Question

Find a point on the surface \(z=2 x^{2}+3 y^{2}\) where the tangent plane is parallel to the plane \(8 x-3 y-z=0\).

Short answer

The point on the surface is \((2, -\frac{1}{2}, \frac{35}{4})\).

Step-by-step solution

Tangent Plane Equation

The equation of a tangent plane to the surface \(z = f(x,y)\) at a point \((x_0, y_0, z_0)\) is given by \(z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\), where \(f_x\) and \(f_y\) are partial derivatives of \(f\) with respect to \(x\) and \(y\). In our case, \(f(x,y) = 2x^2 + 3y^2\).

Calculate Partial Derivatives

Compute the partial derivatives of \(f(x, y)\): \( f_x = \frac{{\partial}}{{\partial x}}(2x^2 + 3y^2) = 4x \) and \( f_y = \frac{{\partial}}{{\partial y}}(2x^2 + 3y^2) = 6y \).

Find Tangent Plane Normal Vector

For the tangent plane to be parallel to the plane \(8x - 3y - z = 0\), their normal vectors must be parallel. The normal vector for the given plane is \( \langle 8, -3, -1 \rangle \) and for the tangent plane from the surface is \( \langle f_x(x_0, y_0), f_y(x_0, y_0), -1 \rangle = \langle 4x_0, 6y_0, -1 \rangle \).

Equating the Normal Vectors

Since the tangent plane normal vector \( \langle 4x_0, 6y_0, -1 \rangle \) must be parallel to \( \langle 8, -3, -1 \rangle \), scale them such that \(4x_0/8 = 6y_0/(-3) = 1\). Solve these equations to find \(x_0\) and \(y_0\).

Solve for \(x_0\) and \(y_0\)

Solving \(4x_0 = 8\) gives \(x_0 = 2\) and solving \(6y_0 = -3\) gives \(y_0 = -\frac{1}{2}\).

Find \(z_0\) on the Surface

Substitute \(x_0 = 2\) and \(y_0 = -\frac{1}{2}\) into the surface equation \(z=2x^2 + 3y^2\) to find \(z_0\). \(z_0 = 2(2)^2 + 3(-\frac{1}{2})^2 = 8 + \frac{3}{4} = \frac{35}{4}\).

Verify Tangent Plane is Parallel

The normal vector for the tangent plane at \((2, -\frac{1}{2}, \frac{35}{4})\) is \( \langle 4(2), 6(-\frac{1}{2}), -1 \rangle = \langle 8, -3, -1 \rangle \), confirming it is parallel to \(\langle 8, -3, -1 \rangle\).

Key concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, especially when dealing with functions of multiple variables. When a function, such as the surface equation given by \(z = 2x^2 + 3y^2\), involves more than one variable, the rate of change of the function with respect to one variable while keeping others constant is given by the partial derivative.
\(f_x\) represents the partial derivative of \(f(x, y)\) with respect to \(x\), calculated by differentiating the function while treating \(y\) as constant. Similarly, \(f_y\) is the partial derivative with respect to \(y\).

To find these, you write:

• \(f_x = \frac{\partial}{\partial x}(2x^2 + 3y^2) = 4x\)
• \(f_y = \frac{\partial}{\partial y}(2x^2 + 3y^2) = 6y\)

These computations give insight into how the surface changes as you move along the \(x\) and \(y\) axes, crucial for determining the behavior of the tangent plane at any given point.

Normal Vector

The normal vector is key when understanding planes in three-dimensional space. This vector is perpendicular to the surface at a given point, which is essential in defining the orientation of the tangent plane.
The normal vector for a plane \(ax + by + cz + d = 0\) is \(\langle a, b, c \rangle\). For the plane \(8x - 3y - z = 0\), this is \(\langle 8, -3, -1 \rangle\).

In the context of the tangent plane to a surface \(z = f(x, y)\) at a point, the normal vector can be derived using the partial derivatives: \(\langle f_x, f_y, -1 \rangle\). For our surface, this would be \(\langle 4x_0, 6y_0, -1 \rangle\).

For two planes to be parallel, their normal vectors must be parallel, meaning they are scalar multiples of each other. In this exercise, equating \(\langle 4x_0, 6y_0, -1 \rangle\) to \(\langle 8, -3, -1 \rangle\) helps find the specific point on the surface where the tangent plane parallels the given plane.

Surface Equation

The surface equation represents the set of all points \((x, y, z)\) that satisfy the equation \(z = f(x, y)\). For this exercise, the surface is defined by \(z = 2x^2 + 3y^2\).

This quadratic equation describes a parabolic surface, which means it will curve upward like a bowl as it extends into space. Understanding this helps visualize why the tangent plane has specific properties at different points.

To find a particular point where the tangent plane is parallel to a given plane, you utilize the partial derivatives to establish the normal vector at any point on this surface. Once the point \((x_0, y_0)\) is determined, we find \(z_0\) using the surface equation: \(z_0 = 2x_0^2 + 3y_0^2\). Substituting the known values \(x_0 = 2\) and \(y_0 = -\frac{1}{2}\), the calculation leads to \(z_0 = \frac{35}{4}\).

Thus, a complete understanding of the surface equation aids in pinning down the precise coordinates \((2, -\frac{1}{2}, \frac{35}{4})\) where the conditions are met.