/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 Let \(\mathbf{u}(t)=2 t^{3} \mat... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(\mathbf{u}(t)=2 t^{3} \mathbf{i}+\left(t^{2}-1\right) \mathbf{j}-8 \mathbf{k}\) 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. $$\mathbf{u}(t) \cdot \mathbf{v}(t)$$

Short Answer

Expert verified
Question: Compute the derivative of the scalar function \(\mathbf{u}(t) \cdot \mathbf{v}(t)\), where \(\mathbf{u}(t) = 2t^3\mathbf{i} + (t^2-1)\mathbf{j} - 8\mathbf{k}\) and \(\mathbf{v}(t) = e^{t}\mathbf{i} + 2e^{-t}\mathbf{j} - e^{2t}\mathbf{k}\). Answer: \(\frac{d}{dt} \left(\mathbf{u}(t) \cdot \mathbf{v}(t)\right) = (6t^2e^t + 2t^3e^t) + (4te^{-t} - 4e^{-t}) - (16e^{2t}).\)

Step by step solution

01

Calculate the dot product

To compute the dot product, we have to multiply the corresponding components of both the vectors and then add them. The dot product is given as: $$\mathbf{u}(t) \cdot \mathbf{v}(t) = (2t^3\mathbf{i} + (t^2-1)\mathbf{j} - 8\mathbf{k}) \cdot (e^{t}\mathbf{i} + 2e^{-t}\mathbf{j} - e^{2t}\mathbf{k}).$$ Then, our dot product becomes: $$\mathbf{u}(t) \cdot \mathbf{v}(t) = (2t^3e^t) + (t^2-1)(2e^{-t}) - (8e^{2t}).$$
02

Find the derivative of the dot product

Now let's find the derivative \(\frac{d}{dt} \left(\mathbf{u}(t) \cdot \mathbf{v}(t)\right)\). We do this by applying the derivative rules to each component: $$\frac{d}{dt} \left(2t^3e^t\right) = 6t^2e^t + 2t^3e^t,$$ $$\frac{d}{dt} \left((t^2-1)(2e^{-t})\right) = 4te^{-t} - 4e^{-t},$$ $$\frac{d}{dt} \left(-8e^{2t}\right) = -16e^{2t}.$$ Combining them all, we get the derivative: $$\frac{d}{dt} \left(\mathbf{u}(t) \cdot \mathbf{v}(t)\right) = (6t^2e^t + 2t^3e^t) + (4te^{-t} - 4e^{-t}) - (16e^{2t}).$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product
Understanding the dot product is crucial for comprehending a range of topics in vector calculus. It's an operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This number can be seen as a projection of one vector onto another, measuring how much one vector goes in the direction of the other.

The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is calculated by multiplying corresponding pairs of their components and summing those products:
\[ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + ... + a_n b_n \]
Where \( a_i \) and \( b_i \) are components of the vectors \( \mathbf{a} \) and \( \mathbf{b} \), respectively. In the given exercise, the vectors have three components, so their dot product is the sum of the products of their \( i \) (x), \( j \) (y), and \( k \) (z) components.
Derivative of Vector Functions
When we talk about the derivative of vector functions, we're extending the concept of differentiation from functions of a single variable to vector functions. A vector function \( \mathbf{v}(t) \) may have components that are functions of a common variable, \( t \), representing different physical quantities like position, velocity, or force over time.

To differentiate a vector function, we separately apply the rules of differentiation to each component function. That is:
\[ \frac{d}{dt} \mathbf{v}(t) = \left( \frac{d}{dt} v_1(t), \frac{d}{dt} v_2(t), ... , \frac{d}{dt} v_n(t) \right) \]
Where \( v_1(t), v_2(t), ... , v_n(t) \) are the component functions of \( \mathbf{v}(t) \). It's akin to differentiating several functions simultaneously, one for each direction in space. In our exercise, we differentiate the dot product of two vector functions, applying product and chain rules as needed.
Exponential Functions
Exponential functions play a central role in calculus due to their unique property of being equal to their own derivative, which makes them particularly easy to differentiate and integrate. The general form of an exponential function is:\[ f(x) = a^x \], where \( a \) is a constant, and for the special case where \( a = e \) (Euler's number, approximately 2.71828), the function \( e^x \) is known as the natural exponential function.

One key characteristic of \( e^x \) is its derivative, which is \( \frac{d}{dx}e^x = e^x \), it has the same value as the function itself at all points. This property is extensively used in the differentiation of vector functions that contain exponential terms. When dealing with vector calculus, the exponential functions often appear in physics and engineering contexts, such as decay rates, growth models, or signal processing, to name a few examples.

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Most popular questions from this chapter

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