, = n\cos ec{\,^{n - 1}}\theta \left( { - \cot \theta .\cos ec\,\theta } \right) - n{\sin ^{n - 1}}\theta .\cos \theta \\\ \,\,\,\,\,\,\, = - n\cot \theta \left( {\cos e{c^n}\,\theta + {{\sin }^{n - 1}}\theta .\sin \theta } \right) \\\ \dfrac{{dy}}{{d\theta }} = - n\cot \theta \left( {\cos e{c^n}\,\theta + {{\sin }^n}\theta } \right)\,\,\,\,\,\, - - - - - - \left( {ii} \right) \\\

Now to find dydx\dfrac{{dy}}{{dx}} , first divide dydθ\dfrac{{dy}}{{d\theta }} and dxdθ\dfrac{{dx}}{{d\theta }} .
⇒dydx=dydθdxdθ=−ncot⁡θ(cos⁡ecn θ+sin⁡nθ)−cot⁡θ(cos⁡ec θ+sin⁡θ) dydx=n(cos⁡ecn θ+sin⁡nθ)(cos⁡ec θ+sin⁡θ)  \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}} = \dfrac{{ - n\cot \theta \left( {\cos e{c^n}\,\theta + {{\sin }^n}\theta } \right)}}{{ - \cot \theta \left( {\cos ec\,\theta + \sin \theta } \right)}} \\\ \dfrac{{dy}}{{dx}} = \dfrac{{n\left( {\cos e{c^n}\,\theta + {{\sin }^n}\theta } \right)}}{{\left( {\cos ec\,\theta + \sin \theta } \right)}} \\\
Now, putting the value of dydx\dfrac{{dy}}{{dx}} in the given equation.
\Rightarrow (x2+y)(dydx)2−n2(y2+4)=0\left( {{x^2} + y} \right){\left( {\dfrac{{dy}}{{dx}}} \right)^2} - {n^2}\left( {{y^2} + 4} \right) = 0
Now, Left hand side is
⇒(x2+y)(dydx)2−n2(y2+4) =((cos⁡ec θ−sin⁡θ)2+4)(n2(cos⁡ec nθ−sin⁡nθ)2(cos⁡ec θ−sin⁡θ)2)−n2((cos⁡ec nθ−sin⁡nθ)2+4) =(cos⁡ec 2θ+sin⁡2θ−2+4)(n2(cos⁡ec nθ−sin⁡nθ)2(cos⁡ec θ−sin⁡θ)2)−n2(cos⁡ec 2nθ+sin⁡2nθ−2+4) =(cos⁡ec 2θ+sin⁡2θ+2cos⁡ec θ.sin⁡θ)n2((cos⁡ec nθ−sin⁡nθ)(cos⁡ec θ−sin⁡θ))2−n2(cos⁡ec 2nθ+sin⁡2nθ+2cos⁡ec nθ.sin⁡nθ) =(cos⁡ec θ+sin⁡θ)2n2((cos⁡ec nθ−sin⁡nθ)2(cos⁡ec θ−sin⁡θ)2)−n2(cos⁡ecn θ+sin⁡nθ)2 =n2(cos⁡ecn θ+sin⁡nθ)2−n2(cos⁡ecn θ+sin⁡nθ)2 =0  \Rightarrow \left( {{x^2} + y} \right){\left( {\dfrac{{dy}}{{dx}}} \right)^2} - {n^2}\left( {{y^2} + 4} \right) \\\ = \left( {{{\left( {\cos ec\,\theta - \sin \theta } \right)}^2} + 4} \right)\left( {{n^2}\dfrac{{{{\left( {\cos ec{\,^n}\theta - {{\sin }^n}\theta } \right)}^2}}}{{{{\left( {\cos ec\,\theta - \sin \theta } \right)}^2}}}} \right) - {n^2}\left( {{{\left( {\cos ec{\,^n}\theta - {{\sin }^n}\theta } \right)}^2} + 4} \right) \\\ = \left( {\cos ec{\,^2}\theta + {{\sin }^2}\theta - 2 + 4} \right)\left( {{n^2}\dfrac{{{{\left( {\cos ec{\,^n}\theta - {{\sin }^n}\theta } \right)}^2}}}{{{{\left( {\cos ec\,\theta - \sin \theta } \right)}^2}}}} \right) - {n^2}\left( {\cos ec{\,^{2n}}\theta + {{\sin }^{2n}}\theta - 2 + 4} \right) \\\ = \left( {\cos ec{\,^2}\theta + {{\sin }^2}\theta + 2\cos ec\,\theta .\sin \theta } \right){n^2}{\left( {\dfrac{{\left( {\cos ec{\,^n}\theta - {{\sin }^n}\theta } \right)}}{{\left( {\cos ec\,\theta - \sin \theta } \right)}}} \right)^2} - {n^2}\left( {\cos ec{\,^{2n}}\theta + {{\sin }^{2n}}\theta + 2\cos ec{\,^n}\theta .{{\sin }^n}\theta } \right) \\\ = {\left( {\cos ec\,\theta + \sin \theta } \right)^2}{n^2}\left( {\dfrac{{{{\left( {\cos ec{\,^n}\theta - {{\sin }^n}\theta } \right)}^2}}}{{{{\left( {\cos ec\,\theta - \sin \theta } \right)}^2}}}} \right) - {n^2}{\left( {\cos e{c^n}\,\theta + {{\sin }^n}\theta } \right)^2} \\\ = {n^2}{\left( {\cos e{c^n}\,\theta + {{\sin }^n}\theta } \right)^2} - {n^2}{\left( {\cos e{c^n}\,\theta + {{\sin }^n}\theta } \right)^2} \\\ = 0 \\\
Hence the left hand side proved equal to right hand side

Note : In this type of questions students often make mistakes while differentiating. Do not directly differentiate yy with respect to xx to get dydx\dfrac{{dy}}{{dx}} as it will be completely incorrect, Hence try to use another variable ” θ\theta ” and then divide them to get dydx\dfrac{{dy}}{{dx}} , Now perform the operation on the given equation very carefully. A simple error in sign can bring you out the wrong result.