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. $$\mathbf{u}(t) \cdot \mathbf{v}(t)$$

Short answer

Answer: The derivative of the dot product of the given vector functions is \(2t - 4t + 3t^2\).

Step-by-step solution

Compute the derivatives of each vector component

First, let's compute the derivatives of the components of both vectors \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\). $$ \frac{d}{dt}\mathbf{u}(t) = \frac{d}{dt}\left\langle 1, t, t^{2}\right\rangle = \left\langle 0, 1, 2t\right\rangle $$ $$ \frac{d}{dt}\mathbf{v}(t) = \frac{d}{dt}\left\langle t^{2},-2 t, 1\right\rangle = \left\langle 2t, -2, 0\right\rangle $$

Compute the dot product of the vectors

Next, let's find the dot product of the vectors \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\). $$ \mathbf{u}(t) \cdot \mathbf{v}(t) = \left\langle 1, t, t^{2}\right\rangle \cdot \left\langle t^{2},-2 t, 1\right\rangle = (1)(t^2) + (t)(-2t) + (t^2)(1) = t^2 - 2t^2 + t^3 $$

Calculate the derivative of the dot product

Finally, let's find the derivative of the dot product with respect to \(t\). $$ \frac{d}{dt}(\mathbf{u}(t) \cdot \mathbf{v}(t)) = \frac{d}{dt}(t^2 - 2t^2 + t^3) = 2t - 4t + 3t^2 $$ Therefore, the derivative of the dot product is: $$ \frac{d}{dt}(\mathbf{u}(t) \cdot \mathbf{v}(t)) = 2t - 4t + 3t^2 $$

Key concepts

Understanding Derivatives

A derivative is a fundamental concept in calculus that measures how a function changes as its input changes. It's essentially the rate of change or the slope of a function at any given point. When dealing with vector functions, we differentiate each component of the vector function separately. For a function like \(\mathbf{u}(t) = \langle 1, t, t^2 \rangle\), the derivative with respect to \(t\) is found by taking the derivative of each component:

• The derivative of \(1\) with respect to \(t\) is \(0\),
• the derivative of \(t\) is \(1\),
• and the derivative of \(t^2\) is \(2t\).

Thus, \(\frac{d}{dt}\mathbf{u}(t) = \langle 0, 1, 2t \rangle\).
This method ensures that each directional change is accurately captured. Differentiating vector components separately is similar to processing multiple single-variable functions individually.

The Dot Product in Vector Calculus

The dot product, also known as the scalar product, is a way of multiplying two vectors to produce a scalar (a single number). It combines the magnitudes of the vectors and the cosine of the angle between them. For vectors \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\), their dot product can be computed by multiplying corresponding components and then summing these products:

• Multiply the x-components: \(1 \cdot t^2\),
• multiply the y-components: \(t \cdot (-2t)\),
• multiply the z-components: \(t^2 \cdot 1\).

This calculation results in: \(t^2 - 2t^2 + t^3\).
The dot product is useful in various applications such as projections and determining angles between vectors. It’s a key operation in vector calculus that simplifies many physical and geometrical problems.

Exploring Vector Functions

Vector functions assign a vector to every number in a domain, usually a subset of real numbers. A vector function might describe a curve in space, providing coordinates along that curve as parameters change. Here, \(\mathbf{u}(t) = \langle 1, t, t^2 \rangle\) and \(\mathbf{v}(t) = \langle t^2, -2t, 1 \rangle\) are examples of vector functions where \(t\) serves as a parameter controlling the position along the path.
Vector functions can represent motion or changes over time in physics, such as the trajectory of an object. By examining each component of the vector function and its corresponding changes (derivatives), we get insights into speed, direction, and forces acting over time. This makes vector functions invaluable in areas like mechanics and electromagnetism.

Applying Differentiation Rules

Differentiation rules provide frameworks for finding derivatives of functions efficiently. The power rule, product rule, and chain rule are all examples.

• The power rule states that for any function \(f(t) = t^n\), \(\frac{d}{dt}t^n = nt^{n-1}\).
• The product rule is essential when differentiating products of two functions: if \(h(t) = f(t) \cdot g(t)\), then \(h'(t) = f'(t)g(t) + f(t)g'(t)\).

In the original exercise, rules like the power rule are applied to each component of vectors, ensuring each change is calculated properly. Understanding and applying these rules simplify complex vector calculus tasks, making it easier to comprehend and predict multidimensional behavior.