Math Problem Statement

Find equations of the tangent plane and normal line to the surface x = y^2 + 5z^2 - 439 at the point (2, 6, -9). Tangent Plane: (make the coefficient of x equal to 1).

Solution

Let's start by solving the problem step-by-step.

We are given the surface equation: x=y2+5z2439x = y^2 + 5z^2 - 439 and asked to find the equations of the tangent plane and the normal line at the point (2,6,9)(2, 6, -9).

1. Tangent Plane Equation:

The general formula for the equation of the tangent plane to a surface F(x,y,z)=0F(x, y, z) = 0 at a point (x0,y0,z0)(x_0, y_0, z_0) is: Fx(x0,y0,z0)(xx0)+Fy(x0,y0,z0)(yy0)+Fz(x0,y0,z0)(zz0)=0F_x(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = 0 Where FxF_x, FyF_y, and FzF_z are the partial derivatives of the surface equation with respect to xx, yy, and zz, respectively.

However, the equation is already in the form x=f(y,z)x = f(y, z), so we only need to compute the partial derivatives of the right-hand side with respect to yy and zz.

  1. Find partial derivatives: xy=y(y2+5z2439)=2y\frac{\partial x}{\partial y} = \frac{\partial}{\partial y}(y^2 + 5z^2 - 439) = 2y xz=z(y2+5z2439)=10z\frac{\partial x}{\partial z} = \frac{\partial}{\partial z}(y^2 + 5z^2 - 439) = 10z

  2. Evaluate these derivatives at the point (2,6,9)(2, 6, -9): xy=2(6)=12\frac{\partial x}{\partial y} = 2(6) = 12 xz=10(9)=90\frac{\partial x}{\partial z} = 10(-9) = -90

Now, the tangent plane has the form: (xx0)=xy(yy0)+xz(zz0)(x - x_0) = \frac{\partial x}{\partial y}(y - y_0) + \frac{\partial x}{\partial z}(z - z_0) Substituting x0=2x_0 = 2, y0=6y_0 = 6, and z0=9z_0 = -9: (x2)=12(y6)90(z+9)(x - 2) = 12(y - 6) - 90(z + 9)

  1. Simplify this equation to make the coefficient of xx equal to 1: x2=12(y6)90(z+9)x - 2 = 12(y - 6) - 90(z + 9)

2. Normal Line Equation:

The direction vector for the normal line is given by the gradient of the function: F=1,xy,xz=1,12,90\nabla F = \left\langle 1, \frac{\partial x}{\partial y}, \frac{\partial x}{\partial z} \right\rangle = \langle 1, 12, -90 \rangle So, the parametric equation for the normal line is: x,y,z=2,6,9+t1,12,90\langle x, y, z \rangle = \langle 2, 6, -9 \rangle + t \langle 1, 12, -90 \rangle

Thus, the equations for the normal line are: x=2+tx = 2 + t y=6+12ty = 6 + 12t z=990tz = -9 - 90t

Would you like more details or clarifications? Here are 5 related questions for further practice:

  1. How would you find the tangent plane if the surface was implicitly given, such as F(x,y,z)=0F(x, y, z) = 0?
  2. Can you generalize the process to find the tangent plane to any surface f(x,y,z)=0f(x, y, z) = 0?
  3. What would happen to the normal line if we chose a different point on the surface?
  4. How does the method change if the surface equation is more complex?
  5. What is the geometric interpretation of the tangent plane and normal line in 3D space?

Tip: Always remember that the direction of the normal line is parallel to the gradient of the surface at a given point.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differentiation
Tangent Plane
Normal Line
Gradient

Formulas

Tangent plane equation: F_x(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = 0
Normal line equation: <x, y, z> = <x_0, y_0, z_0> + t<dF_x, dF_y, dF_z>

Theorems

Gradient theorem
Partial derivatives

Suitable Grade Level

Grades 11-12 or early university level