Question

Use Newton’s method to find the coordinates of the inflection point of the curve \(y = {x^2}\,sinx,\,\,\,0 \le x \le \pi \) , correct to six decimal places.

Short answer

The coordinates of the inflection point of the curve are \(\left( {1.519855,\,\,2.306964} \right)\)

Step-by-step solution

Newton’s formula for nth approximation of root

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)

Finding the coordinates of the infection point

Let, \(f\left( x \right) = y = {x^2}\,sinx,\,\,\,0 \le x \le \pi \)

To find a derivative differentiating equation with respect to x.

\(\begin{aligned}{l}f'\left( x \right) &= \frac{d}{{dx}}\left( {{x^2}\,sinx} \right)\\f'\left( x \right) &= 2x\,sinx + {x^2}cosx\end{aligned}\)

Differentiating again:

\(\begin{aligned}{l}f''\left( x \right) &= \frac{d}{{dx}}\left( {2x\,sinx + {x^2}cosx} \right)\\f''\left( x \right) &= 2\sin x + 2x\cos x + \left( {2x\cos x + {x^2}\left( { - \sin x} \right)} \right)\\f''\left( x \right) &= 4x\cos x + \left( {2 - {x^2}} \right)\sin x\end{aligned}\)

Since we are finding the inflation point of\(f\left( x \right)\), which are zeros of the\(f''\left( x \right)\)

Therefore, for finding the inflection point of\(f\left( x \right)\)we need to find the roots of\(f''\left( x \right)\)using Newton’s method.

\(f''\left( x \right) = 4x\cos x + \left( {2 - {x^2}} \right)\sin x\)

\(\begin{aligned}{l}f'''\left( x \right) &= \frac{d}{{dx}}\left( {4x\cos x + \left( {2 - {x^2}} \right)\sin x} \right)\\f'''\left( x \right) &= 4\cos x + 4x\left( { - \sin x} \right) + \left( { - 2x\sin x + \left( {2 - {x^2}} \right)} \right)\\f'''\left( x \right) &= \left( {6 - {x^2}} \right)\cos x - 6x\sin x\end{aligned}\)

Newton’s formula:

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)

Since we are finding the roots of\(f''\left( x \right)\).

Therefore, the formula will be:

\({x_{n + 1}} = {x_n} - \frac{{f''\left( {{x_n}} \right)}}{{f'''\left( {{x_n}} \right)}}\)

The graph for \(f\left( x \right)\)and \(f''\left( x \right)\) is given below:

From the graph, we can conclude that \(f''\left( x \right)\)has only one root in the interval\(\,0 \le x \le \pi \)which is close to 1.5,

Therefore, taking the initial approximation\({x_1} = 1.5\)

From Newton’s formula

\(\begin{aligned}{l}{x_2} &= {x_1} - \frac{{f''\left( {{x_1}} \right)}}{{f'''\left( {{x_1}} \right)}}\\{x_2} &= 1.5 - \frac{{f''\left( {1.5} \right)}}{{f'''\left( {1.5} \right)}}\\{x_2} &= 1.52009247\end{aligned}\)

Similarly,

\(\begin{aligned}{l}{x_4} &= 1.51985533 - \frac{{f''\left( {1.51985533} \right)}}{{f'''\left( {1.51985533} \right)}}\\{x_4} &= 1.51985530\end{aligned}\)

Since the first 6 digits are not changing so, no need for further calculations.

The root of \(f''\left( x \right)\)is\(x = 1.519855\).

For finding the value of function substituting\(x = 1.519855\)in \(f\left( x \right)\).

Hence, the inflection point of the curve is \(\left( {1.519855,\,\,2.306964} \right)\).