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. $$\mathbf{u}\left(t^{3}\right)$$

Short answer

Answer: The derivative of the function \(\mathbf{u}\left(t^{3}\right)\) is \(\frac{d\mathbf{u}(t^{3})}{dt} = \left\langle 0, 3t^{2}, 6t^{3} \right\rangle\).

Step-by-step solution

Write down the given functions

We are given the following functions: \(\mathbf{u}(t)=\left\langle 1, t, t^{2}\right\rangle\) and \(g(t)=2 \sqrt{t}\). We need to find the derivative of: \(\mathbf{u}\left(t^{3}\right)\)

Apply the Chain Rule

To compute the derivative, we will use the chain rule, which states that if we have a function \(\mathbf{u}(g(t))\), its derivative is given by \(\frac{d\mathbf{u}}{dt} = \frac{d\mathbf{u}}{dg} \cdot \frac{dg}{dt}\). For our problem, \(\frac{d\mathbf{u}}{dg}\) will be the derivative of the vector function \(\mathbf{u}(t)\), and \(\frac{dg}{dt}\) will be the derivative of the scalar function \(g(t)=t^{3}\).

Compute the derivatives of \(\mathbf{u}(t)\) and \(g(t)\)

First, compute the derivative of \(\mathbf{u}(t)\): \(\frac{d\mathbf{u}}{dt}=\left\langle 0, 1, 2t \right\rangle\) Next, compute the derivative of \(g(t) = t^{3}\): \(\frac{dg}{dt} = 3t^{2}\)

Apply the Chain Rule

Now, we are ready to apply the chain rule: \(\frac{d\mathbf{u}(t^{3})}{dt} = \frac{d\mathbf{u}(g(t))}{dt} = \frac{d\mathbf{u}}{dg} \cdot \frac{dg}{dt} = \left\langle 0, 1, 2t \right\rangle \cdot 3t^{2}\)

Calculate the Result

To find the derivative of the vector function, we multiply the scalar function component-wise: \(\frac{d\mathbf{u}(t^{3})}{dt} = \left\langle 0 \cdot 3t^{2}, 1 \cdot 3t^{2}, 2t \cdot 3t^{2} \right\rangle = \left\langle 0, 3t^{2}, 6t^{3} \right\rangle\) So, the derivative of the function \(\mathbf{u}\left(t^{3}\right)\) is: \(\frac{d\mathbf{u}(t^{3})}{dt} = \left\langle 0, 3t^{2}, 6t^{3} \right\rangle\)

Key concepts

Vector Functions

Vector functions are expressions where each component of the vector depends on a variable, like time or another scalar quantity. They are often used to represent paths or flows in spaces of more than one dimension. In our exercise, the vector function \( \mathbf{u}(t) = \langle 1, t, t^2 \rangle \) maps a scalar \( t \) to a vector in three-dimensional space. Each component of the vector \( \langle 1, t, t^2 \rangle \) uniquely corresponds to the respective dimension, making it easy to model trajectories.

• First component is constant: \( 1 \)
• Second component changes linearly: \( t \)
• Third component changes quadratically: \( t^2 \)

Each of these components may describe a different physical property depending on the context, such as position, velocity, or acceleration when considered with respect to time. To understand vector functions better, consider them as a set of scalar functions, one for each component of the vector.

Derivative of Composite Functions

Finding the derivative of composite functions involves using the chain rule, a fundamental calculus tool. A composite function is formed when one function is applied to the result of another, like \( \mathbf{u}(t^3) \). That means we're plugging \( t^3 \) into the vector function \( \mathbf{u}(t) \).The chain rule helps us differentiate composite functions by expressing the derivative of \( \mathbf{u}(g(t)) \) as the product of the derivative of the outer function evaluated at the inner function, \( \frac{d \mathbf{u}}{dg} \), and the derivative of the inner function, \( \frac{dg}{dt} \). Mathematically, it's expressed as:\[ \frac{d\mathbf{u}(g(t))}{dt} = \frac{d\mathbf{u}}{dg} \cdot \frac{dg}{dt} \]Applying this rule, we can tackle more complex derivatives without memorizing endless combinations of functions. The power of the chain rule lies in its simplicity and generality, applicable to countless scenarios in calculus.

Calculus Differentiation

Calculus differentiation focuses on finding the rate at which a function is changing at any given point. It's a crucial tool for analyzing behaviors and trends in a variety of fields such as physics, engineering, and economics.To differentiate the vector function \( \mathbf{u}(t) \), you would perform differentiation on each of its components independently. For example:

• \( \frac{d}{dt}(1) = 0 \)
• \( \frac{d}{dt}(t) = 1 \)
• \( \frac{d}{dt}(t^2) = 2t \)

This results in the derivative vector \( \langle 0, 1, 2t \rangle \), illustrating how each direction in space changes with \( t \). Calculating derivatives like these allows you to find slopes, tangents, and ultimately understand how systems behave dynamically. Differentiation applies universally, so mastering it opens up many paths in both academic and real-world problem-solving scenarios.