Question

Rule By expressing \(\mathbf{u}\) and \(\mathbf{v}\) in terms of their components, prove that $$\frac{d}{d t}(\mathbf{u}(t)+\mathbf{v}(t))=\mathbf{u}^{\prime}(t)+\mathbf{v}^{\prime}(t)$$

Short answer

Question: Prove that the derivative of the sum of two vector functions is equal to the sum of their individual derivatives. Answer: We have proved that given two vector functions \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\), the derivative of their sum is equal to the sum of their individual derivatives, that is, \(\frac{d}{dt}(\mathbf{u}(t)+\mathbf{v}(t))=\mathbf{u}^{\prime}(t)+\mathbf{v}^{\prime}(t)\).

Step-by-step solution

Express the vectors in terms of their components

Let the vector functions \(\mathbf{u}(t)=\langle u_1(t), u_2(t), u_3(t) \rangle\) and \(\mathbf{v}(t)=\langle v_1(t), v_2(t), v_3(t) \rangle\) be in terms of their components.

Derivative of individual vector functions

Calculate the derivative of the individual vector functions \(\mathbf{u}\) and \(\mathbf{v}\) with respect to \(t\). We get: $$\frac{d\mathbf{u}}{dt} = \left\langle \frac{du_1}{dt}, \frac{du_2}{dt}, \frac{du_3}{dt} \right\rangle$$ and $$\frac{d\mathbf{v}}{dt} = \left\langle \frac{dv_1}{dt}, \frac{dv_2}{dt}, \frac{dv_3}{dt} \right\rangle$$

Sum of vector functions

Now we will find the sum of the two vector functions \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\). We get: $$\mathbf{u}(t) + \mathbf{v}(t) = \langle u_1(t)+v_1(t), u_2(t)+v_2(t), u_3(t)+v_3(t) \rangle$$

Derivative of the sum

Next, we will find the derivative of the sum with respect to \(t\). We get: $$\frac{d}{dt}(\mathbf{u}(t) + \mathbf{v}(t)) = \left\langle \frac{d(u_1 + v_1)}{dt}, \frac{d(u_2 + v_2)}{dt}, \frac{d(u_3 + v_3)}{dt} \right\rangle$$ By using the sum rule for derivatives, we have: $$\frac{d}{dt}(\mathbf{u}(t) + \mathbf{v}(t)) = \left\langle \frac{du_1}{dt} + \frac{dv_1}{dt}, \frac{du_2}{dt} + \frac{dv_2}{dt}, \frac{du_3}{dt} + \frac{dv_3}{dt} \right\rangle$$

Compare the results

Now, let's compare the results from the derivative of the sum and the sum of the individual derivatives: $$\mathbf{u}^{\prime}(t) + \mathbf{v}^{\prime}(t) = \left\langle \frac{du_1}{dt} + \frac{dv_1}{dt}, \frac{du_2}{dt} + \frac{dv_2}{dt}, \frac{du_3}{dt} + \frac{dv_3}{dt} \right\rangle$$ Comparing this with the derivative of the sum, we have: $$\frac{d}{dt}(\mathbf{u}(t) + \mathbf{v}(t)) = \left\langle \frac{du_1}{dt} + \frac{dv_1}{dt}, \frac{du_2}{dt} + \frac{dv_2}{dt}, \frac{du_3}{dt} + \frac{dv_3}{dt} \right\rangle$$ Both expressions are equal, so we have proved that: $$\frac{d}{dt}(\mathbf{u}(t)+\mathbf{v}(t))=\mathbf{u}^{\prime}(t)+\mathbf{v}^{\prime}(t)$$

Key concepts

Sum Rule for Derivatives

In calculus, the sum rule for derivatives simplifies the process of finding the derivative of a sum of functions. It states that if you have two differentiable functions, their combined derivative can be found by simply differentiating each function separately and then adding the results. Think of it as splitting the work to make it easier. In mathematical terms, if you have functions \(f(t)\) and \(g(t)\), then the derivative of their sum is:

• \(\frac{d}{dt}(f(t) + g(t)) = \frac{df}{dt} + \frac{dg}{dt}\)

This rule simplifies calculus because it confirms that addition won't complicate how the derivative behaves.
This is very helpful, especially when working with vectors or more complex functions, because it means you can handle each component on its own.

Component Form of Vectors

Vectors are often represented in component form, breaking them down into their parts to make calculations easier. A vector in three-dimensional space, for instance, can be expressed as \(\mathbf{v} = \langle v_1, v_2, v_3 \rangle\).

• This breaks the vector \(\mathbf{v}\) into its components: \(v_1\), \(v_2\), and \(v_3\).

When we deal with vector calculus, we use component form to calculate things like the derivative. For a vector function \(\mathbf{v}(t) = \langle v_1(t), v_2(t), v_3(t) \rangle\), each component can be dealt with separately.
This means evaluating the derivative of the vector is about finding the derivative of each component. It makes complex calculations more manageable and clearer.
Remember, each "part" of a vector moves and changes with time or other parameters separately, adding flexibility to work within physics and engineering.

Vector Calculus

Vector calculus extends ordinary calculus to vector fields, providing tools for analyzing physical forces and phenomena that have both direction and magnitude. Understanding how vectors operate when functions of time or other variables comes from understanding both calculus and the properties of vectors.

• With vector functions like \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\), operations such as addition, subtraction, and taking derivatives maintain the multi-dimensional nature of vectors.

The work with vector calculus often involves derivatives and integrals of vector-valued functions. Calculating a derivative such as \(\frac{d}{dt}(\mathbf{u}(t) + \mathbf{v}(t))\) makes use of principles like the sum rule for derivatives.
Vector fields are key to applications in physics, for denoting fields like gravitational fields, velocity fields in fluid dynamics, and more. It's using vector calculus that these fields are modeled and analyzed, illustrating how calculus of vector functions forms the foundation of understanding how these multi-faceted problems behave in various contexts.