Linear Algebra

Calculus I, II and III

In vector calculus, the derivative of a cross product of two vector functions can be calculated using a rule analogous to the product rule for the derivatives of functions in single-variable calculus. This rule is often referred to as the vector product rule or simply the product rule for vectors. Let's consider two vector functions of a single variable \( t \), denoted as \( \mathbf{A}(t) \) and \( \mathbf{B}(t) \). The cross product of these two vectors is another vector function given by \( \mathbf{C}(t) = \mathbf{A}(t) \times \mathbf{B}(t) \). To find the derivative of \( \mathbf{C}(t) \) with respect to \( t \), we apply the product rule for vectors, which states: \[ \frac{d}{dt}[\mathbf{A}(t) \times \mathbf{B}(t)] = \frac{d\mathbf{A}}{dt} \times \mathbf{B}(t) + \mathbf{A}(t) \times \frac{d\mathbf{B}}{dt} \] Here's a step-by-step solution to calculate the derivative of the cross product \( \mathbf{C}(t) \): Step 1: Differentiate \( \mathbf{A}(t) \) with respect to \( t \) to find \( \frac{d\mathbf{A}}{dt} \). Step 2: Differentiate \( \mathbf{B}(t) \) with respect to \( t \) to find \( \frac{d\mathbf{B}}{dt} \). Step 3: Take the cross product of \( \frac{d\mathbf{A}}{dt} \) with \( \mathbf{B}(t) \). Step 4: Take the cross product of \( \mathbf{A}(t) \) with \( \frac{d\mathbf{B}}{dt} \). Step 5: Add the two cross products from steps 3 and 4 to obtain the derivative of the cross product \( \mathbf{C}(t) \). Let's illustrate this with an example. Suppose we have: \[ \mathbf{A}(t) = \begin{bmatrix} t \\ t^2 \\ t^3 \end{bmatrix}, \quad \mathbf{B}(t) = \begin{bmatrix} \sin(t) \\ \cos(t) \\ e^t \end{bmatrix} \] We want to find the derivative of \( \mathbf{C}(t) = \mathbf{A}(t) \times \mathbf{B}(t) \). Following the steps: Step 1: Differentiate \( \mathbf{A}(t) \): \[ \frac{d\mathbf{A}}{dt} = \begin{bmatrix} 1 \\ 2t \\ 3t^2 \end{bmatrix} \] Step 2: Differentiate \( \mathbf{B}(t) \): \[ \frac{d\mathbf{B}}{dt} = \begin{bmatrix} \cos(t) \\ -\sin(t) \\ e^t \end{bmatrix} \] Step 3: Compute \( \frac{d\mathbf{A}}{dt} \times \mathbf{B}(t) \): \[ \frac{d\mathbf{A}}{dt} \times \mathbf{B}(t) = \begin{bmatrix} 1 \\ 2t \\ 3t^2 \end{bmatrix} \times \begin{bmatrix} \sin(t) \\ \cos(t) \\ e^t \end{bmatrix} = \begin{bmatrix} 2t e^t - 3t^2 \cos(t) \\ -(3t^2 \sin(t) - e^t) \\ 2t \sin(t) - \cos(t) \end{bmatrix} \] Step 4: Compute \( \mathbf{A}(t) \times \frac{d\mathbf{B}}{dt} \): \[ \mathbf{A}(t) \times \frac{d\mathbf{B}}{dt} = \begin{bmatrix} t \\ t^2 \\ t^3 \end{bmatrix} \times \begin{bmatrix} \cos(t) \\ -\sin(t) \\ e^t \end{bmatrix} = \begin{bmatrix} -t^2 e^t - t^3 \sin(t) \\ t^3 \cos(t) - t e^t \\ -t^2 \cos(t) - t \sin(t) \end{bmatrix} \] Step 5: Add the results from steps 3 and 4: \[ \frac{d\mathbf{C}}{dt} = \begin{bmatrix} 2t e^t - 3t^2 \cos(t) \\ -(3t^2 \sin(t) - e^t) \\ 2t \sin(t) - \cos(t) \end{bmatrix} + \begin{bmatrix} -t^2 e^t - t^3 \sin(t) \\ t^3 \cos(t) - t e^t \\ -t^2 \cos(t) - t \sin(t) \end{bmatrix} = \begin{bmatrix} (2t - t^2) e^t - 3t^2 \cos(t) - t^3 \sin(t) \\ t^3 \cos(t) - t e^t - 3t^2 \sin(t) + e^t \\ 2t \sin(t) - \cos(t) - t^2 \cos(t) - t \sin(t) \end{bmatrix} \] This final vector represents the derivative of the cross product \( \mathbf{C}(t) \) with respect to \( t \).

Question

How is the derivative of a cross product calculated in vector calculus?

Solution

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