Problem 14 - Calculus Early Transcendentals

Question

Find an equation of the plane tangent to the following surfaces at the given points. $$y z e^{x z}-8=0 ;(0,2,4) \text { and }(0,-8,-1)$$

Step-by-Step Solution

Answer

In this exercise, the tangent planes to the surface $$yz e^{xz} - 8 = 0$$ at points $$(0,2,4)$$ and $$(0,-8,-1)$$ are given by the equations $$32x + 4y + 8z - 40 = 0$$ and $$8x - y + 8z = 0$$, respectively.

Step 1: Partial Derivatives

First, we find the partial derivatives of the given function with respect to x, y, and z: $$\frac{\partial f}{\partial x} = yz^2 e^{xz}$$ $$\frac{\partial f}{\partial y} = ze^{xz}$$ $$\frac{\partial f}{\partial z} = yze^{xz} + 0$$ Step 2: Evaluate gradient at given points

Step 2: Find Normal Vector

Now, let's find the gradient of the surface at the given points by plugging in the coordinates of \(P_1(0, 2, 4)\) and \(P_2(0, -8, -1)\): At \(P_1(0, 2, 4)\): $$\nabla f_1 = \begin{bmatrix} 2(4)^2 e^{(0)(4)} \\ 4e^{(0)(2)} \\ 2(4)e^{(0)(4)} \end{bmatrix} = \begin{bmatrix} 32 \\ 4 \\ 8 \end{bmatrix}$$ At \(P_2(0, -8, -1)\): $$\nabla f_2 = \begin{bmatrix} (-8)(-1)^2 e^{0(-1)} \\ (-1)e^{0(-8)} \\ (-8)(-1)e^{0(-1)} \end{bmatrix} = \begin{bmatrix} 8 \\ -1 \\ 8 \end{bmatrix}$$ Step 3: Find the equations of tangent planes

Step 3: Point-Normal Form

Next, we use the point-normal form of a plane equation with the gradients as normal vectors: $$\nabla f \cdot (x - x_0, y - y_0, z - z_0) = 0$$ For \(P_1(0, 2, 4)\) and \(\nabla f_1 = \begin{bmatrix} 32 \\ 4 \\ 8 \end{bmatrix}\): $$32(x - 0) + 4(y - 2) + 8(z - 4) = 0$$ $$32x + 4y - 8 + 8z - 32 = 0$$ Then, the equation of the tangent plane at \((0, 2, 4)\) is: $$32x + 4y + 8z - 40 = 0$$ For \(P_2(0, -8, -1)\) and \(\nabla f_2 = \begin{bmatrix} 8 \\ -1 \\ 8 \end{bmatrix}\): $$8(x - 0) - 1(y + 8) + 8(z + 1) = 0$$ $$8x - y - 8 + 8z + 8 = 0$$ Then, the equation of the tangent plane at \((0, -8, -1)\) is: $$8x - y + 8z = 0$$ Thus, the equations of the tangent planes to the surface \(yz e^{xz} - 8 = 0\) at points \((0,2,4)\) and \((0,-8,-1)\) are \(32x + 4y + 8z - 40 = 0\) and \(8x - y + 8z = 0\), respectively.

Key Concepts

Partial Derivatives

Partial derivatives help us understand how a function changes as we vary one of its variables while keeping others fixed. Consider the function given by the equation \( f(x, y, z) = yz e^{xz} - 8 \). To find how this function behaves with respect to each variable individually, we compute partial derivatives:

• For \( x \): We differentiate with respect to \( x \) while treating \( y \) and \( z \) as constants. This gives us \( \frac{\partial f}{\partial x} = yz^2 e^{xz} \).


• For \( y \): We keep \( x \) and \( z \) constant and differentiate with respect to \( y \), resulting in \( \frac{\partial f}{\partial y} = ze^{xz} \).


• For \( z \): Treating \( x \) and \( y \) as constants, we find \( \frac{\partial f}{\partial z} = yze^{xz} \).

Partial derivatives are crucial for constructing the gradient vector, which informs us about the direction in which the function increases the fastest.

Gradient Vector

The gradient vector is a collection of all the partial derivatives of a function. It indicates the direction of the steepest ascent of the surface described by the function. For a three-variable function \( f(x, y, z) \), its gradient is:

\[ abla f = \begin{bmatrix} \frac{\partial f}{\partial x} \ \frac{\partial f}{\partial y} \ \frac{\partial f}{\partial z} \end{bmatrix} \]

For the given function, we calculate the gradient by substituting the partial derivatives we found earlier:

• At point \( P_1(0, 2, 4) \), the gradient is \( abla f_1 = \begin{bmatrix} 32 \ 4 \ 8 \end{bmatrix} \).


• At point \( P_2(0, -8, -1) \), it becomes \( abla f_2 = \begin{bmatrix} 8 \ -1 \ 8 \end{bmatrix} \).

Understanding the gradient at a specific point is essential for determining the equation of the tangent plane, as it acts as the normal vector to the plane.

Point-Normal Form

The point-normal form of a plane helps us define the plane by using a normal vector and a point on the plane. The general equation for a plane in three-dimensional space is \( abla f \cdot (x - x_0, y - y_0, z - z_0) = 0 \). Here, \( abla f \) is the gradient or normal vector, and \((x_0, y_0, z_0)\) is a point on the plane.

Using the points and gradients obtained, we construct the tangent plane equations:

• For \( P_1(0, 2, 4) \) with gradient \( abla f_1 = \begin{bmatrix} 32 \ 4 \ 8 \end{bmatrix} \), the equation is:
  \( 32(x - 0) + 4(y - 2) + 8(z - 4) = 0 \) which simplifies to \( 32x + 4y + 8z - 40 = 0 \).


• For \( P_2(0, -8, -1) \) with gradient \( abla f_2 = \begin{bmatrix} 8 \ -1 \ 8 \end{bmatrix} \), the equation is:
  \( 8(x - 0) - 1(y + 8) + 8(z + 1) = 0 \) simplifying to \( 8x - y + 8z = 0 \).

These plane equations represent the planes tangent to the surface at the given points, reflecting the changes in all three spatial dimensions.