Question

Proof of Product 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 the derivative of the product of a scalar function \(f(t)\) and a vector function \(\mathbf{u}(t)\) is given by \(f^{\prime}(t) \mathbf{u}(t)+f(t) \mathbf{u}^{\prime}(t)\).

Step-by-step solution

Express the vector function in terms of its components

Let \(\mathbf{u}(t)=\langle u_1(t),u_2(t),u_3(t)\rangle\) be a vector function of time where \(u_1(t),u_2(t),\) and \(u_3(t)\) are the components of the vector function.

Find the product of the scalar function and the vector function

We need to find the product of the scalar function \(f(t)\) and the vector function \(\mathbf{u}(t)\): $$f(t)\mathbf{u}(t)= f(t)\langle u_1(t),u_2(t),u_3(t)\rangle = \langle f(t)u_1(t), f(t)u_2(t), f(t)u_3(t)\rangle.$$

Find the derivative of the product with respect to time

Now we will find the derivative of the product of \(f(t)\) and \(\mathbf{u}(t)\) with respect to \(t\): $$\frac{d}{dt}(f(t)\mathbf{u}(t))= \left\langle\frac{d}{dt}(f(t)u_1(t)),\frac{d}{dt}(f(t)u_2(t)),\frac{d}{dt}(f(t)u_3(t))\right\rangle.$$

Apply the Product Rule to each component

We will apply the Product Rule to each component in the expression above: $$\frac{d}{dt}(f(t)\mathbf{u}(t))= \left\langle f^{\prime}(t)u_1(t)+f(t)u_1^{\prime}(t),f^{\prime}(t)u_2(t)+f(t)u_2^{\prime}(t),f^{\prime}(t)u_3(t)+f(t)u_3^{\prime}(t)\right\rangle.$$

Rewrite the expression in vector form

We can now rewrite the expression in vector form: $$\frac{d}{dt}(f(t)\mathbf{u}(t))= f^{\prime}(t)\langle u_1(t),u_2(t),u_3(t)\rangle + f(t)\langle u_1^{\prime}(t),u_2^{\prime}(t),u_3^{\prime}(t)\rangle = f^{\prime}(t)\mathbf{u}(t) + f(t)\mathbf{u}^{\prime}(t).$$ Thus, we have proved that $$\frac{d}{dt}(f(t)\mathbf{u}(t))=f^{\prime}(t) \mathbf{u}(t)+f(t) \mathbf{u}^{\prime}(t).$$

Key concepts

Vector Functions Differentiation

Differentiating vector functions is a fundamental concept in calculus that extends our understanding of derivatives to multi-dimensional spaces. When you have a vector function, such as \( \mathbf{u}(t) \), which depends on a parameter \( t \) and has components that are also functions of \( t \), the differentiation process involves finding the rate of change of each component. Essentially, you treat each component as a separate function and differentiate accordingly.

For example, if \( \mathbf{u}(t) = \langle u_1(t), u_2(t), u_3(t) \rangle \), then the derivative of \( \mathbf{u}(t) \) with respect to \( t \) is found by taking the derivative of each component function. The result is another vector function, \( \mathbf{u}'(t) = \langle u_1'(t), u_2'(t), u_3'(t) \rangle \) which represents the instantaneous rate of change of \( \mathbf{u}(t) \) in each of its respective directions.

Scalar and Vector Function Multiplication

Multiplying a scalar function by a vector function combines a one-dimensional quantity with a multi-dimensional quantity. This operation is carried out by multiplying the scalar function with each component of the vector function individually.

If we have a scalar function \( f(t) \) and a vector function \( \mathbf{u}(t) \) with components \( \langle u_1(t), u_2(t), u_3(t) \rangle \) as given in the example, their product is defined component-wise: \( f(t)\mathbf{u}(t) = \langle f(t)u_1(t), f(t)u_2(t), f(t)u_3(t) \rangle \). What we obtain is a new vector function where each component has been scaled according to the value of \( f(t) \). This process illustrates how scalar multiplication can stretch or compress the vector function in the direction of each of its components.

Derivatives of Vector Functions

Taking the derivative of a vector function is critical in physics and engineering for understanding motion and change in multi-dimensional systems. When finding the derivative of a vector function, we apply the standard differentiation rules to each component of the vector.

If \( \mathbf{u}(t) = \langle u_1(t), u_2(t), u_3(t) \rangle \), then the derivative, denoted by \( \mathbf{u}'(t) \), is computed as \( \mathbf{u}'(t) = \langle u_1'(t), u_2'(t), u_3'(t) \rangle \). The derivative \( \mathbf{u}'(t) \) provides us with a vector that describes how each component of \( \mathbf{u}(t) \) changes with respect to \( t \) separately. The understanding of these derivatives is crucial for analyzing the dynamics of systems in which direction and magnitude change over time.

Applying Product Rule to Components

The Product Rule in calculus is a key rule for differentiating products of functions. It states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. When we have a scalar function and a vector function, we apply the Product Rule to each component of the vector function.

Thus, if we consider the product \( f(t)\mathbf{u}(t) \) where \( f(t) \) is a scalar function and \( \mathbf{u}(t) = \langle u_1(t), u_2(t), u_3(t) \rangle \) is a vector function, we differentiate each component following the Product Rule: \( \frac{d}{dt}(f(t)u_i(t)) = f'(t)u_i(t) + f(t)u_i'(t) \) for \( i = 1, 2, 3 \). This leads to the expression \( f'(t)\mathbf{u}(t) + f(t)\mathbf{u}'(t) \), where each term is a vector representing appropriately scaled rates of change. Through this process, we maintain the relationship between the scalar and vector functions upon differentiation.