Problem 185 - Calculus 3

Question

For the following exercises, find parametric equations for the normal line to the surface at the indicated point. (Recall that to find the equation of a line in space, you need a point on the line, \(P_{0}\left(x_{0}, y_{0}, z_{0}\right),\) and a vector \(\mathbf{n}=\langle a, b, c\rangle \) that is parallel to the line. Then the equation \(x-x_{0}=a t, y-y_{0}=b t, z-z_{0}=c t . )\) $$z=\ln \left(3 x^{2}+7 y^{2}+1\right), P(0,0,0)$$

Step-by-Step Solution

Answer

The parametric equations are \( x = 0 \), \( y = 0 \), \( z = t \).

Step 1: Find the Gradient of the Surface

To determine the direction of the normal line, we first need to compute the gradient of the surface, denoted as \( abla z \). The surface is given by \( z = \ln (3x^2 + 7y^2 + 1) \). The gradient \( abla z \) is a vector composed of the partial derivatives of the function. Thus, \( abla z = \langle \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \rangle \). Compute these:\[ \frac{\partial z}{\partial x} = \frac{6x}{3x^2 + 7y^2 + 1} \]\[ \frac{\partial z}{\partial y} = \frac{14y}{3x^2 + 7y^2 + 1} \]

Step 2: Evaluate the Gradient at the Given Point

Next, we evaluate the gradient at the point \((0,0,0)\). Substitute \(x = 0\) and \(y = 0\) into the partial derivatives:\[ \frac{\partial z}{\partial x}(0, 0) = \frac{6 \cdot 0}{1} = 0 \]\[ \frac{\partial z}{\partial y}(0, 0) = \frac{14 \cdot 0}{1} = 0 \]So at the point \((0,0,0)\), the gradient vector \( abla z \) is \( \langle 0, 0 \rangle \). This implies that we need to consider the full derivative \( abla f = \langle 0, 0, 1\rangle \).

Step 3: Write Normal Line Equations

With a point \( P_0(0, 0, 0) \) on the line and a direction vector \( \mathbf{n} = \langle 0, 0, 1 \rangle \), the parametric equations for the normal line to the surface are:\[ x(t) = 0 + 0 \cdot t = 0 \]\[ y(t) = 0 + 0 \cdot t = 0 \]\[ z(t) = 0 + 1 \cdot t = t \]Thus, the parametric equations of the line are \( x=0 \), \( y=0 \), and \( z=t \).

Key Concepts

Gradient of a Surface

The gradient of a surface is a vector that points in the direction of the steepest ascent of a surface. It's denoted by \(abla z\) in this context. The function given is \(z = \ln(3x^2 + 7y^2 + 1)\).
To find the gradient of such a function, you'll need partial derivatives pertaining to each variable. Here, the gradient becomes \(abla z = \langle \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \rangle\).

Calculating these gives us:

• For \(x\), \(\frac{\partial z}{\partial x} = \frac{6x}{3x^2 + 7y^2 + 1}\)
• For \(y\), \(\frac{\partial z}{\partial y} = \frac{14y}{3x^2 + 7y^2 + 1}\)

Together, these together form a vector describing the direction and strength of the slope at any given point on the surface. This information guides us in tracing the normal line, which is a key concept for graphs involving parametric equations.

Normal Line

The normal line to a surface at a given point is a line that is perpendicular to the tangent plane of that surface at the specified point. In the context of parametric equations, it helps in understanding the orientation of the surface in space.

When dealing with functions like \(z = \ln(3x^2 + 7y^2 + 1)\), the initial step is finding the gradient, as discussed. At the point \((0,0,0)\), the evaluated gradient gives you the direction vector \(\mathbf{n} = \langle 0, 0, 1 \rangle\). This vector is crucial to depict the orientation of the normal line.

Now, using the point and this vector, parametric equations are formulated:

• \(x(t) = 0\)
• \(y(t) = 0\)
• \(z(t) = t\)

This means the normal line rises linearly with \(z\), while \(x\) and \(y\) remain constant.

Partial Derivatives

Partial derivatives are used to measure how a multivariable function changes as each variable is varied independently. In this exercise, for the surface \(z = \ln(3x^2 + 7y^2 + 1)\), we find two crucial partial derivatives.

These derivatives, \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\), provide insight into how the surface's gradient behaves:

• \(\frac{\partial z}{\partial x} = \frac{6x}{3x^2 + 7y^2 + 1}\)
• \(\frac{\partial z}{\partial y} = \frac{14y}{3x^2 + 7y^2 + 1}\)

Both derivatives are computed at \((0, 0)\) to yield \(0\), which implies the direction of steepest ascent is perpendicular to the \(xy\)-plane at this point, an important observation for constructing the normal line.

Direction Vector

A direction vector is a vector that indicates direction without consideration of location. For the parametric equations of a line, this vector guides the line straight through space.

In this exercise, after determining the partial derivatives, we find that the vector \(\langle 0, 0, 1 \rangle\) is the direction vector for our normal line.

This vector tells us that the line moves in the direction where only \(z\) changes and \(x, y\) remain at zero. This consistency presents a clear path for constructing the line using parametric equations, which breaks down to:

• \(x(t) = 0\)
• \(y(t) = 0\)
• \(z(t) = t\)

Thus, by using this direction vector, you create a line that is perpendicular to the tangent of the surface at \((0, 0, 0)\).