Question

Proof of Cross Product Rule Prove that $$\frac{d}{dt}(\mathbf{u}(t)\times \mathbf{v}(t))={\mathbf{u}}^{\prime }(t)\times \mathbf{v}(t)+\mathbf{u}(t)\times {\mathbf{v}}^{\prime }(t)$$ 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 the cross product rule for vector-valued functions: $\frac{d}{dt}(\mathbf{u}(t)\times \mathbf{v}(t))={\mathbf{u}}^{\prime }(t)\times \mathbf{v}(t)+\mathbf{u}(t)\times {\mathbf{v}}^{\prime }(t)$. Answer: To prove the cross product rule, we started by defining the cross product and expressing the derivative of the cross product with respect to time. Using the linearity property of the cross product, we rewrote the derivative expression and separated it into two separate limits. Finally, we used the definition of the derivative for each limit to show that the derivative of the cross product is equal to the sum of the two cross products formed by the derivatives of each vector with respect to time and the original vectors.

Step-by-step solution

Define the Cross Product

To prove the cross product rule, we first need to define the cross product. The cross product of two vectors $\mathbf{u}$ and $\mathbf{v}$, denoted as $\mathbf{u}\times \mathbf{v}$, is a vector whose direction is perpendicular to the plane containing $\mathbf{u}$ and $\mathbf{v}$, and whose magnitude is given by the product of the magnitudes of $\mathbf{u}$ and $\mathbf{v}$ and the sine of the angle between them. The cross product can also be expressed in terms of the components of $\mathbf{u}$ and $\mathbf{v}$ as: $$\mathbf{u}\times \mathbf{v}=\left(\begin{array}{c}{u}_2{v}_3-u_3{v}_2\\ u_3{v}_1-u_1{v}_3\\ u_1{v}_2-u_2{v}_1\end{array}\right)$$

Express the Derivative of the Cross Product

Now we need to find the derivative of the cross product with respect to time $t$. Using the definition of the derivative, we can write this as: $$\frac{d}{dt}(\mathbf{u}(t)\times \mathbf{v}(t))=\underset{h\to 0}{lim}\frac{\mathbf{u}(t+h)\times \mathbf{v}(t+h)-\mathbf{u}(t)\times \mathbf{v}(t)}{h}$$

Use the Linearity Property of Cross Product

The cross product has the property of linearity, which means that it is distributive over addition and compatible with scalar multiplication. That is: $$\mathbf{u}\times (\mathbf{v}+\mathbf{w})=\mathbf{u}\times \mathbf{v}+\mathbf{u}\times \mathbf{w}$$ Using this property, we can rewrite the derivative expression as: $$\frac{d}{dt}(\mathbf{u}(t)\times \mathbf{v}(t))=\underset{h\to 0}{lim}\frac{\mathbf{u}(t+h)\times \mathbf{v}(t+h)-\mathbf{u}(t)\times \mathbf{v}(t+h)+\mathbf{u}(t)\times \mathbf{v}(t+h)-\mathbf{u}(t)\times \mathbf{v}(t)}{h}$$

Separate the Limit

Now we can separate the limit into two separate limits: $$\frac{d}{dt}(\mathbf{u}(t)\times \mathbf{v}(t))=\underset{h\to 0}{lim}\frac{\mathbf{u}(t+h)\times \mathbf{v}(t+h)-\mathbf{u}(t)\times \mathbf{v}(t+h)}{h}+\underset{h\to 0}{lim}\frac{\mathbf{u}(t)\times \mathbf{v}(t+h)-\mathbf{u}(t)\times \mathbf{v}(t)}{h}$$

Use the Definition of the Derivative for Each Limit

Using the definition of the derivative for each limit, we can rewrite the expression as: $$\frac{d}{dt}(\mathbf{u}(t)\times \mathbf{v}(t))={\mathbf{u}}^{\prime }(t)\times \mathbf{v}(t)+\mathbf{u}(t)\times {\mathbf{v}}^{\prime }(t)$$ This shows that the derivative of the cross product is equal to the sum of the two cross products formed by the derivatives of each vector with respect to time and the original vectors. Thus, we have proved the cross product rule for vector-valued functions as required.