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. $$\mathbf{v}\left(e^{t}\right)$$

Short answer

Short Answer: The derivative of \(\mathbf{v}\left(e^{t}\right)\) is given by the vector function \(\mathbf{v}'\left(e^{t}\right) = \left\langle 2e^{2t}, -2e^t, 0 \right\rangle.\)

Step-by-step solution

Compute the derivative of \(\mathbf{v}(t)\)

We must compute the derivative of \(\mathbf{v}(t)\) component-wise, that is, find the derivative of each component of the vector function: - Derivative of \(t^2\) is \(2t\). - Derivative of \(-2t\) is \(-2\). - Derivative of \(1\) is \(0\). Now we can write the derivative of the function \(\mathbf{v}(t)\): \(\mathbf{v}'(t) = \left\langle 2t, -2, 0 \right\rangle\).

Apply the Chain Rule

Now, we must apply the Chain Rule to find the derivative of \(\mathbf{v}\left(e^t\right)\). Using the Chain Rule and noting that the derivative of \(e^t\) is also \(e^t\), we get: $$\mathbf{v}'\left(e^{t}\right) = \left\langle 2e^t, -2, 0 \right\rangle e^{t}.$$ We must now perform entry-wise multiplication for each component. - \(2e^{t} \cdot e^t = 2e^{2t}\) - \(-2 \cdot e^t = -2e^t\) - \(0 \cdot e^t =0\)

Writing the Final Answer

The derivative of \(\mathbf{v}\left(e^{t}\right)\) is given by the vector function: $$\mathbf{v}'\left(e^{t}\right) = \left\langle 2e^{2t}, -2e^t, 0 \right\rangle.$$

Key concepts

Chain Rule Application

When it comes to understanding derivatives of functions, especially composite functions, mastering the Chain Rule is crucial. Essentially, the Chain Rule is a formula to compute the derivative of a composite function. In real-world terms, think of it like unpacking a box within a box. You take the outer box away first (the derivative of the outer function) and then address the contents (the inner function).
For example, if you have a vector function like \(\textbf{v}(e^t)\), you're dealing with a composition of the functions \(\textbf{v}(t)\) and the exponential function \(e^t\). To find the derivative, you have to take the derivative of the outer function \(\textbf{v}(t)\) and multiply it by the derivative of the inner function \(e^t\), which is conveniently also \(e^t\). This operation is essential because it allows you to break down complex derivative problems into simpler parts that are easier to solve.

• To apply Chain Rule: Identify the inner and outer functions.
• Find the derivatives of both the inner and outer functions separately.
• Multiply these derivatives together, maintaining the order.

It's similar to following a recipe step by step to ensure the end result turns out well. Skipping a step or reversing the order might lead to an unsatisfactory outcome just like in cooking!

Component-wise Differentiation

Vector function differentiation can be thought of as a task best tackled one piece at a time through Component-wise Differentiation. When presented with a vector function like \(\textbf{v}(t) = \langle t^2, -2t, 1 \rangle\), it's like having a to-do list where each item needs individual attention. Each component of the vector function is a separate function of \(t\) that can be differentiated on its own.
Imagine each component as a different ingredient in a recipe. You prepare each one separately before combining them to create the final dish. By applying the derivative to each individual component, like taking the derivative of \(t^2\) to get \(2t\), you ensure that each part is handled correctly.

Here's a short guide:

• Write down each component of the vector function.
• Take the derivative of each component as if it were a standalone function.
• Reassemble the derivatives into a new vector, maintaining the original order.

This method is essential for solving vector-related derivative problems accurately and systematically, ensuring that each 'ingredient' or component contributes correctly to the 'recipe' or the final vector function derivative.

Derivative of Exponential Functions

Exponential functions, such as \(e^t\), play a pivotal role in various fields like economics, physics, and biology. Understanding the Derivative of Exponential Functions is like learning the concept of growth or decay in these systems. The base of the natural exponential function is \(e\), which is approximately 2.71828, and represents continuous growth. Any exponential function's rate of change is proportional to its current value, which is a unique and powerful property.
When you take the derivative of \(e^t\), the rate of increase is the same as the value of the function at any point \(t\) – which is why the derivative of \(e^t\) is \(e^t\) itself.

• The derivative of \(e^t\) is \(e^t\), indicating constant growth rate.
• The rate of change of the exponential function remains consistent across all values of \(t\).

This concept is vital when dealing with compound interest in finance, radioactive decay in physics, or natural growth in biology. The derivative of exponential functions is also critical when working on problems involving vector functions where an exponential function is part of the composition, necessitating the use of the Chain Rule for proper differentiation.