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Chapter 11: Problem 4

Compute \(\mathbf{r}^{\prime \prime}(t)\) when \(\mathbf{r}(t)=\left\langle t^{10}, 8 t, \cos t\right\rangle\)

Short Answer

Expert verified
Answer: The second derivative of the vector function is \(\mathbf{r}''(t) = \left\langle 90t^{8}, 0, -\cos t \right\rangle\).

Step by step solution

01

Differentiate \(\mathbf{r}(t)\) once to get \(\mathbf{r}'(t)\)

First, we need to find the first derivative of \(\mathbf{r}(t)\). To do this, differentiate each component with respect to \(t\). Let's compute the first derivative of \(\mathbf{r}(t)\): - For the first component: \(\frac{d(t^{10})}{dt} = 10 t^{9}\) - For the second component: \(\frac{d(8t)}{dt} = 8\) - For the third component: \(\frac{d(\cos t)}{dt} = -\sin t\) Therefore, \(\mathbf{r}'(t) = \left\langle 10t^{9}, 8, -\sin t \right\rangle\).
02

Differentiate \(\mathbf{r}'(t)\) once more to get \(\mathbf{r}''(t)\)

Now, we need to find the second derivative of \(\mathbf{r}(t)\). To do this, differentiate each component of \(\mathbf{r}'(t)\) with respect to \(t\). Let's compute the second derivative of \(\mathbf{r}(t)\): - For the first component: \(\frac{d(10t^{9})}{dt} = 90 t^{8}\) - For the second component: \(\frac{d(8)}{dt} = 0\) - For the third component: \(\frac{d(-\sin t)}{dt} = -\cos t\) So, \(\mathbf{r}''(t) = \left\langle 90t^{8}, 0, -\cos t \right\rangle\). Our final answer is \(\mathbf{r}''(t) = \left\langle 90t^{8}, 0, -\cos t \right\rangle\).

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Most popular questions from this chapter

Direction angles and cosines Let \(\mathbf{v}=\langle a, b, c\rangle\) and let \(\alpha, \beta\) and \(\gamma\) be the angles between \(\mathbf{v}\) and the positive \(x\) -axis, the positive \(y\) -axis, and the positive \(z\) -axis, respectively (see figure). a. Prove that \(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\) b. Find a vector that makes a \(45^{\circ}\) angle with \(\mathbf{i}\) and \(\mathbf{j}\). What angle does it make with \(\mathbf{k} ?\) c. Find a vector that makes a \(60^{\circ}\) angle with i and \(\mathbf{j}\). What angle does it make with k? d. Is there a vector that makes a \(30^{\circ}\) angle with \(\mathbf{i}\) and \(\mathbf{j} ?\) Explain. e. Find a vector \(\mathbf{v}\) such that \(\alpha=\beta=\gamma .\) What is the angle?

Prove the following vector properties using components. Then make a sketch to illustrate the property geometrically. Suppose \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are vectors in the \(x y\) -plane and a and \(c\) are scalars. $$(\mathbf{u}+\mathbf{v})+\mathbf{w}=\mathbf{u}+(\mathbf{v}+\mathbf{w})$$

Let $$\mathbf{u}(t)=2 t^{3} \mathbf{i}+\left(t^{2}-1\right) \mathbf{j}-8 \mathbf{k} \text { and } \mathbf{v}(t)=e^{t} \mathbf{i}+2 e^{-t} \mathbf{j}-e^{2 t} \mathbf{k}$$ Compute the derivative of the following functions. $$\mathbf{v}(\sqrt{t})$$

Compute the following derivatives. $$\frac{d}{d t}\left(\left(t^{3} \mathbf{i}+6 \mathbf{j}-2 \sqrt{t} \mathbf{k}\right) \times\left(3 t \mathbf{i}-12 t^{2} \mathbf{j}-6 t^{-2} \mathbf{k}\right)\right)$$

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 \(\mathbb{R}^{3}\).
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