Question

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

Short answer

Question: Prove the given formula: $$ \frac{d}{d t}(f(t) \mathbf{u}(t))=f^{\prime}(t) \mathbf{u}(t)+f(t) \mathbf{u}^{\prime}(t) $$

Step-by-step solution

Express \(\mathbf{u}\) in terms of its components

Let \(\mathbf{u}(t) = \langle u_1(t), u_2(t), u_3(t) \rangle\), then \(f(t)\mathbf{u}(t) = \langle f(t)u_1(t), f(t)u_2(t), f(t)u_3(t) \rangle\).

Differentiate the product \(f(t)\mathbf{u}(t)\) with respect to \(t\)

To find the derivative of the product, we will differentiate each component with respect to \(t\): $$ \frac{d}{dt} (f(t)\mathbf{u}(t)) = \frac{d}{dt} \langle f(t)u_1(t), f(t)u_2(t), f(t)u_3(t) \rangle = \langle \frac{d}{dt}(f(t)u_1(t)), \frac{d}{dt}(f(t)u_2(t)), \frac{d}{dt}(f(t)u_3(t)) \rangle $$

Apply the product rule in each component of the derivative

Now, we will apply the product rule in each component of the derivative: $$ \frac{d}{dt}(f(t)\mathbf{u}(t)) = \langle f^{\prime}(t)u_1(t) + f(t)u_1^{\prime}(t), f^{\prime}(t)u_2(t) + f(t)u_2^{\prime}(t), f^{\prime}(t)u_3(t) + f(t)u_3^{\prime}(t) \rangle $$

Rewrite the derivative in vector form

The resulting derivative can be rewritten in vector form as follows: $$ \frac{d}{dt}(f(t)\mathbf{u}(t)) = f^{\prime}(t) \langle u_1(t), u_2(t), u_3(t) \rangle + f(t) \langle u_1^{\prime}(t), u_2^{\prime}(t), u_3^{\prime}(t) \rangle $$ Which is equivalent to: $$ \frac{d}{dt}(f(t)\mathbf{u}(t)) = f^{\prime}(t) \mathbf{u}(t) + f(t) \mathbf{u}^{\prime}(t) $$ Thus, the given formula is proven to be true: $$ \frac{d}{d t}(f(t) \mathbf{u}(t))=f^{\prime}(t) \mathbf{u}(t)+f(t) \mathbf{u}^{\prime}(t) $$

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with differentiation and integration of vector fields, primarily in 3-dimensional Euclidean space. It extends calculus, which is the study of change defined for scalars, to vector fields, which can represent quantities that have both magnitude and direction. Some common applications of vector calculus include studying electromagnetic fields, fluid flow, and other phenomena in physics.
In vector calculus, vectors are often expressed in terms of their components. For instance, a vector function \( \mathbf{u}(t) \) can be represented by components like \( \langle u_1(t), u_2(t), u_3(t) \rangle \), where each \( u_i(t) \) is a scalar function. Vectors can thus be manipulated by performing operations on their components, which allows for complex calculations and analyzes.
When dealing with vector calculus, it's essential to understand how to apply calculus rules to vectors, which includes differentiation and integration. These operations are foundational for solving problems involving forces, velocity, and fields in 3D space.

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate of change of a function with respect to a variable. For scalar functions, the derivative measures how the function value changes as its input changes. The concept extends to vector-valued functions which map one or more variables to a vector output.
For a product of a scalar and a vector function, like \( f(t) \mathbf{u}(t) \), derivatives play a crucial role. Here, \( f(t) \) is a scalar function, while \( \mathbf{u}(t) \) is a vector function. The derivative of their product incorporates both the rate of change of \( f(t) \) and the change in each component of \( \mathbf{u}(t) \).

• The derivative of \( f(t) \mathbf{u}(t) \) gives us insights into how the product changes as a whole, by considering how each part contributes separately.
• You apply product differentiation to handle such problems, using rules like the product rule.

Understanding derivatives of vector-valued functions is crucial for fields such as physics and engineering, where multi-dimensional problems are common.

Differentiation of Vector-Valued Functions

The differentiation of vector-valued functions extends the idea of derivatives from scalar functions to vectors. A vector-valued function is one that takes one or more variables and returns a vector. When differentiating these functions, each component has to be differentiated separately.
Differentiation of a product involving vector-valued functions uses the product rule, just like scalar calculus. The rule states that the derivative of a product is the derivative of the first function multiplied by the second function plus the first function multiplied by the derivative of the second function. For a product \( f(t) \mathbf{u}(t) \), this becomes:

• First differentiate \( f(t) \), multiplying it by \( \mathbf{u}(t) \).
• Then, multiply \( f(t) \) by the derivative of \( \mathbf{u}(t) \) and add the results.

Mathematically, it's expressed as \( \frac{d}{d t}(f(t) \mathbf{u}(t)) = f^{\prime}(t) \mathbf{u}(t) + f(t) \mathbf{u}^{\prime}(t) \).
Mastering differentiation of vector-valued functions enables analysts to model and predict behaviors of systems where multiple changing dimensions interact, such as in mechanics and vector field analysis.