Mathematics-I (3110014)

Winter-2019

Question-3c

BE | Semester-1 Winter-2019 | 17-01-2020

Q3) (c)

Find the maximum and minimum distance from the point $(1, 2, 2)$ to the sphere $x^{2} + y^{2} + z^{2} = 36$ .

$Let (x, y, z) be any point on the sphere . Its distance D from the point (1, 2, 2) is,$

$D = \sqrt{{(x - 1)}^{2} + {(y - 2)}^{2} + {(z - 2)}^{2}}$

$D^{2} = f (x, y, z) = {(x - 1)}^{2} + {(y - 2)}^{2} + {(z - 2)}^{2} . . . . . . . . . [1]$

$Let, φ (x, y, z) = x^{2} + y^{2} + z^{2} - 36 . . . . . . . . . [2]$

$Now, Auxiliary equation is,$

$f (x, y, z) + λ \cdot φ (x, y, z) = 0$

${(x - 1)}^{2} + {(y - 2)}^{2} + {(z - 2)}^{2} + λ \cdot (x^{2} + y^{2} + z^{2} - 36) = 0 . . . . . . . . . [3]$

$Taking partial differentiation of Eq . [3] with respect to x,$

$⟹ 2 (x - 1) + λ \cdot (2 x) = 0$

$⟹ λ \cdot (2 x) = - 2 (x - 1)$

$⟹ λ = \frac{1 - x}{x} = \frac{1}{x} - 1$

$Taking partial differentiation of Eq . [3] with respect to y,$

$⟹ 2 (y - 2) + λ \cdot (2 y) = 0$

$⟹ λ \cdot (2 y) = - 2 (y - 2)$

$⟹ λ = \frac{2 - y}{y} = \frac{2}{y} - 1$

$Taking partial differentiation of Eq . [3] with respect to z,$

$⟹ 2 (z - 2) + λ \cdot (2 z) = 0$

$⟹ λ \cdot (2 z) = - 2 (z - 2)$

$⟹ λ = \frac{2 - z}{z} = \frac{2}{z} - 1$

$Comparing values of λ,$

$\frac{1}{x} - 1 = \frac{2}{y} - 1$

$⟹ \frac{1}{x} = \frac{2}{y}$

$⟹ y = 2 x$

$\frac{1}{x} - 1 = \frac{2}{z} - 1$

$⟹ \frac{1}{x} = \frac{2}{z}$

$⟹ z = 2 x$

$Substituting the values of y and z in given eq . of sphere,$

$x^{2} + y^{2} + z^{2} = 36$

$\Rightarrow x^{2} + 4 x^{2} + 4 x^{2} = 36$

$\Rightarrow 9 x^{2} = 36$

$\Rightarrow x^{2} = 4$

$\Rightarrow x = \pm 2$

$Now,$

$y = 2 x = 2 (\pm 2) = \pm 4$

$z = 2 x = 2 (\pm 2) = \pm 4$

$So, Stationary Point are (2, 4, 4) & (- 2, - 4, - 4) .$

$Putting Values for x, y & z to find distance,$

$D = \sqrt{{(x - 1)}^{2} + {(y - 2)}^{2} + {(z - 2)}^{2}} = \sqrt{{(2 - 1)}^{2} + {(4 - 2)}^{2} + {(4 - 2)}^{2}} = 3$

$D = \sqrt{{(x - 1)}^{2} + {(y - 2)}^{2} + {(z - 2)}^{2}} = \sqrt{{(- 2 - 1)}^{2} + {(- 4 - 2)}^{2} + {(- 4 - 2)}^{2}} = 9$

$So, Minimum distance is 3 and maximum distance is 9 .$

Mathematics-I (3110014)

Q3) (c)

Find the maximum and minimum distance from the point $(1, 2, 2)$ to the sphere $x^{2} + y^{2} + z^{2} = 36$ .

Questions

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Mathematics-I (3110014)

Q3) (c)

Find the maximum and minimum distance from the point 1,2,2 to the sphere x2+y2+z2=36.

Questions

Go to Question Paper

Find the maximum and minimum distance from the point $(1, 2, 2)$ to the sphere $x^{2} + y^{2} + z^{2} = 36$ .