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

Answer: The derivative of the vector function \(\mathbf{v}(e^t)\) is \(\left\langle 2e^{2t}, -2e^t, 0\right\rangle\).

Step-by-step solution

Recall the chain rule for derivative

The chain rule states that \((f(g(t)))' = f'(g(t)) \cdot g'(t)\). We will apply this rule to each component of the vector function.

Find the derivative of the first component

First, we will differentiate the first component of \(\mathbf{v}(t)\), which is \(t^2\). The derivative of this component is \(2t\). Now, apply the chain rule to get the derivative of the first component of \(\mathbf{v}(e^t)\). We have: $$\frac{d}{dt}(e^{t^2}) = 2e^t \cdot e^t = 2e^{2t}$$

Find the derivative of the second component

Next, we will differentiate the second component of \(\mathbf{v}(t)\), which is \(-2t\). The derivative of this component is \(-2\). Now, apply the chain rule to get the derivative of the second component of \(\mathbf{v}(e^t)\). We have: $$\frac{d}{dt}(-2e^t) = -2e^t$$

Find the derivative of the third component

Finally, we will differentiate the third component of \(\mathbf{v}(t)\), which is \(1\). The derivative of this component is \(0\). So, the derivative of the third component of \(\mathbf{v}(e^t)\) is also \(0\).

Combine the derivatives of all components

Now we will combine the derivatives of all three components to get the final derivative of the vector function \(\mathbf{v}(e^t)\). We have: $$\frac{d}{dt}\mathbf{v}(e^t) = \left\langle 2e^{2t}, -2e^t, 0\right\rangle$$

Key concepts

Vector Calculus

Vector calculus is a field of mathematics that focuses on differentiating and integrating vector fields, typically in three-dimensional space. Unlike regular calculus, which deals with scalar functions, vector calculus deals with functions that have more than one component. This means you will often work with vectors that possess both magnitude and direction. In our exercise, we have a vector function, \( \mathbf{v}(t) = \left\langle t^2, -2t, 1 \right\rangle \), each expressing a change in different dimensions over time

• Components of a Vector: These are the individual parts (e.g., \(t^2\), \(-2t\), and \(1\)) that comprise the vector function. Each component is essentially a function itself.
• Differentiation: This technique applies the principles of calculus to each independent component of the vector.
• Chain Rule in Vectors: When differentiating vector functions with chains of composition, such as \( \mathbf{v}(e^t) \), the chain rule is extended to apply individually to each component.

Applying the chain rule in vector calculus allows us to understand how composite functions change, and examining each component helps in understanding how the overall vector field behaves.

Derivative

One of the most fundamental concepts in calculus, the derivative, represents how a function changes as its input changes, that is, it gives the function's rate of change at any given point. In our exercise involving vector calculus, we are asked to find the derivative of \( \mathbf{v}(t) \) components recursively:

• First Component: The derivative of \(t^2\) is \(2t\). By applying the chain rule when looking at \( \mathbf{v}(e^t) \), it is transformed into \(2e^{2t}\).
• Second Component: The derivative of \(-2t\) is \(-2\). Following the chain rule, it becomes \(-2e^t\).
• Third Component: The derivative of a constant (\(1\)) is \(0\).

The combination of these differentiated components gives us the derivative of the whole vector function, showing how each dimension of the vector changes with time.

Exponentiation

Exponentiation refers to the operation of raising a base number, often referring to the exponential function with base Euler's number \(e\), to a power. An exponential function such as \(e^t\) changes its rate of increase or decrease continuously and is a key element in calculus problems, especially those involving derivatives and integration.

• Impact on Differentiation: When differentiating an exponential function in terms of a different variable, chain rules are often applied. In our problem, when differentiating \(e^{t^2}\), we multiply by the derivative of \(t^2\), displaying the power of exponential derivatives.
• Exponential Growth: The derivative of \(e^t\) is particularly interesting as it equals itself, \(e^t\). In the context of our exercise, each component multiplied by the exponential growth factor reflects how rapidly the original vector function \( \mathbf{v}(t) \) grows or shrinks over time.

Understanding graduation in exponential functions is crucial for analyzing processes that involve growth or decay across time dimensions, especially in vector calculus where these changes occur in multiple dimensions simultaneously.