Question

Looking ahead- tangent planes 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 set 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: Find the gradient of the function f(x, y, z) given by f(x, y, z) = −x^2 − y^2 − z^2, and the equation of the tangent plane to the surface at the point P(2, 2, 2). Answer: The gradient of the function f(x, y, z) at any point (x, y, z) is given by (-2x, -2y, -2z). At the point P(2, 2, 2), the gradient is (-4, -4, -4). The equation of the tangent plane to the surface at the point P(2, 2, 2) is -4(x - 2) - 4(y - 2) - 4(z - 2) = 0, which simplifies to -4x - 4y - 4z + 12 = 0.

Step-by-step solution

Find the gradient of function f(x,y,z)

In order to find the gradient of the given function, we need to compute its partial derivatives with respect to x, y, and z and then create a vector from the resulting expressions. The gradient of a function is given by: $$\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)$$ Compute partial derivatives: $$\frac{\partial f}{\partial x} = -y z$$ $$\frac{\partial f}{\partial y} = -x z$$ $$\frac{\partial f}{\partial z} = -x y$$ The gradient of f is: $$\nabla f= \left(-y z, -xz, -xy\right)$$

Evaluate the gradient at P(a, b, c)

Now that we have the gradient of our function f, we can calculate its value at the given point P(2,2,2): $$\nabla f_{(2, 2, 2)} = (-2(2)(2), -2(2)(2), -2(2)(2)) = (-8, -8, -8)$$

Determine the tangent plane equation

Now that we have evaluated the gradient vector, we can use it as the normal vector for the tangent plane. Let \((x, y, z)\) be the general point on the tangent plane, and \((a, b, c)\) be the known point P(2,2,2). The equation of the tangent plane can be found using the dot product of the normal vector with the vector formed by the general point and known point, i.e., \((x-2, y-2, z-2)\). So, the equation of the tangent plane is: $$(x-2, y-2, z-2) \cdot (-8, -8, -8) = 0$$ This simplifies to: $$-8(x-2) -8(y-2) -8(z-2) = 0$$

Simplify the tangent plane equation

Finally, we can simplify the tangent plane equation: $$-8x +16 -8y +16 -8z +16 = 0$$ Now combine constant terms: $$-8x -8y -8z +48 = 0$$ The equation of the tangent plane at point P(2, 2, 2) is: $$-8x -8y -8z +48 = 0$$

Key concepts

Gradient

In multivariable calculus, the concept of the gradient plays a crucial role. For a function of three variables, like in our exercise, the gradient is a vector that points in the direction of the greatest rate of increase of the function. It comprises the partial derivatives of the function with respect to each variable:

• The partial derivative with respect to x tells us how the function changes as we vary x while keeping y and z constant.
• The partial derivative with respect to y reveals the rate of change as y varies, holding x and z steady.
• The partial derivative with respect to z illustrates how the function changes as z changes, with x and y constant.

The gradient (\( abla f \)) is expressed as:\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \]In our solution, the gradient of the function \( f(x, y, z) = 8 - xyz \)is \((-yz, -xz, -xy)\), which informs us about its behavior around any point \( (x, y, z) \).

Partial Derivatives

Partial derivatives allow us to understand how a multivariable function changes with respect to one variable at a time, keeping the others fixed. They are fundamental in computing the gradient vector. Let's highlight how:

• For the partial derivative \( \frac{\partial f}{\partial x} \), treat y and z as constants and differentiate normally with respect to x.
• Similarly, for \( \frac{\partial f}{\partial y} \), hold x and z constant, differentiating with respect to y.
• For \( \frac{\partial f}{\partial z} \), consider x and y as constants and differentiate with respect to z.

In our case, with \( f(x, y, z) = 8 - xyz \), the partial derivatives are:

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

These derivatives are the building blocks of the gradient and are essential for understanding the local behavior of the function.

Level Surfaces

A level surface is essentially a slice of three-dimensional space, where a multivariable function equals a constant value. In simpler terms, if you think of a mountain range mapped to a function, a level surface could be considered as a curve of constant elevation (level).
For a function \( f(x, y, z) \), setting it equal to a constant like zero def(\( f(x, y, z) = 0 \)) describes a surface in this space. This collection of points creates what we call the level surface.
In our exercise, the level surface is defined by \( f(x, y, z) = 8 - x y z = 0 \). Points on this surface yield special geometric and analytical information about the behavior at any given point, such as the tangent plane.

Normal Vector

The normal vector is a fundamental concept for understanding the geometry of surfaces. For a given point on a surface, the normal vector is perpendicular, or orthogonal, to the tangent plane at that point. This orthogonal relationship makes the normal vector especially useful:

• It's derived directly from the gradient of the function at the point of interest.
• The normal vector remains constant in all directions defining orthogonal space to the surface.

In our particular problem, the gradient vector \( abla f = (-yz, -xz, -xy) \)evaluated at \( P(2, 2, 2) \)becomes \((-8, -8, -8)\).
This vector acts as the normal vector to the tangent plane at that point. The equation of the tangent plane is constructed by utilizing this normal vector, emphasizing its role in capturing the essence of the surface's orientation at \((2, 2, 2)\).