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By writing a \(\cdot \mathbf{b}\) in terms of components prove that the product rule for differentiation applies to the dot product of two vectors; that is, \(\frac{d}{d t}(\mathbf{a} \cdot \mathbf{b})=\frac{d \mathbf{a}}{d t} \cdot \mathbf{b}+\mathbf{a} \cdot \frac{d \mathbf{b}}{d t}\).

Short Answer

Expert verified
The differentiation of a dot product follows \(\frac{d}{dt}(\mathbf{a} \cdot \mathbf{b}) = \frac{d\mathbf{a}}{dt} \cdot \mathbf{b} + \mathbf{a} \cdot \frac{d\mathbf{b}}{dt}\).

Step by step solution

01

Express Dot Product in Terms of Components

Let the vectors \( \mathbf{a}(t) = \langle a_1(t), a_2(t), a_3(t) \rangle \) and \( \mathbf{b}(t) = \langle b_1(t), b_2(t), b_3(t) \rangle \). The dot product \( \mathbf{a} \cdot \mathbf{b} \) is expressed as: \[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3. \] This expression uses the component-wise multiplication of the vector components.
02

Differentiate the Dot Product with Respect to Time

Differentiate the dot product \( \mathbf{a} \cdot \mathbf{b} \) with respect to time \( t \):\[ \frac{d}{dt} ( \mathbf{a} \cdot \mathbf{b} ) = \frac{d}{dt} (a_1b_1 + a_2b_2 + a_3b_3). \] Use the product rule for each term: \[ \frac{d}{dt} (a_ib_i) = \frac{da_i}{dt}b_i + a_i\frac{db_i}{dt}. \]
03

Apply Product Rule to Each Component

Applying the product rule to each term, we have:- \( \frac{d}{dt}(a_1b_1) = \frac{da_1}{dt}b_1 + a_1\frac{db_1}{dt}\)- \( \frac{d}{dt}(a_2b_2) = \frac{da_2}{dt}b_2 + a_2\frac{db_2}{dt}\)- \( \frac{d}{dt}(a_3b_3) = \frac{da_3}{dt}b_3 + a_3\frac{db_3}{dt}\)
04

Sum All Component Derivatives

Add the results from the component-wise differentiation:\[ \frac{d}{dt} ( \mathbf{a} \cdot \mathbf{b} ) = ( \frac{da_1}{dt}b_1 + a_1\frac{db_1}{dt} ) + ( \frac{da_2}{dt}b_2 + a_2\frac{db_2}{dt} ) + ( \frac{da_3}{dt}b_3 + a_3\frac{db_3}{dt} ). \] Group the results into two sums:\[ ( \frac{da_1}{dt}b_1 + \frac{da_2}{dt}b_2 + \frac{da_3}{dt}b_3 ) + ( a_1\frac{db_1}{dt} + a_2\frac{db_2}{dt} + a_3\frac{db_3}{dt} ). \]
05

Simplify the Expression

Recognize these as two separate dot products:- \( \frac{d\mathbf{a}}{dt} \cdot \mathbf{b} = \frac{da_1}{dt}b_1 + \frac{da_2}{dt}b_2 + \frac{da_3}{dt}b_3 \)- \( \mathbf{a} \cdot \frac{d\mathbf{b}}{dt} = a_1\frac{db_1}{dt} + a_2\frac{db_2}{dt} + a_3\frac{db_3}{dt} \) Thus, \[ \frac{d}{dt} ( \mathbf{a} \cdot \mathbf{b} ) = \frac{d\mathbf{a}}{dt} \cdot \mathbf{b} + \mathbf{a} \cdot \frac{d\mathbf{b}}{dt}. \] This confirms that the product rule for differentiation applies to the dot product of vectors.

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Key Concepts

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

Vector Calculus
Vector calculus is a powerful mathematical framework that extends calculus, which is usually applied to single-variable functions, to functions involving vectors.
Vectors are quantities that have both magnitude and direction, such as velocity or force.
With vector calculus, we can analyze how vectors change over time, which is crucial in physics and engineering. One of the operations in vector calculus is the dot product, which is the sum of the products of the corresponding components of two vectors.
This operation helps measure things like the angle between two vectors or projecting one vector onto another.
By understanding and manipulating the dot product, we can explore how these vectors interact and connect in space. In vector calculus, we routinely perform differentiations and integrations involving vectors to solve complex problems.
Differentiation tells us how a vector changes at an instant, while integration helps us understand the total influence over an interval.
With these tools, scientists and engineers can model the behavior of physical systems precisely.
Product Rule for Differentiation
The product rule is a fundamental principle in calculus that describes how to differentiate the product of two functions.
In simple terms, if you have two functions multiplied together, the derivative of this product is given by taking the derivative of each function individually and combining these results. When it comes to vectors, we can apply the product rule to their dot product to find out how this product changes over time.
This is particularly important because vectors often represent quantities like displacement or velocity, which change as time progresses. For the product rule applied to the dot product of vectors, it states:
  • Differentiating the dot product of two vectors \( \mathbf{a}(t) \) and \( \mathbf{b}(t) \), gives:
  • \( \frac{d}{dt}(\mathbf{a} \cdot \mathbf{b}) = \frac{d\mathbf{a}}{dt} \cdot \mathbf{b} + \mathbf{a} \cdot \frac{d\mathbf{b}}{dt} \)
This means, effectively, that we treat each component of the vectors separately and apply the product rule to these individual terms.
This ensures an accurate representation of how the entire vector expression behaves as it changes over time.
Component-Wise Multiplication
Instead of dealing with entire vectors at once, it's often helpful to break them down into their components.
This is especially true when working with dot products, where each component of one vector multiplies with the corresponding component of another. For a vector \( \mathbf{a}(t) = \langle a_1(t), a_2(t), a_3(t) \rangle \), its dot product with another vector \( \mathbf{b}(t) = \langle b_1(t), b_2(t), b_3(t) \rangle \) is expressed component-wise as:
  • \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)
Each term involves the multiplication of corresponding components and sums these products.
By differentiating each of these terms individually using the product rule, you can observe changes in the dot product over time.
Breaking vectors down in this manner simplifies the differentiation process and makes it more intuitive. Thus, component-wise multiplication not only aids in simplifying calculations but also provides deeper insights into the structure of vector operations.

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Most popular questions from this chapter

An interesting one-dimensional system is the simple pendulum, consisting of a point mass \(m\), fixed to the end of a massless rod (length \(l\) ), whose other end is pivoted from the ceiling to let it swing freely in a vertical plane, as shown in Figure \(4.26 .\) The pendulum's position can be specified by its angle \(\phi\) from the equilibrium position. (It could equally be specified by its distance \(s\) from equilibrium \(-\) indeed \(s=l \phi-\) but the angle is a little more convenient.) (a) Prove that the pendulum's potential energy (measured from the equilibrium level) is \(U(\phi)=m g l(1-\cos \phi)\). Write down the total energy \(E\) as a function of \(\phi\) and \(\dot{\phi}\). (b) Show that by differentiating your expression for \(E\) with respect to \(t\) you can get the equation of motion for \(\phi\) and that the equation of motion is just the familiar \(\Gamma=I \alpha\) (where \(\Gamma\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration \(\ddot{\phi}\) ). (c) Assuming that the angle \(\phi\) remains small throughout the motion, solve for \(\phi(t)\) and show that the motion is periodic with period \(\tau_{\mathrm{o}}=2 \pi \sqrt{l / g}\).

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