Question

Proof of Cross Product Rule Prove that $$\frac{d}{d t}(\mathbf{u}(t) \times \mathbf{v}(t))=\mathbf{u}^{\prime}(t) \times \mathbf{v}(t)+\mathbf{u}(t) \times \mathbf{v}^{\prime}(t)$$ There are two ways to proceed: Either express \(\mathbf{u}\) and \(\mathbf{v}\) in terms of their three components or use the definition of the derivative.

Short answer

Question: Prove that for any two differentiable vector functions \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\), we have $$\frac{d}{dt}(\mathbf{u}(t) \times \mathbf{v}(t)) = \mathbf{u}^{\prime}(t) \times \mathbf{v}(t) + \mathbf{u}(t) \times \mathbf{v}^{\prime}(t)$$ Solution: We need to follow these steps to prove the given equation: 1. Write the cross product in terms of components. 2. Find the time derivatives of \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\). 3. Calculate the cross products involving the derivatives. 4. Add the cross products from Step 3 together. 5. Find the derivative of the cross product; it should match the sum of the cross products from Step 4, proving the given equation.

Step-by-step solution

Write the cross product in terms of components

First, let's express \(\mathbf{u}\) and \(\mathbf{v}\) as: $$ \mathbf{u}(t)=\langle u_1(t), u_2(t), u_3(t)\rangle \\ \mathbf{v}(t)=\langle v_1(t), v_2(t), v_3(t)\rangle $$ The cross product, \(\mathbf{u}(t) \times \mathbf{v}(t)\), is given by: $$ \mathbf{u}(t) \times \mathbf{v}(t)=\langle u_2(t)v_3(t)-u_3(t)v_2(t), u_3(t)v_1(t)-u_1(t)v_3(t), u_1(t)v_2(t)-u_2(t)v_1(t)\rangle $$

Find the time derivatives of \(\mathbf{u}\) and \(\mathbf{v}\)

Next, we need to find the derivatives of \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) with respect to time \(t\). We have: $$ \mathbf{u}^{\prime}(t) = \langle \frac{d u_1(t)}{dt}, \frac{d u_2(t)}{dt}, \frac{d u_3(t)}{dt} \rangle \\ \mathbf{v}^{\prime}(t) = \langle \frac{d v_1(t)}{dt}, \frac{d v_2(t)}{dt}, \frac{d v_3(t)}{dt} \rangle $$

Calculate the cross products involving the derivatives

Now, we need to find the cross products of \(\mathbf{u}^{\prime}(t)\) and \(\mathbf{v}(t)\), and of \(\mathbf{u}(t)\) and \(\mathbf{v}^{\prime}(t)\). These are given by: $$ \mathbf{u}^{\prime}(t) \times \mathbf{v}(t) = \langle \frac{d u_2(t)}{dt}v_3(t)-\frac{d u_3(t)}{dt}v_2(t), \frac{d u_3(t)}{dt}v_1(t)-\frac{d u_1(t)}{dt}v_3(t), \frac{d u_1(t)}{dt}v_2(t)-\frac{d u_2(t)}{dt}v_1(t) \rangle $$ and $$ \mathbf{u}(t) \times \mathbf{v}^{\prime}(t) = \langle u_2(t)\frac{d v_3(t)}{dt} - u_3(t)\frac{d v_2(t)}{dt}, u_3(t)\frac{d v_1(t)}{dt} - u_1(t)\frac{d v_3(t)}{dt}, u_1(t)\frac{d v_2(t)}{dt} - u_2(t)\frac{d v_1(t)}{dt} \rangle $$

Add the cross products from Step 3 together

Now, we need to add the cross products from Step 3 together to get the sum of the cross products: $$ \mathbf{u}^{\prime}(t) \times \mathbf{v}(t) + \mathbf{u}(t) \times \mathbf{v}^{\prime}(t) = \langle \frac{d u_2(t)}{dt}v_3(t) - \frac{d u_3(t)}{dt}v_2(t)+u_2(t)\frac{d v_3(t)}{dt} - u_3(t)\frac{d v_2(t)}{dt}, \\ \frac{d u_3(t)}{dt}v_1(t) - \frac{d u_1(t)}{dt}v_3(t)+u_3(t)\frac{d v_1(t)}{dt} - u_1(t)\frac{d v_3(t)}{dt}, \\ \frac{d u_1(t)}{dt}v_2(t) - \frac{d u_2(t)}{dt}v_1(t)+u_1(t)\frac{d v_2(t)}{dt} - u_2(t)\frac{d v_1(t)}{dt} \rangle $$

Derivative of cross product

Finally, we find the derivative of the cross product \(\mathbf{u}(t) \times \mathbf{v}(t)\): $$ \frac{d}{dt}(\mathbf{u}(t) \times \mathbf{v}(t)) = \langle \frac{d [u_2(t)v_3(t) - u_3(t)v_2(t)]}{dt}, \frac{d [u_3(t)v_1(t) - u_1(t)v_3(t)]}{dt}, \frac{d [u_1(t)v_2(t) - u_2(t)v_1(t)]}{dt} \rangle $$ By applying the product rule, we get the same terms as we have in the sum of cross products from Step 4. This completes the proof: $$ \frac{d}{dt}(\mathbf{u}(t) \times \mathbf{v}(t)) = \mathbf{u}^{\prime}(t) \times \mathbf{v}(t) + \mathbf{u}(t) \times \mathbf{v}^{\prime}(t) $$

Key concepts

Vector Calculus

Vector Calculus is a branch of mathematics that deals with vector fields and differential operators. Unlike regular calculus which deals mainly with scalars, vector calculus involves multi-dimensional vectors that have both direction and magnitude. This is particularly useful for real-world physics applications, like fluid flow or electromagnetic fields.

In vector calculus, you often work with vector functions - these are functions where each input is a vector, and each output is also a vector. Examples of vector functions include position, velocity, and acceleration in physics.

Key operations in vector calculus include:

• Gradient: Measures the rate and direction of change in a scalar field.
• Divergence: Measures the magnitude of a source or sink at a given point in a vector field.
• Curl: Measures the rotation of a vector field.

Understanding these operations helps you apply mathematical models to solve complex physics and engineering problems.

Product Rule in Vector Calculus

The Product Rule is a basic rule in calculus, adapted for vectors in vector calculus. When you differentiate the product of two functions, the product rule helps you find the derivative effectively. For vector functions like vectors \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\), the rule adapts to account for cross products.

When dealing with the cross product of two vector functions, \(\mathbf{u}(t) \times \mathbf{v}(t)\), the derivative can be expressed using the product rule for vectors:

• Differentiate \(\mathbf{u}(t)\), while keeping \(\mathbf{v}(t)\) constant.
• Differentiate \(\mathbf{v}(t)\), while keeping \(\mathbf{u}(t)\) constant.

Apply the cross product to both results, and then sum them up:\[ \frac{d}{dt} (\mathbf{u}(t) \times \mathbf{v}(t)) = \mathbf{u}^{\prime}(t) \times \mathbf{v}(t) + \mathbf{u}(t) \times \mathbf{v}^{\prime}(t) \]This principle allows for simplified calculations when handling the product of vector functions in physical systems.

Derivatives of Vector Functions

Derivatives of vector functions describe how the function's output changes with respect to an input variable, often time. For vector functions, calculating derivatives can become more involved since you handle multiple dimensions.

If \(\mathbf{u}(t)\) is a vector function expressed as \(\mathbf{u}(t) = \langle u_1(t), u_2(t), u_3(t) \rangle \), then its derivative, \(\mathbf{u}^{\prime}(t)\), is:\[ \mathbf{u}^{\prime}(t) = \langle \frac{du_1(t)}{dt}, \frac{du_2(t)}{dt}, \frac{du_3(t)}{dt} \rangle \]

• Each component is differentiated separately.
• This method is applicable for any vector function with a well-defined split into components.

Such derivatives are foundational in physics, impacting concepts like velocity and acceleration as derivatives of position. They provide insight into dynamic systems and are essential in fields ranging from engineering to economics.