Question

Find the linear approximation of the function\(f(x,y,z) = \sqrt {{x^2} + {y^2} + {z^2}} \) at \((3,2,6)\)and use it to approximate \(\sqrt {{{(3.02)}^2} + {{(1.97)}^2} + {{(5.99)}^2}} \)..

Short answer

The linear approximation at \((3,2,6)\)is given by \(\sqrt {{x^2} + {y^2} + {z^2}} \)\( = \frac{{3x}}{7} + \frac{{2y}}{7} + \frac{{6z}}{7}\)..

Step-by-step solution

Step-1(Equation used in finding the Linear Approximation )

The equation use to find Approximation is:-

\(L(x,y) = f(a,b,c) + {f_x}(a,b,c)(x - a) + {f_y}(a,b,c)(y - b) + {f_z}(a,b,c)(z - c)\),where \({f_x}\),\({f_y}\)and \({f_z}\)are partial derivatives..

Step-2(Calculation of  Partial Derivatives)

\({f_x} = \frac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\)

\({f_x}(3,2,6) = \frac{3}{{\sqrt {{3^2} + {2^2} + {6^2}} }} = \frac{3}{7}\)

Now, the Partial derivative of \({f_y}(x,y,z)\)

\({f_y} = \frac{y}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\)

\({f_y}(3,2,6) = \frac{2}{{\sqrt {{3^2} + {2^2} + {6^2}} }} = \frac{2}{7}\)

Now, the Partial derivative of \({f_z}(x,y,z)\)

\({f_z} = \frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\)

\({f_z}(3,2,6) = \frac{6}{{\sqrt {{3^2} + {2^2} + {6^2}} }} = \frac{6}{7}\)

Partial Derivative exist and continuous at given point.Therefore,Linearization is Possible.

Step-3(Calculation of Linearization of a function)

\(f(3,2,6) = \sqrt {{3^2} + {2^2} + {6^2}} = 7\)

\(f(3,2,6)\)\( = 7\)

\(L(x,y) = f(a,b,c) + {f_x}(a,b,c)(x - a) + {f_y}(a,b,c)(y - b) + {f_z}(a,b,c)(z - c)\)

\(L(3,2,6) = \frac{{3x}}{7} + \frac{{2y}}{7} + \frac{{6z}}{7} + 7 - \frac{9}{7} - \frac{4}{7} - \frac{{36}}{7}\)

\(f(x,y,z) = \frac{{3x}}{7} + \frac{{2y}}{7} + \frac{{6z}}{7}\)

\(\sqrt {{x^2} + {y^2} + {z^2}} \)=\( = \frac{{3x}}{7} + \frac{{2y}}{7} + \frac{{6z}}{7}\)

\(\sqrt {{{(3.02)}^2} + {{(1.97)}^2} + {{(5..99)}^2}} \)\( = \frac{{3(3.02)}}{7} + \frac{{2(1.97)}}{7} + \frac{{6(5.99)}}{7}\)

\( = 6.9914\)