Question

Let \(\mathbf{u}(t)=\left\langle 1, t, t^{2}\right\rangle, \mathbf{v}(t)=\left\langle t^{2},-2 t, 1\right\rangle\) and \(g(t)=2 \sqrt{t} .\) Compute the derivatives of the following functions. $$\mathbf{u}(t) \times \mathbf{v}(t)$$

Short answer

Answer: The derivative of the cross product of these vector functions is \(\frac{d\mathbf{w}}{dt} = \left\langle 6t^2 + 1, 4t^3, -2 - 3t^2 \right\rangle\).

Step-by-step solution

Compute the cross product

To find the cross product \(\mathbf{w}(t) = \mathbf{u}(t) \times \mathbf{v}(t)\), use the following formula for cross product in a 3-dimensional space: $${\mathbf{w}}(t)=\biggl\langle\biggl(\frac{\partial \mathbf{u}}{\partial y}\frac{\partial \mathbf{v}}{\partial z} - \frac{\partial \mathbf{u}}{\partial z}\frac{\partial \mathbf{v}}{\partial y}\biggr),\biggl(\frac{\partial \mathbf{u}}{\partial z}\frac{\partial \mathbf{v}}{\partial x} - \frac{\partial \mathbf{u}}{\partial x}\frac{\partial \mathbf{v}}{\partial z}\biggr),\biggl(\frac{\partial \mathbf{u}}{\partial x}\frac{\partial \mathbf{v}}{\partial y} - \frac{\partial \mathbf{u}}{\partial y}\frac{\partial \mathbf{v}}{\partial x}\biggr) \biggr\rangle$$ Plug in the given vectors \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\): $$\mathbf{w}(t) = \left\langle (t)(1) - (t^2)(-2t), (t^2)(t^2) - (1)(1), (1)(-2t) - (t)(t^2) \right\rangle$$ Now simplify the expression: $$\mathbf{w}(t) = \left\langle 2t^3 + t, t^4 - 1, -2t - t^3\right\rangle$$

Compute the derivative of each component

Now find the derivative of each component of \(\mathbf{w}(t)\) with respect to \(t\): $$\frac{d\mathbf{w}}{dt} = \left\langle \frac{d}{dt}(2t^3 + t), \frac{d}{dt}(t^4 - 1), \frac{d}{dt}(-2t - t^3) \right\rangle$$ Differentiate each component with respect to \(t\): $$\frac{d\mathbf{w}}{dt} = \left\langle 6t^2 + 1, 4t^3, -2 - 3t^2 \right\rangle$$ So, the derivative of the cross product \(\mathbf{u}(t) \times \mathbf{v}(t)\) is $$\frac{d\mathbf{w}}{dt} = \left\langle 6t^2 + 1, 4t^3, -2 - 3t^2 \right\rangle$$.

Key concepts

Vector Calculus

Vector calculus is a fascinating branch of mathematics that deals with vector fields and operations on vectors. It is the natural extension of traditional calculus, but instead of dealing with just real numbers, vector calculus allows us to explore vectors in space, their magnitudes, directions, and how they change over time or space.
A key focus of vector calculus is studying various operations that can be performed on vectors, such as the dot product, cross product, and differentiation. All of these operations help us model and analyze physical phenomena like fluid flow, electromagnetic fields, and forces in three-dimensional space.
In this exercise, vector calculus is used to find the derivative of the cross product of two vector functions. Understanding how vectors change through differentiation is crucial in fields like physics and engineering.

Cross Product

The cross product is a vital operation in vector calculus, particularly for vectors in three-dimensional space. It combines two vectors to create a third, unique vector that is perpendicular to the plane containing the initial vectors. This perpendicular vector is often used to determine the orientation of a system or the rotational force (torque) exerted by the vectors.
To calculate the cross product of two vectors, \ \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \ \) and \ \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \ \), you use the formula:

• \( \mathbf{a} \times \mathbf{b} = \left\langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \right\rangle \)

In the problem at hand, vectors \ \( \mathbf{u}(t) \) and \ \( \mathbf{v}(t) \) are given and their cross product generates a new vector function, whose components are computed through this formula. Understanding the cross product helps us interpret phenomena where orientation matters, such as determining the direction of an angular vector.

Differentiation

Differentiation is a fundamental concept in calculus that involves finding how a function changes as its input changes. When it comes to vector functions like the one in this problem, differentiation is performed component-wise. This means that each part of the vector is treated separately, and its derivative is calculated independently.
For a vector \ \( \mathbf{w}(t) = \langle w_1(t), w_2(t), w_3(t) \rangle \ \), the derivative \ \( \frac{d\mathbf{w}(t)}{dt} \ \) is calculated as:

• \( \left\langle \frac{dw_1}{dt}, \frac{dw_2}{dt}, \frac{dw_3}{dt} \right\rangle \)

In the solution provided, each component of the cross product vector is differentiated with respect to time \ \( t \ \). This step reveals how the orientation and length of the resulting vector change over time, which is crucial for applications like motion analysis in physics.

3-Dimensional Space

Three-dimensional space, or 3D space, is the physical universe's domain, where objects have length, width, and height. In mathematics, we describe vectors and their operations in this space using coordinate systems, usually \ \( x, y, \ \) and \ \( z \ \) axes.
Operations like the cross product are peculiar to three-dimensional space due to the intrinsic nature of having three axes, allowing a new vector to be perpendicular to the original pair of vectors. Understanding 3D space is essential for visualizing and solving problems in vector calculus, making the representation of forces, fields, and positions more intuitive.
In the exercise, vectors \ \( \mathbf{u}(t) \) and \ \( \mathbf{v}(t) \) are presented in 3D space, illustrating how they can be operated upon using vector calculus methodologies, such as cross products and differentiation, to produce results with meaningful physical interpretations.