Q14Mathematics I

Question

If $u = sin^{- 1} (\frac{x + y}{x + y})$ , show that $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{1}{2} tan u$ .

Answer 1

In multivariable calculus, the gradient vector of a scalar function $F (x, y, z)$ evaluated at a specific point on the level surface $F (x, y, z) = 0$ provides a vector that is absolutely orthogonal (perpendicular) to the tangent plane at that exact point. This gradient vector is defined as the normal vector $n$ .

Step 1: Calculating the Gradient Vector

The surface is implicitly defined as $F (x, y, z) = x^{3} + y^{3} + 3 x y z - 3 = 0$ . We compute the general gradient vector $\nabla F$ using partial derivatives:

\nabla F = ⟨ \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} ⟩

$\frac{\partial F}{\partial x} = 3 x^{2} + 3 y z$
$\frac{\partial F}{\partial y} = 3 y^{2} + 3 x z$
$\frac{\partial F}{\partial z} = 3 x y$

We now strictly evaluate these partial derivatives precisely at the given coordinate point $P (1, 2, - 1)$ :

X-component: $3 (1)^{2} + 3 (2) (- 1) = 3 - 6 = - 3$
Y-component: $3 (2)^{2} + 3 (1) (- 1) = 12 - 3 = 9$
Z-component: $3 (1) (2) = 6$

Thus, the exact normal vector to the surface at point $P$ is $n = ⟨ - 3, 9, 6 ⟩$ .

Step 2: Deriving the Tangent Plane Equation

The standard mathematical equation for a plane passing strictly through a point $(x_{0}, y_{0}, z_{0})$ with a normal vector $⟨ A, B, C ⟩$ is given by the dot product formulation:

A (x - x_{0}) + B (y - y_{0}) + C (z - z_{0}) = 0

We substitute our derived normal vector and given point coordinates:

- 3 (x - 1) + 9 (y - 2) + 6 (z - (- 1)) = 0

We can divide the entire equation strictly by $- 3$ to elegantly simplify the coefficients:

1 (x - 1) - 3 (y - 2) - 2 (z + 1) = 0

x - 1 - 3 y + 6 - 2 z - 2 = 0 ⟹ x - 3 y - 2 z + 3 = 0

Note: My algebraic expansion yields $+ 3 = 0$ . The summary stated $+ 7 = 0$ . Let me recheck. $- 1 + 6 - 2 = + 3$ . Yes, $+ 3 = 0$ is algebraically correct. I will correct the explanation to reflect this.

Step 3: Deriving the Normal Line Equation

The strict symmetric equations for a 3D line passing through $(x_{0}, y_{0}, z_{0})$ with direction vector $⟨ A, B, C ⟩$ (which is our normal vector) are:

\frac{x - x _{0}}{A} = \frac{y - y _{0}}{B} = \frac{z - z _{0}}{C}

Substituting the raw normal vector $⟨ - 3, 9, 6 ⟩$ :

\frac{x - 1}{- 3} = \frac{y - 2}{9} = \frac{z + 1}{6}

Dividing all denominators by the common factor $- 3$ yields the simplified normal line equation:

\frac{x - 1}{1} = \frac{y - 2}{- 3} = \frac{z + 1}{- 2}

Q14Mathematics I

Question

Answer

Step 1: Calculating the Gradient Vector

Step 2: Deriving the Tangent Plane Equation

Step 3: Deriving the Normal Line Equation