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. $$\mathbf{v}(g(t))$$

Short answer

The derivative of the given function $\mathbf{v}(g(t))$ is: $$\frac{d}{dt}(\mathbf{v}(g(t))) = \left\langle 4, -2t^{-\frac{1}{2}}, 0\right\rangle$$

Step-by-step solution

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

Firstly, let's compute the derivative of the vector function \(\mathbf{v}(t)\). To do that, we need to find the derivative of its components with respect to \(t\). Recall that: $$\mathbf{v}(t) = \left\langle t^{2}, -2t, 1\right\rangle$$ Let's find the derivatives of each component: 1. \(\frac{d}{dt}(t^2) = 2t\) 2. \(\frac{d}{dt}(-2t) = -2\) 3. \(\frac{d}{dt}(1) = 0\) Now that we've found the derivatives of each component, we can find \(\mathbf{v}'(t)\) as follows: $$\mathbf{v}'(t) = \left\langle 2t, -2, 0\right\rangle$$

Compute the derivative of \(g(t)\)

Now, let's find the derivative of the scalar function \(g(t)\). Recall that: $$g(t) = 2\sqrt{t}$$ To find its derivative, we need to find the derivative of the function with respect to \(t\): $$g'(t) = \frac{d}{dt}(2\sqrt{t}) = 2 \cdot \frac{d}{dt}(t^{\frac{1}{2}}) = 2 \cdot \frac{1}{2}t^{-\frac{1}{2}} = t^{-\frac{1}{2}}$$

Derivative of \(\mathbf{v}(g(t))\) using the chain rule

We have found the derivatives of both \(\mathbf{v}(t)\) and \(g(t)\). Now, we can apply the chain rule to find the derivative of the composite function \(\mathbf{v}(g(t))\). The chain rule states: $$\frac{d}{dt}(\mathbf{v}(g(t))) = \mathbf{v}'(g(t)) \cdot g'(t)$$ Substituting the values of \(\mathbf{v}'(t)\) and \(g'(t)\) calculated above, we get: $$\frac{d}{dt}(\mathbf{v}(g(t))) = \left\langle 2g(t), -2, 0\right\rangle \cdot t^{-\frac{1}{2}}$$ Now, substitute the expression of \(g(t)\) back into the formula: $$\frac{d}{dt}(\mathbf{v}(g(t))) = \left\langle 2(2\sqrt{t}), -2, 0\right\rangle \cdot t^{-\frac{1}{2}}$$ Finally, we can simplify the equation by multiplying the components of the vector by \(t^{-\frac{1}{2}}\): $$\frac{d}{dt}(\mathbf{v}(g(t))) = \left\langle 4\sqrt{t}t^{-\frac{1}{2}}, -2t^{-\frac{1}{2}}, 0\right\rangle = \left\langle 4, -2t^{-\frac{1}{2}}, 0\right\rangle$$ Thus, the derivative of the given function \(\mathbf{v}(g(t))\) is: $$\frac{d}{dt}(\mathbf{v}(g(t))) = \left\langle 4, -2t^{-\frac{1}{2}}, 0\right\rangle$$

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields, which are essentially functions that assign a vector to every point in space. It's commonly used in physics and engineering to model phenomena such as fluid flow and electromagnetic fields. In this context, functions like \( \mathbf{u}(t) \) and \( \mathbf{v}(t) \) assign vectors to each time \( t \). These vector functions represent space curves, where each component of the function, such as \( t^2 \) or \( -2t \), changes with time. Understanding vector calculus involves:

• Grasping how these vectors can change or evolve, which requires understanding derivatives of vector functions.
• Learning how to perform operations like addition, subtraction, and differentiation on these vectors.

In calculations, vectors are often represented in a component form to make them easier to manipulate.Functions \( \mathbf{u} \) and \( \mathbf{v} \) are essential in describing trajectories or paths traced by moving particles in a given timeframe.

Differentiation

Differentiation refers to the process of finding a derivative, which tells us how a function changes as its input changes. When we differentiate vector functions, we individually find the derivatives of each component of the vector. For a vector function \( \mathbf{v}(t) = \langle t^2, -2t, 1 \rangle \), differentiation is performed component-wise:

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

The result, \( \mathbf{v}'(t) = \langle 2t, -2, 0 \rangle \), represents how the vector changes at each point \( t \). The process of differentiation is fundamental to analyzing and understanding the behavior of functions, providing insights into rates of change and slopes of curves.

Composite Functions

Composite functions involve functions within functions. In this exercise, we have a composite function \( \mathbf{v}(g(t)) \), where \( g(t) \) is another function nested within \( \mathbf{v} \). Understanding these requires using the chain rule, a key differentiation technique. The chain rule for derivatives states:

• If you have a composite function \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \).

When applying this to \( \mathbf{v}(g(t)) \), we first differentiate \( \mathbf{v} \) and then multiply by the derivative of \( g(t) \).This strategy allows us to find the derivative with respect to \( t \) even when the function comprises nested layers. By doing so, we can understand how changes in \( t \) affect the entire composite function, capturing the intricate dynamics of interconnected systems and processes.