Chapter 6: Problem 6

Find a vector normal to the surface \(x^{2}+y^{2}-z=0\) at the point (3,4,25) . Find the equations of the tangent plane and normal line to the surface at that point.

Short Answer

Expert verified

Gradient vector at (3,4,25) is (6,8,-1). Tangent plane equation: 6x+8y-z=25. Normal line equations: x=3+6t, y=4+8t, z=25-t.

Step by step solution

Find the gradient vector

Given the surface equation is \[ x^2 + y^2 - z = 0 \] Calculate the gradient vector \( abla F \) of \( F(x, y, z) = x^2 + y^2 - z \). The gradient vector \( abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \).

Compute partial derivatives

Find the partial derivatives of \( F(x, y, z) \): \[ \frac{\partial F}{\partial x} = 2x \] \[ \frac{\partial F}{\partial y} = 2y \] \[ \frac{\partial F}{\partial z} = -1 \]

Evaluate the gradient at the given point

Substitute \((x, y, z) = (3, 4, 25)\) into the gradient: \[ abla F(3, 4, 25) = (2 \cdot 3, 2 \cdot 4, -1) = (6, 8, -1) \]. This is the normal vector to the surface at the point (3,4,25).

Find the equation of the tangent plane

The equation of the tangent plane at the point \( (x_0, y_0, z_0) \) is: \[ (abla F)\cdot((x - x_0), (y - y_0), (z - z_0)) = 0 \] Using the normal vector \( abla F(3, 4, 25) = (6, 8, -1) \),and the point (3,4,25), the tangent plane equation is: \[ 6(x-3) + 8(y-4) - (z-25) = 0 \] Simplifying:\[ 6x + 8y - z = 25 \]

Find the parametric equations of the normal line

The normal line through the point \( (3, 4, 25) \) with direction ratios \( (6, 8, -1) \): The parametric equations of the normal line are: \[ x = 3 + 6t \] \[ y = 4 + 8t \] \[ z = 25 - t \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Vector

In vector calculus, the normal vector to a surface at a given point is a crucial concept. It is perpendicular to the tangent plane of the surface at that point. For the surface described by the equation \(x^2 + y^2 - z = 0\), the gradient vector at any point gives the normal vector to the surface at that point. To find this, compute the partial derivatives of the function \(F(x, y, z) = x^2 + y^2 - z\). This yields the components of the normal vector as follows:

\(\frac{\partial F}{\partial x} = 2x\)
\(\frac{\partial F}{\partial y} = 2y\)
\(\frac{\partial F}{\partial z} = -1\)

At the point (3, 4, 25), substituting the values into the partial derivatives, we get the normal vector \((6, 8, -1)\). This vector is perpendicular to the surface at the given point.

Gradient Vector

The gradient vector is an essential tool in vector calculus. It is a vector that points in the direction of the greatest rate of increase of the function. For the function \(F(x, y, z) = x^2 + y^2 - z\), the gradient vector \(abla F\) is calculated by taking the partial derivatives of F with respect to x, y, and z. This provides the components of the gradient vector:

\(\frac{\partial F}{\partial x} = 2x\)
\(\frac{\partial F}{\partial y} = 2y\)
\(\frac{\partial F}{\partial z} = -1\)

By substituting the coordinates of the point (3, 4, 25) into these partial derivatives, we get the gradient vector \((6, 8, -1)\). This vector is not only the gradient but also the normal vector to the surface at that point.

Tangent Plane

The tangent plane to a surface at a point provides a linear approximation of the surface around that point. To determine the equation of the tangent plane for the surface \(x^2 + y^2 - z = 0\) at the point (3, 4, 25), use the gradient vector \((6, 8, -1)\). The equation of the tangent plane is given by:\[(abla F) \bullet (x - x_0, y - y_0, z - z_0) = 0\]Substituting \( abla F = (6, 8, -1) \)\ and the point \( (3, 4, 25) \), we get:\[6(x-3) + 8(y-4) - (z-25) = 0\]Simplify this to:\[6x + 8y - z = 25\]This is the equation of the tangent plane at the given point.

Normal Line

The normal line to a surface at a given point is a line that passes through the point and is perpendicular to the tangent plane. For the surface \(x^2 + y^2 - z = 0\) at the point (3, 4, 25), the normal line has direction ratios given by the components of the normal vector \((6, 8, -1)\). The parametric equations of the normal line can be written as:

\(x = 3 + 6t\)
\(y = 4 + 8t\)
\(z = 25 - t\)

These equations describe the normal line that passes through the point \((3, 4, 25)\) and follows the direction of the normal vector.

Partial Derivatives

Partial derivatives represent the rate of change of a function with respect to one variable while keeping other variables constant. For the function \(F(x, y, z) = x^2 + y^2 - z\), the partial derivatives with respect to x, y, and z are:

\(\frac{\partial F}{\partial x} = 2x\)
\(\frac{\partial F}{\partial y} = 2y\)
\(\frac{\partial F}{\partial z} = -1\)

By substituting the coordinates \((3, 4, 25)\) into these expressions, we find the values of these partial derivatives, yielding the components of the gradient vector:\((6, 8, -1)\). Partial derivatives are essential in calculating the gradient vector and, consequently, the normal vector to the surface at a given point.