Problem 39 - Essential Calculus Early Transcendentals

Question

\(39-44=\) Find the derivative of the vector function. $$ \mathbf{r}(t)=\left\langle t \sin t, t^{2}, t \cos 2 t\right\rangle $$

Step-by-Step Solution

Answer

The derivative of the vector function is \(\mathbf{r}'(t) = \langle \sin t + t \cos t, 2t, \cos 2t - 2t \sin 2t \rangle\).

Step 1: Identify the Vector Function Components

The vector function is given as \(\mathbf{r}(t) = \langle t \sin t, t^2, t \cos 2t \rangle\). This vector function has three components: \(x(t) = t \sin t\), \(y(t) = t^2\), and \(z(t) = t \cos 2t\).

Step 2: Differentiate the First Component

To find the derivative of the first component \(x(t) = t \sin t\), we use the product rule: \((uv)' = u'v + uv'\). Let \(u = t\) and \(v = \sin t\). Then, \(u' = 1\) and \(v' = \cos t\). The derivative is \(\frac{d}{dt}[t \sin t] = (1)(\sin t) + (t)(\cos t) = \sin t + t \cos t\).

Step 3: Differentiate the Second Component

The second component is \(y(t) = t^2\). The derivative is straightforward: \(\frac{d}{dt}[t^2] = 2t\).

Step 4: Differentiate the Third Component

To find the derivative of the third component \(z(t) = t \cos 2t\), we again use the product rule. Let \(u = t\) and \(v = \cos 2t\). Then, \(u' = 1\) and \(v' = -2 \sin 2t\) (applying the chain rule). The derivative is \(\frac{d}{dt}[t \cos 2t] = (1)(\cos 2t) + (t)(-2 \sin 2t) = \cos 2t - 2t \sin 2t\).

Step 5: Combine the Derivatives

Combine all the derived components to find the derivative of the vector function \(\mathbf{r}'(t) = \langle \frac{d}{dt}[t \sin t], \frac{d}{dt}[t^2], \frac{d}{dt}[t \cos 2t] \rangle\). This gives us \(\mathbf{r}'(t) = \langle \sin t + t \cos t, 2t, \cos 2t - 2t \sin 2t \rangle\).

Key Concepts

Product Rule in Derivatives

When we encounter the task of differentiating a product of functions, such as in the vector function component \(x(t) = t \sin t\), we employ the product rule. This rule is pivotal in calculus for handling derivatives of function products. The product rule is expressed as \((uv)' = u'v + uv'\).
In this formula:

• \(u\) and \(v\) are functions of \(t\).
• \(u'\) and \(v'\) are the derivatives of \(u\) and \(v\), respectively.

Let's break it down step-by-step for \(x(t)\):

• Let \(u = t\) and \(v = \sin t\).
• The derivative of \(u\), \(u'\), is \(1\).
• The derivative of \(v\), \(v'\), is \(\cos t\).

Applying the product rule: \[\frac{d}{dt}[t \sin t] = (1)(\sin t) + (t)(\cos t) = \sin t + t \cos t.\]
This result gives us the derivative of the first component in the vector function, illustrating the power of the product rule in simplifying derivative calculations.

Chain Rule in Calculus

The chain rule is essential when dealing with composite functions, where a function is nested within another. In the context of our vector function, the chain rule is applied when differentiating \(z(t) = t \cos 2t\). This situation requires us to find the derivative of a function like \(\cos 2t\), which itself contains a simple chain: a cosine function of \(2t\).
The chain rule formula is: \((f(g(t)))' = f'(g(t)) \cdot g'(t)\).

• Here, \(f\) is the outer function (e.g., \(\cos\)), and \(g(t)\) is the inner function (e.g., \(2t\)).
• We first differentiate the outer function with respect to the inner function.
• Then, multiply this result by the derivative of the inner function.

For \(\cos 2t\):

• Outer function \(f(g) = \cos(g)\); derivative \(f'(g) = -\sin(g)\).
• Inner function \(g(t) = 2t\); derivative \(g'(t) = 2\).
• Therefore, \((\cos 2t)' = -\sin(2t) \cdot 2 = -2 \sin 2t\).

Combining with the product rule for \(z(t)\), the derivative becomes \(\cos 2t - 2t \sin 2t\).
This part of the derivative showcases how seamlessly the chain rule integrates with the product rule to handle derivatives involving more complex compositions.

Differentiation of Vector Functions

Vector functions often describe the motion of objects in space, represented with components that need to be differentiated individually. Understanding how to differentiate these vector functions richens our toolset in vector calculus. The primary step is to differentiate each component separately.
Given a vector function \(\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle\):

• We compute the derivative of each component: \(\mathbf{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle\).
• For \(x(t) = t \sin t\), as seen, \(x'(t) = \sin t + t \cos t\).
• For \(y(t) = t^2\), a simple power function, \(y'(t) = 2t\).
• For \(z(t) = t \cos 2t\), we applied both the product and chain rules to find \(z'(t) = \cos 2t - 2t \sin 2t\).

Compiling these results forms the derivative: \[\mathbf{r}'(t) = \langle \sin t + t \cos t, 2t, \cos 2t - 2t \sin 2t \rangle.\]
Each component tells part of the story of the vector's changing direction and magnitude. With calculus, we decipher how these components evolve with time, enhancing our comprehension of movements across different fields including physics and engineering.