Question

Find the derivative of the given function. $$ r(t)=\left\langle t^{2}+1, t-1\right\rangle \cdot\langle\sin t, 2 t+5\rangle $$

Short answer

The derivative is \( r'(t) = 2t \sin t + (t^2 + 1) \cos t + 4t + 3 \).

Step-by-step solution

Identify the vectors

The given function is the dot product of two vectors. These vectors are \( \mathbf{a}(t) = \langle t^2 + 1, t - 1 \rangle \) and \( \mathbf{b}(t) = \langle \sin t, 2t + 5 \rangle \). We need to differentiate their dot product.

Write down the dot product formula

The dot product \( r(t) = \mathbf{a}(t) \cdot \mathbf{b}(t) \) is calculated as: \[ r(t) = (t^2 + 1)(\sin t) + (t - 1)(2t + 5) \] Now, we need to find the derivative \( r'(t) \).

Differentiate the dot product

Using the derivative of a product formula, \((uv)' = u'v + uv'\), we find the derivatives of each product:1. Differentiating \((t^2 + 1)\sin t\): - \( u = t^2 + 1 \), \( u' = 2t \) - \( v = \sin t \), \( v' = \cos t \) - The derivative is: \( (2t)(\sin t) + (t^2 + 1)(\cos t)\).2. Differentiating \((t - 1)(2t + 5)\): - \( u = t - 1 \), \( u' = 1 \) - \( v = 2t + 5 \), \( v' = 2 \) - The derivative is: \( (1)(2t + 5) + (t - 1)(2)\).

Combine the derivatives

Combine the derivatives from step 3:\[ r'(t) = (2t \sin t + (t^2 + 1) \cos t) + ((2t + 5) + 2(t - 1))\]Simplify the expression:\[ r'(t) = 2t \sin t + (t^2 + 1) \cos t + 2t + 5 + 2t - 2\]\[ r'(t) = 2t \sin t + (t^2 + 1) \cos t + 4t + 3\]

Final Expression

The derivative of \( r(t) \) is:\[ r'(t) = 2t \sin t + (t^2 + 1) \cos t + 4t + 3 \]

Key concepts

Dot Product Differentiation

In the realm of calculus, finding the derivative of a dot product involves clear understanding of differentiation rules and vector algebra. When dealing with vector functions, such as in our exercise, the dot product differentiation requires considering how each vector component contributes to the function.
The core of this process involves taking two vector functions, say \( \mathbf{a}(t) \) and \( \mathbf{b}(t) \), and computing their dot product. The differentiation of this dot product, \( (\mathbf{a}(t) \cdot \mathbf{b}(t))' \), relies heavily on the product rule of calculus because the dot product is, inherently, a product.
Understanding this allows us to apply the differentiation process as we evaluate each part of the vector relationship. This operation becomes more intuitive with practice, yet remains a fundamental technique in vector calculus that is applicable in physics and engineering problems.

Vector Functions

Vector functions extend the idea of functions to multi-component outputs, where each component can vary with respect to a parameter, often time \( t \). Consider our vector functions \( \mathbf{a}(t) = \langle t^2 + 1, t - 1 \rangle \) and \( \mathbf{b}(t) = \langle \sin t, 2t + 5 \rangle \). These describe paths or fields with respect to \( t \).
Each vector function consists of smaller scalar functions, one per component. The manipulation and computation with these vectors involve coordinating the components through operations like addition, multiplication by scalars, and importantly, differentiation.
The challenge is handling each component according to its individual rules but also considering the function as a whole entity. Vector calculus, thus, becomes a game of orchestration - managing and combining multiple dimensions into cohesive results.

Product Rule in Calculus

The product rule is a foundational principle for differentiation, particularly when dealing with products of differentiable functions. Formulated as \( (uv)' = u'v + uv' \), it provides a means to find derivatives of products by separately differentiating each factor.
In our worked solution, the product rule is applied to differentiate both \( (t^2 + 1)\sin t \) and \( (t - 1)(2t + 5) \). Here’s how it plays out:

• For \( u = t^2 + 1 \), \( u' = 2t \); and \( v = \sin t \), \( v' = \cos t \), yielding \( (2t)(\sin t) + (t^2 + 1)(\cos t) \).
• For \( u = t - 1 \), \( u' = 1 \); and \( v = 2t + 5 \), \( v' = 2 \), rendering \( (1)(2t + 5) + (t - 1)(2) \).

This technique becomes even more critical in vector differentiation, where each component of the vector might be a product of functions themselves.