Question

Let $$\mathbf{u}(t)=2 t^{3} \mathbf{i}+\left(t^{2}-1\right) \mathbf{j}-8 \mathbf{k} \text { and } \mathbf{v}(t)=e^{t} \mathbf{i}+2 e^{-t} \mathbf{j}-e^{2 t} \mathbf{k}$$ Compute the derivative of the following functions. $$\mathbf{v}(\sqrt{t})$$

Short answer

Question: Compute the derivative of the vector function \(\mathbf{v}(\sqrt{t})\) with respect to \(t\), given that \(\mathbf{v}(t)= e^{t} \mathbf{i} + 2e^{-t} \mathbf{j} - e^{2t} \mathbf{k}\). Answer: The derivative of \(\mathbf{v}(\sqrt{t})\) with respect to \(t\) is given by: $$\frac{d\mathbf{v}}{dt}(\sqrt{t}) = \frac{e^{\sqrt{t}}}{2\sqrt{t}}\mathbf{i} - \frac{e^{-\sqrt{t}}}{\sqrt{t}} \mathbf{j} - \frac{e^{2\sqrt{t}}}{\sqrt{t}} \mathbf{k}$$

Step-by-step solution

1. Rewrite the vector function with \(\sqrt{t}\)

First, let's rewrite the given vector function \(\mathbf{v}(t)\) in terms of \(\sqrt{t}\). We get: $$\mathbf{v}(\sqrt{t}) = e^{\sqrt{t}} \mathbf{i} + 2e^{-\sqrt{t}} \mathbf{j} - e^{2\sqrt{t}} \mathbf{k}$$

2. Differentiate each component of the vector function

Now that we have rewritten the vector function in terms of \(\sqrt{t}\), we can differentiate each component with respect to \(t\). We get: $$\frac{d\mathbf{v}}{dt}(\sqrt{t}) = \frac{d}{dt}\left(e^{\sqrt{t}} \mathbf{i}\right) + \frac{d}{dt}\left(2e^{-\sqrt{t}} \mathbf{j}\right) + \frac{d}{dt}\left(-e^{2\sqrt{t}} \mathbf{k}\right)$$

3. Apply the chain rule for each component

To differentiate each component, we will use the chain rule. The chain rule states that if a function \(y\) depends on another function \(u\) which in turn depends on \(t\), then the derivative of \(y\) with respect to \(t\) is given by: $$\frac{dy}{dt} = \frac{dy}{du}\frac{du}{dt}$$ Applying the chain rule for each component, we get: For the \(\mathbf{i}\) component: $$\frac{d}{dt}\left(e^{\sqrt{t}}\right) = \frac{d}{d\sqrt{t}}\left(e^{\sqrt{t}}\right) \cdot \frac{d\sqrt{t}}{dt} = e^{\sqrt{t}} \cdot \frac{1}{2\sqrt{t}} = \frac{e^{\sqrt{t}}}{2\sqrt{t}}\mathbf{i}$$ For the \(\mathbf{j}\) component: $$\frac{d}{dt}\left(2e^{-\sqrt{t}}\right) = 2\left( \frac{d}{d\sqrt{t}}\left(e^{-\sqrt{t}}\right) \cdot \frac{d\sqrt{t}}{dt} \right) = 2\left(-e^{-\sqrt{t}} \cdot \frac{1}{2\sqrt{t}}\right) = -\frac{e^{-\sqrt{t}}}{\sqrt{t}} \mathbf{j}$$ For the \(\mathbf{k}\) component: $$\frac{d}{dt}\left(-e^{2\sqrt{t}}\right) = - \frac{d}{d\sqrt{t}}\left(e^{2\sqrt{t}}\right) \cdot \frac{d\sqrt{t}}{dt} = -2e^{2\sqrt{t}} \cdot \frac{1}{2\sqrt{t}} = -\frac{e^{2\sqrt{t}}}{\sqrt{t}} \mathbf{k}$$

4. Combine the derivatives into a single vector function

Finally, we combine the derivatives of each component into a single vector function. This gives us the derivative of \(\mathbf{v}(\sqrt{t})\) with respect to \(t\): $$\frac{d\mathbf{v}}{dt}(\sqrt{t}) = \frac{e^{\sqrt{t}}}{2\sqrt{t}}\mathbf{i} - \frac{e^{-\sqrt{t}}}{\sqrt{t}} \mathbf{j} - \frac{e^{2\sqrt{t}}}{\sqrt{t}} \mathbf{k}$$

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus that is used to find the derivative of composite functions. When a function y is dependent on a variable u, which itself is a function of another variable t, the chain rule allows us to differentiate y with respect to t. In mathematical terms, if y = f(u) and u = g(t), then the derivative of y with respect to t is given by the formula:\[\frac{{dy}}{{dt}} = \frac{{dy}}{{du}} \frac{{du}}{{dt}}\]Essentially, we multiply the derivative of y with respect to u by the derivative of u with respect to t. This process is what allows us to handle complex functions that involve compositions of multiple functions, such as vector functions that depend on another variable as shown in the exercise provided.To apply the chain rule effectively, one should follow these steps:

• Identify the outer function and the inner function.
• Differentiate the outer function with respect to the inner function.
• Differentiate the inner function with respect to the original variable.
• Multiply both derivatives to get the final result.

The chain rule proves to be an incredibly useful tool, especially in vector calculus where functions can often be compositions of scalar functions, as seen in our exercise.

Vector Calculus

Vector calculus is an extension of calculus that deals with vector-valued functions. These functions, unlike scalar functions that output a single value, output vectors that have both magnitude and direction. Vector calculus encompasses various operations including differentiation and integration of vector fields.A vector function, like \(\textbf{v}(t)\), assigns a vector to every point in its domain, which in many cases is a subset of real numbers. A fundamental aspect of vector calculus is the ability to differentiate these functions with respect to a scalar variable like t. This process helps in understanding how a vector function changes at each point in space with respect to time or another independent variable.When dealing with vector calculus, certain rules from scalar calculus still apply, yet they must be extended to accommodate the direction and multiple components of vectors. For example, when taking derivatives, each component of a vector function is differentiated individually. This component-wise differentiation can appear more complex due to the multiple parts of a vector function but is essential in understanding the behavior of physical systems in fields such as physics and engineering.

Derivative of Vector Functions

The derivative of vector functions is determined by differentiating each of the vector's components separately with respect to the variable of interest. In the context of the exercise, we're interested in how the vector function \(\textbf{v}(\textbf{t})\) changes as the variable t changes. Importantly, if the components of the vector function are themselves functions of another function of t (as with \(\textbf{v}(\(\sqrt{t}\))\)), we must utilize the chain rule.Each component of the vector function is treated like a standard function for differentiation purposes. This means we can apply rules like the power rule, product rule, and the aforementioned chain rule to find the component's derivative.For example, to differentiate the \(\textbf{i}\) component of the vector function \(\textbf{v}(\(\sqrt{t}\))\), we first consider the component as a function of \(\sqrt{t}\) and then apply the chain rule to differentiate with respect to t. This process is repeated for each component of the vector, finally combining the derivatives of the components to form the derivative of the entire vector function.With these clear steps, even complex-looking vector function differentiations become manageable, and once mastered, it can greatly aid in the analysis of various problems in science and engineering where vector functions are used to express quantities such as velocity, acceleration, and force.