Question

Show that the inflection points of the curve \(y = x\sin x\) lie on the curve \({y^2}\left( {{x^2} + 4} \right) = 4{x^2}\).

Short answer

It is proved that inflection points lie on this curve.

Step-by-step solution

Given data

The given function \({y^2}\left( {{x^2} + 4} \right) = 4{x^2}\).

Concept of Differentiation

Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables.

Calculate the value of \({y^{\prime \prime }}\)

Calculate the value of \({y^{\prime \prime }}\):

\(\begin{aligned}{c}y &= x\sin x\\{y^\prime } &= \frac{{d\left( x \right)}}{{dx}}\sin x + x\cos x\\{y^\prime } &= \sin x + x\cos x\end{aligned}\)

Similarly, calculate further:

\(\begin{aligned}{c}{y^{\prime \prime }} &= \cos x + \frac{{d\left( x \right)}}{{dx}}\cos x + x( - \sin x)\\{y^{\prime \prime }} &= \cos x + \cos x - x\sin x\\{y^{\prime \prime }} &= 2\cos x - x\sin x\end{aligned}\)

Inflection points will be where \({y^{\prime \prime }} = 0\)

\(\begin{aligned}{c}2\cos x - x\sin x &= 0\\2\cos x &= x\sin x\end{aligned}\)

Note that since \(y = x\sin x\), then this says that at the inflection points \(y = 2\cos x\).

Square both the sides and simplify the expression

The strategy is to turn \(2\cos x = x\sin x\) into an equation with just \(x\) and \(y\), (the one given). First square both sides.

\(\begin{aligned}{c}{(2\cos x)^2} &= {(x\sin x)^2}\\4{\cos ^2}x &= {x^2}{\sin ^2}x\\4{\cos ^2}x &= {x^2}\left( {1 - {{\cos }^2}x} \right)\;\;\;\left( {{{\cos }^2}x + {{\sin }^2}x = 1} \right)\\4{\cos ^2}x &= {x^2} - {x^2}{\cos ^2}x\end{aligned}\)

Similarly, calculate further:

\(\begin{aligned}{c}4{\cos ^2}x + {x^2}{\cos ^2}x &= {x^2}\\{\cos ^2}x \cdot \left( {{x^2} + 4} \right) &= {x^2}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} Multiply{\kern 1pt} {\kern 1pt} {\kern 1pt} both{\kern 1pt} {\kern 1pt} {\kern 1pt} sides{\kern 1pt} {\kern 1pt} {\kern 1pt} by{\kern 1pt} {\kern 1pt} {\kern 1pt} 4\\4{\cos ^2}x \cdot \left( {{x^2} + 4} \right) &= 4{x^2}\end{aligned}\)

Since \(y = x\sin x = 2\cos x\) at the inflection points, and \({(2\cos x)^2} = {y^2}\) then

\(\begin{aligned}{c}{(2\cos x)^2} \cdot \left( {{x^2} + 4} \right) &= 4{x^2}{y^2}\left( {{x^2} + 4} \right)\\{(2\cos x)^2} \cdot \left( {{x^2} + 4} \right) &= 4{x^2}\end{aligned}\)

The inflection points lie on this curve.

It is proved that inflection points lie on this curve.