Question

Find the direction of the line normal to the surface \(x^{2} y+y^{2} z+z^{2} x+1=0\) at the point \((1,2,-1)\), Write the equations of the tangent plane and normal line at this point.

Short answer

The normal line direction is (5, -3, 2), the tangent plane equation is \(5x - 3y + 2z + 3 = 0\), and the normal line equation is \( x = 1 + 5t \', y = 2 - 3t \', z = -1 + 2t \).

Step-by-step solution

Determine the Gradient Vector

To find the normal direction to the surface, calculate the gradient \(abla f\) of \( f(x, y, z) = x^2 y + y^2 z + z^2 x + 1 \). The gradient is given by the partial derivatives: \(abla f = abla (x^2 y + y^2 z + z^2 x + 1)\). Compute the partial derivatives \( \frac{abla f}{abla x}\), \( \frac{abla f}{abla y}\), and \( \frac{abla f}{abla z}\).

Compute Partial Derivatives

Find the partial derivatives at the given point (1, 2, -1): \( \frac{abla f}{abla x} = 2xy + z^2 = 2(1)(2) + (-1)^2 = 4 + 1 = 5 \), \( \frac{abla f}{abla y} = x^2 + 2yz = (1)^2 + 2(2)(-1) = 1 - 4 = -3 \), and \( \frac{abla f}{abla z} = y^2 + 2zx = (2)^2 + 2(-1)(1) = 4 - 2 = 2 \). The gradient vector is then \( abla f (1, 2, -1) = (5, -3, 2) \).

Direction of the Normal Line

The direction vector of the normal line is given by the gradient \( abla f (1, 2, -1) = (5, -3, 2) \).

Equation of the Tangent Plane

Using the point (1,2,-1) and the normal vector (5, -3, 2), form the equation of the tangent plane: \(5(x-1) - 3(y-2) + 2(z+1) = 0\). Simplify to get \( 5x - 5 - 3y + 6 + 2z + 2 = 0 \) or \( 5x - 3y + 2z + 3 = 0 \).

Equation of the Normal Line

The equation of the normal line can be written using the point (1,2,-1) and the direction vector (5,-3,2): \( x = 1 + 5t \', \ y = 2 - 3t \', \ z = -1 + 2t \).

Key concepts

Partial Derivatives

In multivariable calculus, partial derivatives measure how a function changes as each individual variable changes, while the others remain constant. To find the gradient vector, compute partial derivatives for each variable of the function.