Question

Ask you to verify parts of Theorem \(11.2 .4 .\) In each let \(f(t)=t^{3}, \vec{r}(t)=\left\langle t^{2}, t-1,1\right\rangle\) and \(\vec{s}(t)=\) \(\left\langle\sin t, e^{t}, t\right\rangle .\) Compute the various derivatives as indicated. Simplify \(\vec{r}(t) \cdot \vec{s}(t),\) then find its derivative; show this is the same as \(\vec{r}^{\prime}(t) \cdot \vec{s}(t)+\vec{r}(t) \cdot \vec{s}^{\prime}(t)\)

Short answer

The derivative matches the sum of the individual terms.

Step-by-step solution

Compute Dot Product

The dot product of vectors \( \vec{r}(t) \) and \( \vec{s}(t) \) is calculated as follows: \[ \vec{r}(t) \cdot \vec{s}(t) = (t^2)(\sin t) + (t-1)(e^t) + (1)(t) = t^2 \sin t + te^t - e^t + t \]

Simplify the Dot Product

Simplify the expression \[ \vec{r}(t) \cdot \vec{s}(t) = t^2 \sin t + te^t - e^t + t \]. The expression is already simplified.

Differentiate the Dot Product

To differentiate \[ \vec{r}(t) \cdot \vec{s}(t) = t^2 \sin t + te^t - e^t + t \], apply the sum rule to find the derivative: \[ \frac{d}{dt}(t^2 \sin t) + \frac{d}{dt}(te^t) - \frac{d}{dt}(e^t) + \frac{d}{dt}(t) \]. This equals \[ 2t \sin t + t^2 \cos t + e^t(t + 1) - e^t + 1 \].

Compute Derivatives of the Vectors

Now find \( \vec{r}^{\prime}(t) \) and \( \vec{s}^{\prime}(t) \): \[ \vec{r}^{\prime}(t) = \langle 2t, 1, 0 \rangle \] and \[ \vec{s}^{\prime}(t) = \langle \cos t, e^t, 1 \rangle \].

Compute Each Term of the Theorem Expression

Compute \( \vec{r}^{\prime}(t) \cdot \vec{s}(t) \): \[ \vec{r}^{\prime}(t) \cdot \vec{s}(t) = (2t)(\sin t) + (1)(e^t) + (0)(t) = 2t \sin t + e^t \].Compute \( \vec{r}(t) \cdot \vec{s}^{\prime}(t) \): \[ \vec{r}(t) \cdot \vec{s}^{\prime}(t) = (t^2)(\cos t) + (t-1)(e^t) + (1)(1) = t^2 \cos t + te^t - e^t + 1 \].

Sum the Theorem Expression Terms

Add the two results from the previous step: \[ (2t \sin t + e^t) + (t^2 \cos t + te^t - e^t + 1) \]. This simplifies to \[ 2t \sin t + t^2 \cos t + e^t(t + 1) - e^t + 1 \].This matches the derivative of the dot product.

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics that focuses on vector fields and operations that involve them. Vectors are quantities that have both magnitude and direction. In calculus, we use vectors to understand how things like velocity or force change over time and space.
When dealing with functions that turn input values into vectors, vector calculus helps us compute derivatives and integrals in a multi-dimensional framework.

• The dot product, also known as the scalar product, is a specific operation that results in a scalar and combines two vectors.
• The dot product is calculated as the product of the magnitudes of the two vectors and the cosine of the angle between them.
• It can be used to calculate work done by a force or to determine the angle between vectors.

In our exercise, we looked at vectors \( \vec{r}(t) \) and \( \vec{s}(t) \), where \( t \) represents an independent parameter, to compute values like dot products and their derivatives.

Product Rule

The product rule is a fundamental rule of differentiation in calculus applied to products of functions.
It states that the derivative of the product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function.
In mathematical terms, if \( u(t) \) and \( v(t) \) are two differentiable functions, then the derivative of their product \( u(t)v(t) \) is:

• \( \frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t) \).

In the exercise, this principle was used to verify Theorem 11.2.4 by exploring how it applies to the dot product of vectors. We applied the product rule to the expressions involving our vectors \( \vec{r}(t) \) and \( \vec{s}(t) \) and found that the principle holds true when dealing with their dot product.

Derivative Calculation

Derivative calculation is an essential part of calculus that determines how a function changes as its input changes. In vector calculus, derivatives can help us understand how vector fields change over time and space.
The derivative of a vector function is found by differentiating each component of the vector separately.
Steps to calculate derivatives:

• Differentiation of scalar functions: Use standard rules like power rule, product rule, and chain rule.
• Differentiation of vector functions: Differentiate each component and arrange them into a new vector.

In the exercise, we calculated derivatives for both \( \vec{r}(t) \) and \( \vec{s}(t) \), yielding \( \vec{r}'(t) = \langle 2t, 1, 0 \rangle \) and \( \vec{s}'(t) = \langle \cos t, e^t, 1 \rangle \).
These derivatives were crucial for verifying the theorem related to the dot product.

Theorem Verification

Theorem verification involves validating a mathematical statement using logical reasoning and calculations. In calculus, it's crucial to ensure that results are consistent with established theorems.
For our specific exercise, Theorem 11.2.4 claims a certain property about the derivative of the dot product of two vector functions.
Steps to verify:

• Calculate the dot product of the vector functions.
• Differentiate this dot product.
• Calculate derivatives separately and apply the product rule.
• Verify if the differentiated dot product equals the sum of the products of the derivatives, as predicted by the theorem.

In the original solution, both paths - direct differentiation of the dot product and sum of product of derivatives - led to the same result, thereby verifying the theorem.