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Chapter 11: Problem 12

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: Find the derivative of the vector-valued function \(\mathbf{r}(t)=\tan(t)\mathbf{i}+\sec(t)\mathbf{j}+\cos^2(t)\mathbf{k}\). Answer: The derivative of the given vector-valued function 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, we need to find the derivative of \(\tan (t)\). We will use the well-known differentiation rule: \(\frac{d(\tan x)}{dx} = \sec^2 x\). Differentiating \(\tan (t)\) with respect to t: $$\frac{d}{dt}(\tan t \mathbf{i}) = \sec^2 t \mathbf{i}$$
02

- Differentiate the j component

To differentiate the j component, we need to find the derivative of \(\sec(t)\). Recall that \(\sec(x) = \frac{1}{\cos(x)}\), so we can also differentiate \(\frac{1}{\cos(x)}\) with respect to x. Using the chain rule: \(\frac{d}{dx}(\frac{1}{\cos{x}})= \frac{\sin{x}}{\cos^2{x}} = \sec(x) \tan(x)\) So differentiating \(\sec (t)\) with respect to t: $$\frac{d}{dt}(\sec t \mathbf{j}) = \sec t \tan t \mathbf{j}$$
03

- Differentiate the k component

Now let's differentiate the k component. We need to find the derivative of \(\cos^2(t)\). Apply the chain rule to differentiate the composition of functions \(u(t) = \cos(t)\) and \(v(u) = u^2\): $$\frac{d}{dt}(\cos^2 t \mathbf{k}) = 2 \cos t \cdot (-\sin t) \mathbf{k} = -2\cos t\sin t \mathbf{k}$$
04

- Combine the components

We have found the derivatives of the i, j, and k components. Now, we just need to combine them to find the derivative of the given vector-valued function \(\mathbf{r}(t)\): $$\frac{d\mathbf{r}}{dt} = \sec^2 t \mathbf{i} + \sec t \tan t \mathbf{j} -2\cos t\sin t \mathbf{k}$$

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Most popular questions from this chapter

Let \(\mathbf{u}(t)=\left\langle 1, t, t^{2}\right\rangle, \mathbf{v}(t)=\left\langle t^{2},-2 t, 1\right\rangle\) and \(g(t)=2 \sqrt{t}\). Compute the derivatives of the following functions. $$\mathbf{v}(g(t))$$

Cauchy-Schwarz Inequality The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) (because \(|\cos \theta| \leq 1\) ). This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Triangle Inequality Consider the vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{u}+\mathbf{v}\) (in any number of dimensions). Use the following steps to prove that \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}|\) a. Show that \(|\mathbf{u}+\mathbf{v}|^{2}=(\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+\) \(2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2}\) b. Use the Cauchy-Schwarz Inequality to show that \(|\mathbf{u}+\mathbf{v}|^{2} \leq(|\mathbf{u}|+|\mathbf{v}|)^{2}\) c. Conclude that \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}|\) d. Interpret the Triangle Inequality geometrically in \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3}\).

Let \(\mathbf{u}(t)=\left\langle 1, t, t^{2}\right\rangle, \mathbf{v}(t)=\left\langle t^{2},-2 t, 1\right\rangle\) and \(g(t)=2 \sqrt{t}\). Compute the derivatives of the following functions. $$\mathbf{v}\left(e^{t}\right)$$

Compute the indefinite integral of the following functions. $$\mathbf{r}(t)=t e^{t} \mathbf{i}+t \sin t^{2} \mathbf{j}-\frac{2 t}{\sqrt{t^{2}+4}} \mathbf{k}$$

Proof of Sum Rule By expressing \(\mathbf{u}\) and \(\mathbf{v}\) in terms of their components, prove that $$\frac{d}{d t}(\mathbf{u}(t)+\mathbf{v}(t))=\mathbf{u}^{\prime}(t)+\mathbf{v}^{\prime}(t)$$
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