91Ó°ÊÓ

Vector functions are functions that take one or more variables and return a vector, which has both a direction and a magnitude. These functions are commonly used in physics and engineering to describe motion or other vector quantities. In our case, we have vector functions \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\), which are defined in terms of the parameter \(t\).

• \(\mathbf{u}(t) = 2t^3 \mathbf{i} + (t^2 - 1) \mathbf{j} - 8 \mathbf{k}\)
• \(\mathbf{v}(t) = e^{t} \mathbf{i} + 2e^{-t} \mathbf{j} - e^{2t} \mathbf{k}\)

Each of these vector functions can be broken into components aligned along the three-dimensional coordinate axes represented by the unit vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\). The challenge often lies in how these vector functions change over a given parameter, demanding an understanding of the chain rule when external functions such as \(\sqrt{t}\) are involved.

Computing the derivative of a vector function requires applying principles that extend from regular derivatives of scalar functions to vectors. When calculating the derivative at a certain point, you look at how each component of the vector changes separately. In this exercise, we're tasked with finding the derivative of \(\mathbf{v}(\sqrt{t})\) using the chain rule.

• The i-component: Differentiate \(e^{\sqrt{t}} \mathbf{i}\) to obtain \(\frac{e^{\sqrt{t}}}{2\sqrt{t}} \mathbf{i}\).
• The j-component: Differentiate \(2e^{-\sqrt{t}} \mathbf{j}\) to obtain \(-\frac{e^{-\sqrt{t}}}{\sqrt{t}} \mathbf{j}\).
• The k-component: Differentiate \(-e^{2\sqrt{t}} \mathbf{k}\) to obtain \(-\frac{e^{2\sqrt{t}}}{\sqrt{t}} \mathbf{k}\).

Adding these differentiated components together gives us the rate of change of the entire vector function at \(\sqrt{t}\). Understanding how these component derivatives contribute to the whole helps students see the larger picture of vector calculus in action.

Composite functions involve the nesting of one function inside another, such as \(\mathbf{v}(\sqrt{t})\), where \(\sqrt{t}\) is substituted into function \(\mathbf{v}(t)\). This nesting is where the chain rule shines. The chain rule gives a method to differentiate composite functions by pulling the derivative of the outer function with respect to the inner function, and multiplying by the derivative of the inner function.

• For \(e^{\sqrt{t}}\), the outer function is \(e^u\) with \(u = \sqrt{t}\).
• The derivative \(f'(u) = e^u\) helps derive the result when multiplied by \(\frac{d}{dt} \sqrt{t} = \frac{1}{2\sqrt{t}}\).

This approach is repeated for each component vector, highlighting the power and elegance of the chain rule in differentiating complex constructs. Tackle each component with the chain rule and then sum to find the total derivative. This methodology helps break down seemingly daunting calculus problems into manageable steps.