Question

Consider the following surfaces \(f(x, y, z)=0,\) which may be regarded as a level surface of the function \(w=f(x, y, z) . A\) point \(P(a, b, c)\) on the surface is also given. a. Find the (three-dimensional) gradient of \(f\) and evaluate it at \(P\). b. The heads of all vectors orthogonal to the gradient with their tails at \(P\) form a plane. Find an equation of that plane (soon to be called the tangent plane). $$f(x, y, z)=8-x y z=0 ; P(2,2,2)$$

Short answer

Question: Determine the equation of the tangent plane to the level surface defined by the function \(f(x, y, z) = 8 - xyz\) at the point \(P(2, 2, 2)\). Answer: The equation for the tangent plane is given by \(x + y + z = 6\).

Step-by-step solution

Find the gradient of function f.

To find the gradient of the function \(f(x, y, z) = 8 - xyz\), we need to find its partial derivatives with respect to \(x\), \(y\), and \(z\). We compute these partial derivatives as follows: $$ \nabla f(x, y, z) = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \rangle $$ Taking the partial derivatives, we find: $$ \frac{\partial f}{\partial x} = -yz, \quad \frac{\partial f}{\partial y} = -xz, \quad \frac{\partial f}{\partial z} = -xy $$ So, the gradient of the function is: $$ \nabla f(x, y, z) = \langle -yz, -xz, -xy \rangle $$

Evaluate the gradient at P.

With the gradient found, we need to evaluate it at the point \(P(2, 2, 2)\): $$ \nabla f(2, 2, 2) = \langle -2 \cdot 2, -2 \cdot 2, -2 \cdot 2 \rangle = \langle -4, -4, -4 \rangle $$

Find an equation for the tangent plane.

As we have found the gradient vector at point P, we can now find the equation of the tangent plane. Recall that the tangent plane is orthogonal to the gradient vector. Hence, its normal vector is the gradient vector itself. Given the gradient vector \(\langle -4, -4, -4 \rangle\) and the point \(P(2, 2, 2)\), we can use the point-normal form of the equation for the plane: $$\begin{aligned}&(-4)(x - 2) + (-4)(y - 2) + (-4)(z - 2) = 0\\&-4(x-2)-4(y-2)-4(z-2)=0\end{aligned}$$ Now we can simplify the equation for the tangent plane: $$\begin{aligned}-4x + 8 - 4y + 8 - 4z + 8 &= 0\\-4x - 4y - 4z &= -24\end{aligned}$$ Finally, we can divide by -4 to get the equation of the plane in its simplest form: $$x + y + z = 6$$ This is the equation for the tangent plane.

Key concepts

Gradient

The concept of the gradient is central in multivariable calculus. Think of it as a vector that points in the direction of the greatest rate of increase of a function. For a function of three variables, like our function \( f(x, y, z) = 8 - xyz \), the gradient vector has three components, each being the partial derivative of the function with respect to one of the variables. This is represented as:

• \( \frac{\partial f}{\partial x} \) - Partial derivative with respect to \( x \)
• \( \frac{\partial f}{\partial y} \) - Partial derivative with respect to \( y \)
• \( \frac{\partial f}{\partial z} \) - Partial derivative with respect to \( z \)

The gradient vector is written as \( abla f = \langle -yz, -xz, -xy \rangle \), signifying how fast and in which direction the function changes concerning each variable.
Evaluating this at a specific point, like \( P(2,2,2) \), gives the specific direction of the steepest ascent at that point.

Tangent Plane

When you have a surface, such as a level surface of a function, at any given point on the surface, there exists a plane that 'just touches' the surface at that point without slicing through it. This is known as the tangent plane.
It helps to imagine the tangent plane as the mathematical equivalent of a flat surface like paper that is gently laid over a curved surface, matching it perfectly at the point of contact.
The tangent plane at a point is orthogonal (perpendicular) to the gradient vector at that same point. Therefore, to find the equation of this plane, we can use the gradient vector obtained previously: \( \langle -4, -4, -4 \rangle \).
The equation is derived using the normal vector (which is the gradient vector) and the known point, resulting in the equation:

• \(x + y + z = 6\)

This equation describes the plane that lies tangent to the surface at point \( P(2,2,2) \).

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They measure how a function changes as only one of its input variables changes, holding all other input variables constant.
This is similar to how regular derivatives show the rate of change of a function with respect to one variable in single-variable calculus.
When working with functions of multiple variables, such as \( f(x, y, z) \), the partial derivative \( \frac{\partial f}{\partial x} \) gives the change in \( f \) when \( x \) changes, but \( y \) and \( z \) stay the same.
In our original exercise, we calculated:

• \( \frac{\partial f}{\partial x} = -yz \)
• \( \frac{\partial f}{\partial y} = -xz \)
• \( \frac{\partial f}{\partial z} = -xy \)

These partial derivatives are crucial as they compose the gradient, pointing to the most significant rate of increase.

Level Surface

The term 'level surface' in multivariable calculus refers to the set of points that satisfy an equation of the form \( f(x, y, z) = c \), where \( c \) is a constant.
This concept is similar to contour lines on a map that show places with a constant altitude.
For our problem, the surface described by \( f(x, y, z) = 0 \) represents a specific cutting section of the three-dimensional graph of \( f \).
Level surfaces are significant in visualizing how functions behave in three-dimensional space and they help when solving equations involving vectors and planes.
By understanding these surfaces, you can gain insights into the connections between different points in the function’s domain, like predicting how the surface might interact with a tangent plane or how to orient the gradient vector effectively.