RE: LeoThread 2025-10-14 17-41

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Part 1/5:

Understanding the Proof of the Derivative Rule for Vector Functions

In this detailed exploration, we analyze the proof of a fundamental rule in calculus related to vector functions: the derivative of a sum of two vector functions equals the sum of their derivatives. This concept is essential for understanding how derivatives operate within the context of vector-valued functions, which are frequently used in physics, engineering, and advanced mathematics.

The Basic Formula

The core statement under examination is:

[

\frac{d}{dt}(u(t) + v(t)) = u'(t) + v'(t)

]

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Part 2/5:

This rule asserts that when we differentiate the sum of two vector functions (u(t)) and (v(t)), the operation distributes over addition, similar to the rule for scalar functions. Proving this involves breaking down the vector functions into their component functions and applying established calculus rules.

Defining the Vector Functions and Their Components

Let’s consider two vector functions, each composed of three real-valued component functions:

[

u(t) = \begin{bmatrix} u_1(t) \ u_2(t) \ u_3(t) \end{bmatrix}, \quad v(t) = \begin{bmatrix} v_1(t) \ v_2(t) \ v_3(t) \end{bmatrix}

]

Here, (u_1(t), u_2(t), u_3(t)) and (v_1(t), v_2(t), v_3(t)) are real-valued functions of (t).

Summing the Component Functions

The sum of these vector functions is:

[

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Part 3/5:

u(t) + v(t) = \begin{bmatrix} u_1(t) + v_1(t) \ u_2(t) + v_2(t) \ u_3(t) + v_3(t) \end{bmatrix}

]

Differentiating this sum requires applying the derivative component-wise:

[

\frac{d}{dt} [u(t) + v(t)] = \begin{bmatrix} \frac{d}{dt} (u_1(t) + v_1(t)) \ \frac{d}{dt} (u_2(t) + v_2(t)) \ \frac{d}{dt} (u_3(t) + v_3(t)) \end{bmatrix}

]

Applying Derivative Rules to Components

Since each component function is a real-valued function, standard differentiation rules apply. Specifically:

[

\frac{d}{dt} (u_i(t) + v_i(t)) = u_i'(t) + v_i'(t)

]

for (i = 1, 2, 3).

Substituting these into the vector derivative, we obtain:

[

\frac{d}{dt} [u(t) + v(t)] = \begin{bmatrix} u_1'(t) + v_1'(t) \ u_2'(t) + v_2'(t) \ u_3'(t) + v_3'(t) \end{bmatrix}

]

which can be expressed as:

[

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Part 4/5:

u'(t) + v'(t)

]

Vector Formulation and Finalized Proof

Thus, the derivative of the sum of two vector functions is the sum of their derivatives. This aligns with the properties of derivatives for scalar functions and extends naturally to vectors. The process involves component-wise differentiation and concatenating the results back into a vector form:

[

\boxed{

\frac{d}{dt} [u(t) + v(t)] = u'(t) + v'(t)

}

]

This proof hinges on the linearity of differentiation and the structure of vector functions, where each component acts independently under the differentiation operation.

Key Takeaways

  • The derivative of a sum of vector functions equals the sum of their derivatives.
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Part 5/5:

  • The proof relies on expressing vector functions in component form and applying the basic differentiation rules to each component.

  • This property confirms that vector calculus behaves similarly to scalar calculus in terms of linearity and component-wise operations.

Conclusion

Understanding this fundamental rule simplifies the process of differentiating complex vector functions in practice. It also reinforces the importance of component-wise analysis in vector calculus, ensuring that operations familiar from scalar calculus extend naturally to higher dimensions. This principle is foundational for tackling more advanced topics involving derivatives of vector fields, parametric curves, and multi-dimensional differential equations.

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