Question

Rule 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 that for a scalar function \(f(t)\) and a vector function \(\mathbf{u}(t)\), the derivative of their product is given by \(\frac{d}{dt}(f(t)\mathbf{u}(t)) = f'(t) \mathbf{u}(t) + f(t) \mathbf{u}'(t) \).

Step-by-step solution

1. Express the vector function in terms of components

Let's represent the vector function \(\mathbf{u}(t)\) in terms of its components. Suppose that \(\mathbf{u}(t)\) has \(n\) components, then we can write it as: $$\mathbf{u}(t) = u_1(t)\mathbf{i} + u_2(t)\mathbf{j} + \cdots + u_n(t)\mathbf{k}$$ Where \(u_1(t), u_2(t), \dots, u_n(t)\) are scalar functions of \(t\), and \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are the standard unit vectors in the respective directions.

2. Find the derivative of the vector function and the scalar function

Now, let's find the derivatives of each component of the vector function and the scalar function with respect to \(t\). We have: $$\frac{d}{dt}(f(t)) = f'(t)$$ and $$\frac{d}{dt}(u_i(t)) = u_i'(t)$$ for \(i=1, 2, \dots, n\).

3. Calculate the derivative of the product of scalar function and vector function

To calculate the derivative of the product of the scalar function \(f(t)\) and the vector function \(\mathbf{u}(t)\), we apply the product rule for each component. We have: $$\frac{d}{dt}(f(t)\mathbf{u}(t)) = \frac{d}{dt}(f(t) u_i(t))\mathbf{i} + \frac{d}{dt}(f(t) u_i(t))\mathbf{j} + \cdots + \frac{d}{dt}(f(t) u_n(t))\mathbf{k}$$ Using the product rule for differentiation, we get: $$\frac{d}{dt}(f(t)\mathbf{u}(t)) = (f'(t) u_i(t) + f(t) u_i'(t))\mathbf{i} + (f'(t) u_i(t) + f(t) u_i'(t))\mathbf{j} + \cdots + (f'(t) u_n(t) + f(t) u_n'(t))\mathbf{k}$$

4. Write the result in the given form

Now, notice that we can write each term in the form of the sum of scalar product of \(f'(t)\) with \(\mathbf{u}(t)\) and \(f(t)\) with \(\mathbf{u}'(t)\) as follows: $$\frac{d}{dt}(f(t)\mathbf{u}(t)) = f'(t)(u_i(t)\mathbf{i} + u_i(t)\mathbf{j} + \cdots + u_n(t)\mathbf{k}) + f(t)(u_i'(t)\mathbf{i} + u_i'(t)\mathbf{j} + \cdots + u_n'(t)\mathbf{k})$$ Which can be simplified to: $$\frac{d}{dt}(f(t)\mathbf{u}(t)) = f'(t) \mathbf{u}(t) + f(t) \mathbf{u}'(t)$$ Thus, the given equation is proved.

Key concepts

Vector Functions

A vector function describes a quantity that has both magnitude and direction, and varies with another parameter, typically time \(t\). It can be expressed as a combination of scalar functions multiplied by unit vectors. For example, the vector function \(\mathbf{u}(t)\) can be written in terms of its components:

• \(u_1(t)\mathbf{i}\)
• \(u_2(t)\mathbf{j}\)
• \(u_n(t)\mathbf{k}\)

Here, \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are unit vectors in their respective directions and each component \(u_i(t)\) is a scalar function dependent on \(t\). Breaking a vector function into components helps in solving problems involving differentiation and integration as each component can be dealt with separately.
Understanding vector functions is essential for calculating motions and forces in physics, and is a vital part of multivariable calculus.

Derivative of Scalar Functions

Taking the derivative of a scalar function refers to finding its rate of change with respect to a variable, commonly time or another independent parameter. Given a function \(f(t)\), its derivative, denoted as \(f'(t)\), represents how \(f\) changes as \(t\) changes.

• If \(f(t) = t^2\), then \(f'(t) = 2t\).
• If \(f(t) = \sin(t)\), then \(f'(t) = \cos(t)\).

The concept can also be applied to each component of a vector function. For example, if \(u_i(t)\) is a part of the vector function, its derivative \(u_i'(t)\) describes how that specific part changes with time. This idea is fundamental in physics to understand how quantities evolve over time.

Differentiation

Differentiation is the process of finding the derivative of a function, which measures how a function changes as its input changes. It is the cornerstone of calculus and essential for analyzing variable relationships.Differentiation can be applied to both scalar and vector functions. When differentiating a product of two functions, like a scalar function \(f(t)\) and a vector function \(\mathbf{u}(t)\), the product rule is used: \[\frac{d}{dt}(f(t) \mathbf{u}(t)) = f'(t) \mathbf{u}(t) + f(t) \mathbf{u}'(t)\] This rule helps decompose the complex behavior of changes happening in product functions into simpler parts. Understanding differentiation allows us to find instantaneous rates of change in real-world contexts.

Calculus

Calculus is the mathematical study of continuous change and includes fundamental concepts like differentiation and integration. It's used to solve problems involving rate of change and accumulation of quantities. Through differentiation, calculus helps us discover how changing one quantity affects another. The principles used in the exercise show how calculus bridges scalar and vector functions.

• It allows engineers to calculate stress and strain in materials.
• Physicists can determine velocity and acceleration from position functions.

Understanding calculus creates a framework to approach complex problems by simplifying them into manageable solutions with precise mathematical techniques.