Differentiating a dot product of two vector functions involves a rule that mirrors the familiar product rule from ordinary calculus. If you have two vector functions, u(t) and v(t), which depend on a scalar variable t (often representing time), the derivative of their dot product with respect to t is given by:
$\frac{d}{dt} [\mathbf{u}(t) \cdot \mathbf{v}(t)] = \mathbf{u}'(t) \cdot \mathbf{v}(t) + \mathbf{u}(t) \cdot \mathbf{v}'(t)$
This formula is remarkably similar to the product rule for scalar functions, where you differentiate the first term, dot it with the second, then add the first term dotted with the derivative of the second.
The Dot Product Differentiation Rule
The rule states that the derivative of the dot product of two vector functions is the dot product of the derivative of the first vector function with the second vector function, plus the dot product of the first vector function with the derivative of the second vector function.
Let's break down the components of this essential formula:
- u(t) and v(t): These are vector functions, meaning each component of the vector depends on the scalar variable t. For example, u(t) could be
<x(t), y(t), z(t)>. - u'(t) and v'(t): These represent the derivatives of the vector functions with respect to t. To find the derivative of a vector function, you simply differentiate each of its components individually with respect to t. For instance, if u(t) =
<x(t), y(t), z(t)>, then u'(t) =<x'(t), y'(t), z'(t)>. - ⋅: This symbol denotes the dot product operation. Remember that the dot product of two vectors results in a scalar value.
- $\frac{d}{dt}$: This indicates differentiation with respect to the scalar variable t.
Understanding the Formula
The beauty of this rule lies in its consistency with the scalar product rule, making it intuitive to remember and apply. When you differentiate a dot product, the output is a scalar function, just as the original dot product of two vector functions is a scalar function. This is because the operations involved (u'(t) ⋅ v(t) and u(t) ⋅ v'(t)) each yield a scalar, and the sum of two scalars is another scalar.
Why This Rule Works
The rule can be rigorously derived by expressing the vector functions in their component forms. For instance, if u(t) = <u_x(t), u_y(t), u_z(t)> and v(t) = <v_x(t), v_y(t), v_z(t)>, their dot product is:
u(t) ⋅ v(t) = u_x(t)v_x(t) + u_y(t)v_y(t) + u_z(t)v_z(t)
Applying the standard product rule to each term of this scalar sum and then re-grouping the terms demonstrates the vector product rule. Each component-wise differentiation (d/dt [u_x(t)v_x(t)]) follows the scalar product rule (u_x'(t)v_x(t) + u_x(t)v_x'(t)). Summing these up leads directly to the vector dot product rule.
Step-by-Step Differentiation
To differentiate a dot product of two vector functions u(t) and v(t):
- Find the derivative of the first vector function: Calculate u'(t) by differentiating each component of u(t) with respect to t.
- Find the derivative of the second vector function: Calculate v'(t) by differentiating each component of v(t) with respect to t.
- Perform the first dot product: Calculate u'(t) ⋅ v(t).
- Perform the second dot product: Calculate u(t) ⋅ v'(t).
- Add the results: Sum the two scalar results from steps 3 and 4 to obtain the final scalar derivative.
Practical Example
Let's differentiate the dot product of two simple 3D vector functions:
- u(t) =
<t, t^2, 1> - v(t) =
<t^3, t, 1>
Here's how we apply the rule:
-
Find the derivatives of the individual vector functions:
- u'(t) = $\frac{d}{dt} <t, t^2, 1>$ =
<1, 2t, 0> - v'(t) = $\frac{d}{dt} <t^3, t, 1>$ =
<3t^2, 1, 0>
- u'(t) = $\frac{d}{dt} <t, t^2, 1>$ =
-
Apply the dot product differentiation formula:
$\frac{d}{dt} [\mathbf{u}(t) \cdot \mathbf{v}(t)] = \mathbf{u}'(t) \cdot \mathbf{v}(t) + \mathbf{u}(t) \cdot \mathbf{v}'(t)$ -
Calculate each dot product term:
| Term | Calculation | Result |
|---|---|---|
| u'(t) ⋅ v(t) | <1, 2t, 0> ⋅ <t^3, t, 1> = (1 t^3) + (2t t) + (0 * 1) = t^3 + 2t^2 |
t^3 + 2t^2 |
| u(t) ⋅ v'(t) | <t, t^2, 1> ⋅ <3t^2, 1, 0> = (t 3t^2) + (t^2 1) + (1 * 0) = 3t^3 + t^2 |
3t^3 + t^2 |
| Sum of results | (t^3 + 2t^2) + (3t^3 + t^2) | 4t^3 + 3t^2 |
Therefore, the derivative of the dot product $\mathbf{u}(t) \cdot \mathbf{v}(t)$ is 4t^3 + 3t^2.
Applications in Science and Engineering
Understanding how to differentiate dot products is fundamental in many areas of physics, engineering, and computer graphics:
- Mechanics: Calculating the rate of change of kinetic energy (which often involves a dot product of velocity with itself), power, or work done by a variable force.
- Electromagnetism: Analyzing the rate of change of magnetic flux or electric potential.
- Fluid Dynamics: Describing the flow and interaction of fluids.
- Computer Graphics: Simulating light interactions or object movement.
The ability to differentiate vector operations like the dot product is a cornerstone of vector calculus, providing powerful tools for analyzing dynamic systems.
