Question

Find the gradient of \(f(x, y, z)=\sin \sqrt{x^{2}+y^{2}+z^{2}}\). Show that the gradient always points directly toward the origin or directly away from the origin.

Short answer

The gradient points in the direction of the position vector (x, y, z) and is oriented depending on \( \cos(u) \).

Step-by-step solution

Calculate the partial derivatives

To find the gradient of the function \( f(x, y, z) = \sin(\sqrt{x^2 + y^2 + z^2}) \), we need to compute its partial derivatives with respect to \( x \), \( y \), and \( z \). We set \( u = \sqrt{x^2 + y^2 + z^2} \) so that \( f(x, y, z) = \sin(u) \). The chain rule gives\[ \frac{\partial f}{\partial x} = \cos(u) \cdot \frac{\partial u}{\partial x} = \cos(u) \cdot \frac{x}{u} \] because \( u = \sqrt{x^2 + y^2 + z^2} \) and \( \frac{\partial u}{\partial x} = \frac{x}{u} \).

Repeat for other variables

Similarly, for \( y \) and \( z \), we have:\[ \frac{\partial f}{\partial y} = \cos(u) \cdot \frac{y}{u} \]\[ \frac{\partial f}{\partial z} = \cos(u) \cdot \frac{z}{u} \]

Form the Gradient Vector

The gradient \( abla f \) is formed by combining these partial derivatives:\[ abla f = \left( \cos(u) \cdot \frac{x}{u}, \cos(u) \cdot \frac{y}{u}, \cos(u) \cdot \frac{z}{u} \right) \]

Show the direction of the gradient

Notice that the gradient vector can be written as:\[ abla f = \frac{\cos(u)}{u} \cdot (x, y, z) \]Here, \( (x, y, z) \) is the position vector from the origin. Thus, the gradient vector \( abla f \) is a scalar multiple of \( (x, y, z) \), showing that it points in the direction of \( (x, y, z) \) or opposite depending on the sign of \( \cos(u) \).

Conclusion about the gradient direction

The gradient \( abla f \) always points either directly away from the origin or towards it because it is parallel to the position vector \( (x, y, z) \). If \( \cos(u) > 0 \), the gradient points away from the origin; if \( \cos(u) < 0 \), it points toward the origin.

Key concepts

Partial Derivatives

Partial derivatives are key in multivariable calculus, as they reveal how a function changes with respect to changes in each variable, while keeping the others constant. For a function of multiple variables like \( f(x, y, z) \), a partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), measures the rate of change of the function as \( x \) varies, with \( y \) and \( z \) held fixed.
In the given problem, to find the gradient, we calculate the partial derivatives of \( f \) with respect to \( x \), \( y \), and \( z \). Each of these derivatives shows how the sine of the distance from the origin changes as you move along each axis. This is a powerful tool for understanding the behavior of functions in higher dimensions.

• Partial derivatives help us explore how changes in each separate variable affect the function.
• They are fundamental in forming the gradient vector, which tells us the direction and rate of fastest increase.

Chain Rule

The chain rule is a critical technique in calculus used to differentiate composite functions. When dealing with a function like \( f(x, y, z) = \sin(\sqrt{x^2 + y^2 + z^2}) \), the chain rule allows us to find the derivative by breaking the differentiation into simpler steps.
In this problem, we set \( u = \sqrt{x^2 + y^2 + z^2} \) so that our function becomes \( \sin(u) \). The chain rule helps us differentiate \( \sin(u) \), a function of \( u \), by first differentiating with respect to \( u \) and then multiplying by the derivative of \( u \) with respect to each variable \( x, y, z \). This is crucial for finding each partial derivative.

• The chain rule simplifies differentiation of complex composite functions.
• It breaks down a complex differentiation process into steps involving simpler functions.
• In multivariable contexts, it integrates well with partial derivative calculations.

Directional Vector

A directional vector indicates the direction in which a function increases most rapidly at a given point. In the context of the gradient of a function, the gradient itself serves as a powerful directional vector.
The exercise demonstrates that the gradient vector \( abla f \) not only shows the direction of greatest increase but also aligns with the position vector \( (x, y, z) \) or its opposite. This linkage is shown through the scalar multiplication of \( (x, y, z) \) by \( \frac{\cos(u)}{u} \) in the gradient expression. This alignment means the gradient inherently points towards or away from the origin, depending on the sign of \( \cos(u) \).

• The gradient as a directional vector points towards the most significant rate of increase.
• Its alignment conveys additional geometric insights, especially in spatial contexts.

Position Vector

The position vector is a fundamental geometric construct, representing the location of a point in space relative to the origin. It is expressed in the form \( (x, y, z) \), indicating directions along the x, y, and z axes.
In this particular exercise, the gradient vector is found to be parallel to the position vector. This means the direction indicated by the position vector also determines the potential directions of the gradient vector, pointing either towards or away, based on the scalar multiplication.
Being parallel means that the gradient vector shares the same direction as the position vector, making it clear how geometric properties of vectors intersect with calculus concepts.

• Position vectors provide a spatial reference from the origin.
• In calculations like these, they help visualize the implications of mathematical operations.