Problem 48
Question
Compute the derivatives of the vector-valued functions. $$\mathbf{r}(t)=\tan (2 t) \mathbf{i}+\sec (2 t) \mathbf{j}+\sin ^{2}(t) \mathbf{k}$$
Step-by-Step Solution
Verified Answer
\( \mathbf{r}'(t) = 2 \sec^2(2t) \mathbf{i} + 2 \sec(2t)\tan(2t) \mathbf{j} + \sin(2t) \mathbf{k} \)
1Step 1: Differentiate the i-component
The i-component of the vector function is \( an(2t)\). To differentiate this, we use the chain rule. The derivative of \(\tan(u)\) with respect to \(u\) is \(\sec^2(u)\), and for \(u=2t\), we have:\[ \frac{d}{dt} \tan(2t) = 2 \sec^2(2t) \]
2Step 2: Differentiate the j-component
The j-component is \(\sec(2t)\). Differentiating \(\sec(u)\) gives us \(\sec(u)\tan(u)\) and applying the chain rule for \(u=2t\) yields:\[\frac{d}{dt} \sec(2t) = 2 \sec(2t)\tan(2t)\]
3Step 3: Differentiate the k-component
The k-component is \(\sin^2(t)\). We apply the chain rule here as well. The derivative of \(u^2\) is \(2u\), so:\[\frac{d}{dt} \sin^2(t) = 2 \sin(t) \cos(t) = \sin(2t)\]This uses the identity \(2\sin(t)\cos(t) = \sin(2t)\).
4Step 4: Combine the results
Now, we combine all the differentiated components back into the vector:\[\mathbf{r}'(t) = \left(2 \sec^2(2t)\right) \mathbf{i} + \left(2 \sec(2t)\tan(2t)\right) \mathbf{j} + \sin(2t) \mathbf{k}\]
Key Concepts
Vector-Valued FunctionsDerivativesChain Rule
Vector-Valued Functions
Vector-valued functions are functions that return a vector as their output. They are written in the form \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \), where \( f(t), g(t), \) and \( h(t) \) are scalar functions. These functions map a single input (usually a real number \( t \)) to a vector in three-dimensional space.
Imagine a vector-valued function as a path a flying insect might take through space, described at every moment by a vector whose direction and magnitude can change. The components \( f(t), g(t), \text{ and } h(t) \) tell you how far along the \( x \), \( y \), and \( z \)-axes the insect has flown at time \( t \).
Vector-valued functions are hugely important in physics and engineering because they can describe anything that requires three-dimensional consideration, like fluid flow, orbits of planets, or in this instance, derivatives of motion.
Imagine a vector-valued function as a path a flying insect might take through space, described at every moment by a vector whose direction and magnitude can change. The components \( f(t), g(t), \text{ and } h(t) \) tell you how far along the \( x \), \( y \), and \( z \)-axes the insect has flown at time \( t \).
Vector-valued functions are hugely important in physics and engineering because they can describe anything that requires three-dimensional consideration, like fluid flow, orbits of planets, or in this instance, derivatives of motion.
Derivatives
The derivative of a vector-valued function is a vector that represents the rate of change of the function with respect to the input variable. Technically, it involves differentiating each component of the vector separately. If \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \), then the derivative \( \mathbf{r}'(t) \) is given by:
The derivative provides the velocity or rate of change vector of the original function. Practically, for each component \( f(t), g(t), \) and \( h(t) \), you determine how fast and in what manner they change over time.
In the example, each component is differentiated separately. The formulas used, such as the derivative of \( \tan(u) \) being \( \sec^2(u) \), come from standard calculus rules. This allows us to understand the dynamic behavior of the vector's trajectory over time.
- \( f'(t) \mathbf{i} + g'(t) \mathbf{j} + h'(t) \mathbf{k} \)
The derivative provides the velocity or rate of change vector of the original function. Practically, for each component \( f(t), g(t), \) and \( h(t) \), you determine how fast and in what manner they change over time.
In the example, each component is differentiated separately. The formulas used, such as the derivative of \( \tan(u) \) being \( \sec^2(u) \), come from standard calculus rules. This allows us to understand the dynamic behavior of the vector's trajectory over time.
Chain Rule
The chain rule is a fundamental theorem used to differentiate composite functions. If you have a function dependent on another function, the chain rule tells you how its derivative is found by relating the rates of change of both functions involved.
In more formal terms, for functions \( y(u) \) and \( u(t) \), the derivative \( \frac{dy}{dt} \) is computed as:
This rule is especially useful for handling expressions where the function being differentiated depends on another variable function, much like our exercise.
In the differentiation of \( \tan(2t) \) and \( \sec(2t) \) from the exercise, the chain rule was crucial. It involved finding the derivative of the outer function while multiplying it by the derivative of the inside function \((2t)\). This extra step is what modifies the derivative to suit the multi-layered nature of such functions. It's an essential tool in vector calculus for solving complex derivatives.
In more formal terms, for functions \( y(u) \) and \( u(t) \), the derivative \( \frac{dy}{dt} \) is computed as:
- \( \frac{dy}{du} \times \frac{du}{dt} \)
This rule is especially useful for handling expressions where the function being differentiated depends on another variable function, much like our exercise.
In the differentiation of \( \tan(2t) \) and \( \sec(2t) \) from the exercise, the chain rule was crucial. It involved finding the derivative of the outer function while multiplying it by the derivative of the inside function \((2t)\). This extra step is what modifies the derivative to suit the multi-layered nature of such functions. It's an essential tool in vector calculus for solving complex derivatives.
Other exercises in this chapter
Problem 46
Compute the derivatives of the vector-valued functions. $$\mathbf{r}(t)=t e^{t} \mathbf{i}+t \ln (t) \mathbf{j}+\sin (3 t) \mathbf{k}$$
View solution Problem 47
Compute the derivatives of the vector-valued functions. $$\mathbf{r}(t)=\frac{1}{t+1} \mathbf{i}+\arctan (t) \mathbf{j}+\ln t^{3} \mathbf{k}$$
View solution Problem 49
Compute the derivatives of the vector-valued functions. $$\mathbf{r}(t)=3 \mathbf{i}+4 \sin (3 t) \mathbf{j}+t \cos (t) \mathbf{k}$$
View solution Problem 50
Compute the derivatives of the vector-valued functions. $$\mathbf{r}(t)=t^{2} \mathbf{i}+t e^{-2 t} \mathbf{j}-5 e^{-4 t} \mathbf{k}$$
View solution