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Chapter 14: Problem 14

Differentiate the following functions. $$\mathbf{r}(t)=\tan t \mathbf{i}+\sec t \mathbf{j}+\cos ^{2} t \mathbf{k}$$

Short Answer

Expert verified
Question: Determine the derivative of the given vector function with respect to t: \(\mathbf{r}(t) = (\tan t \mathbf{i}+\sec t \mathbf{j}+\cos ^{2} t \mathbf{k})\). Answer: The derivative of the given vector function with respect to t is \(\frac{d\mathbf{r}}{dt} = \sec^2 t \mathbf{i}+\sec t \tan t\mathbf{j}-2\cos t\sin t\mathbf{k}\).

Step by step solution

01

Differentiate the i-component

To differentiate the i-component (\(\tan t\)) with respect to t, recall that the derivative of \(\tan t\) is \(\sec^2 t\). So, $$\frac{d}{dt}(\tan t \mathbf{i}) = \sec^2 t \mathbf{i}$$
02

Differentiate the j-component

To differentiate the j-component (\(\sec t\)) with respect to t, recall that the derivative of \(\sec t\) is \(\sec t \tan t\). So, $$\frac{d}{dt}(\sec t \mathbf{j}) = \sec t \tan t\mathbf{j}$$
03

Differentiate the k-component

To differentiate the k-component (\(\cos^2 t\)) with respect to t, we use the chain rule. Recall the chain rule states that if we have a function \(u(t)\) and its derivative \(\frac{du}{dt}\), then the derivative of \(u^2\) is \(2u\frac{du}{dt}\). In this case, \(u(t) = \cos t\) and \(\frac{du}{dt} = -\sin t\). Apply the chain rule, we have: $$\frac{d}{dt}(\cos^2 t \mathbf{k}) = 2\cos t (-\sin t)\mathbf{k}=-2\cos t\sin t\mathbf{k}$$
04

Combine the component derivatives

Now that we have the derivatives of each component, we can combine them to find the derivative of the entire vector function \(\mathbf{r}(t)\). We do this by simply adding the i, j, and k components, we have: $$\frac{d\mathbf{r}}{dt}=\frac{d}{dt}(\tan t \mathbf{i}+\sec t \mathbf{j}+\cos ^{2} t \mathbf{k})=\sec^2 t \mathbf{i}+\sec t \tan t\mathbf{j}-2\cos t\sin t\mathbf{k}$$

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Key Concepts

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

Derivative of Trigonometric Functions
In calculus, understanding the derivatives of trigonometric functions is fundamental. Let's look at the derivatives for sine and cosine: the derivative of \(\sin(t)\) is \(\cos(t)\), and the derivative of \(\cos(t)\) is \( -\sin(t) \). However, the problem requires knowledge of the derivatives of \(\tan(t)\) and \(\sec(t)\).

The derivative of \(\tan(t)\) is \(\sec^2(t)\), as seen in the differentiation of the i-component for vector function \(\mathbf{r}(t)\). Similarly, the derivative of \(\sec(t)\) is \(\sec(t)\tan(t)\), which is demonstrated while differentiating the j-component. It's crucial to memorize these derivatives as they are used frequently in calculus.

For the exercises involving such trigonometric functions, these key derivatives play an essential role. By understanding how these derivatives are derived, students can confidently tackle a wide variety of problems requiring trigonometric derivatives.
Chain Rule
The chain rule is a powerful differentiation technique in calculus. It allows us to find the derivative of a composite function. If we have a composite function \(y = f(g(x))\), its derivative \(\frac{dy}{dx}\) can be found using the chain rule as \(f'(g(x))g'(x)\).

In the context of our exercise, we use the chain rule to differentiate \(\cos^2(t)\), which is a function of a function: \(u(t) = \cos(t)\) is our inner function, and \(u^2\) is the outer function. According to the chain rule, the derivative is \(-2\cos(t)\sin(t)\) as computed in step 3.

Remember the practical application of the chain rule: always distinguish the 'outer layer' and the 'inner layer' of the function. This allows a systematic approach to differentiating more complex expressions, especially when they involve trigonometric functions.
Vector Calculus
Vector calculus extends the rules of calculus to vector functions. A vector function, like \(\mathbf{r}(t)\), is composed of individual vector components which are functions of the same variable. Each component is differentiated independently, as shown in the steps of our given exercise.

When differentiating vector functions, it's crucial to apply the rules of differentiation to each component function. In this exercise, each component of \(\mathbf{r}(t)\) corresponds to a trigonometric function whose derivative we already know or can find using rules such as the chain rule.

To find the derivative of the entire vector function, we differentiate each component (\(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\)) separately and then combine them as shown in step 4. This results in the full expression for \(\frac{d\mathbf{r}}{dt}\). Vector calculus is a fundamental part of physics and engineering and mastering it is essential for describing and analyzing physical phenomena involving vectors in motion.

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Most popular questions from this chapter

Find the function \(\mathrm{r}\) that satisfies the given conditions. $$\mathbf{r}^{\prime}(t)=\left\langle e^{2 t}, 1-2 e^{-t}, 1-2 e^{t}\right\rangle ; \quad \mathbf{r}(0)=\langle 1,1,1\rangle$$

Consider the following curves. a. Graph the curve. b. Compute the curvature. c. Graph the curvature as a ficnction of the parameter. d. Identify the points (if any) at which the curve has a maximum or minimum curvature. e. Verify that the graph of the curvature is consistent with the graph of the curve. $$\mathbf{r}(t)=\left\langle\frac{t^{2}}{2}, \frac{t^{3}}{3}\right\rangle, \text { for } t>0$$

Consider the curve. $$\mathbf{r}(t)=\langle a \cos t+b \sin t, c \cos t+d \sin t, e \cos t+f \sin t\rangle$$ where \(a, b, c, d, e,\) and \(f\) are real numbers. It can be shown that this curve lies in a plane. Graph the following curve and describe it. $$r(t)=(2 \cos t+2 \sin t) i+(-\cos t+2 \sin t) \mathbf{j}+(\cos t-2 \sin t) \mathbf{k}$$

Time of flight, range, height Derive the formulas for time of flight, range, and maximum height in the case that an object is launched from the initial position \(\left\langle 0, y_{0}\right\rangle\) above the horizontal ground with initial velocity \(\left|\mathbf{v}_{0}\right|\langle\cos \alpha, \sin \alpha\rangle\).

Relationship between \(\mathbf{r}\) and \(\mathbf{r}^{\prime}.\) Consider the curve \(\mathbf{r}(t)=\langle\sqrt{t}, 1, t\rangle,\) for \(t>0 .\) Find all points on the curve at which \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) are orthogonal.
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