Question

Find equations of (a) the tangent plane and (b) the normal line to the given surface at the specified point.

\(2{(x - 2)^2} + {(y - 1)^2} + {(z - 3)^2} = 10,\quad (3,3,5)\)

Short answer

(a) The equation of the tangent plane to the surface \(2{(x - 2)^2} + {(y - 1)^2} + {(z - 3)^2} = 10\) at the point \((3,3,5)\) is \(x + y + z = 11\).

(b) The equation of the normal line to the surface \(2{(x - 2)^2} + {(y - 1)^2} + {(z - 3)^2} = 10\) at the point \((3,3,5)\) is \(\frac{{x - 3}}{4} = \frac{{y - 3}}{4} = \frac{{z - 5}}{4}\).

Step-by-step solution

The tangent plane and normal to the level surface

The tangent plane to the level surface at the point\(P\left( {{x_0},{y_0},{z_0}} \right)\)is defined as\({F_x}\left( {{x_0},{y_0},{z_0}} \right)\left( {x - {x_0}} \right) + {F_y}\left( {{x_0},{y_0},{z_0}} \right)\left( {y - {y_0}} \right) + {F_z}\left( {{x_0},{y_0},{z_0}} \right)\left( {z - {z_0}} \right) = 0\)

The normal to the level surface at the point\(P\left( {{x_0},{y_0},{z_0}} \right)\)is defined as\(\begin{aligned}{l}\frac{{\left( {x - {x_0}} \right)}}{{{F_x}\left( {{x_0}{y_0},{z_0}} \right)}} = \frac{{\left( {y - {y_0}} \right)}}{{{F_y}\left( {{x_0},{y_0},{z_0}} \right)}}\\ = \frac{{(z - z0)}}{{{F_z}\left( {{x_0},{y_0},{z_0}} \right)}}\eta \end{aligned}\).

The tangent plane

The given surface is, \(2{(x - 2)^2} + {(y - 1)^2} + {(z - 3)^2} = 10\).

Let the surface function be, \(F(x,y,z) = 2{(x - 2)^2} + {(y - 1)^2} + {(z - 3)^2} - 10\). (1)

The equation of the tangent plane to the given surface at the point \((3,3,5)\) is defined by, \({F_x}(3,3,5)(x - 3) + {F_y}(3,3,5)(y - 3) + {F_z}(3,3,5)(z - 5) = 0.\) (2)

Take partial derivative with respect to \(x\) at the point \((3,3,5)\) in the equation (1), \({F_x}(x,y,z) = 2{(x - 2)^2} + {(y - 1)^2} + {(z - 3)^2} - 10\)

\(\begin{aligned}{l} = 2(2(x - 2)(1)) + 0 + 0 - 0\\ = 4x - 8\end{aligned}\)

The value of \({F_x}(x,y,z)\) at the point \((3,3,5)\) is,

\({F_x}(3,3,5) = 4(3) - 8 = 4\)

Thus, the value of \({F_x}(3,3,5) = 4\).

Take partial derivative with respect to \(y\) at the point \((3,3,5)\) in the equation (1),

\(\begin{aligned}{l}{F_y}(x,y,z) = 2{(x - 2)^2} + {(y - 1)^2} + {(z - 3)^2} - 10\\ = 0 + 2(y - 1)(1) + 0 - 0\\ = 2y - 2\end{aligned}\)

The value of \({F_y}(x,y,z)\) at the point \((3,3,5)\) is,

\(\begin{aligned}{l}{F_y}(3,3,5) = 2(3) - 2\\ = 4\end{aligned}\)

Thus, the value of \({F_y}(3,3,5) = 4\).

Take partial derivative with respect to \(z\) at the point \((3,3,5)\) in the equation (1),

\(\begin{aligned}{l}{F_z}(x,y,z) = 2{(x - 2)^2} + {(y - 1)^2} + {(z - 3)^2} - 10\\ = 0 + 0 + 2(z - 3)(1) - 0\\ = 2z - 6\end{aligned}\)

The value of \({F_z}(x,y,z)\) at the point \((3,3,5)\) is,

\({F_z}(3,3,5) = 2(5) - 6\)

\( = 4\)

Thus, the value of \({F_z}(3,3,5) = 4\).

Substitute the respective values in the equation (2) and obtain,

\(\begin{aligned}{l}4(x - 3) + 4(y - 3) + 4(z - 5) = 04x - 12 + 4y - 12 + 4z - 20\\ = 04x + 4y + 4z - 44\\ = 0x + y + z - 11\\ = 0\end{aligned}\)

Thus, the equation of the tangent plane to the surface \(2{(x - 2)^2} + {(y - 1)^2} + {(z - 3)^2} = 10\) at the point \((3,3,5)\) is \(x + y + z = 11\).

The normal line to the surface

(b)

The given surface function is, \(2{(x - 2)^2} + {(y - 1)^2} + {(z - 3)^2} = 10\).

Let the surface function be, \(F(x,y,z) = 2{(x - 2)^2} + {(y - 1)^2} + {(z - 3)^2} - 10 = 0\).

The normal line is defined as, \(\frac{{(x - 3)}}{{{F_x}(3,3,5)}} = \frac{{(y - 3)}}{{{F_y}(3,3,5)}} = \frac{{(z - 5)}}{{{F_2}(3,3,5)}} \cdot \) (3)

From part (a),

The value of \({F_x}(3,3,5) = 4,{F_y}(3,3,5) = 4\) and \({F_z}(3,3,5) = 4\).

Substitute the respective values in the equation (3) and obtain the equation of normal line.

\(\frac{{(x - 3)}}{4} = \frac{{(y - 3)}}{4} = \frac{{(z - 5)}}{4}\)

Thus, the equation of normal line to the surface \(2{(x - 2)^2} + {(y - 1)^2} + {(z - 3)^2} = 10\) at the point \((3,3,5)\) is \(\frac{{x - 3}}{4} = \frac{{y - 3}}{4} = \frac{{z - 5}}{4}\).