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Find \(\mathbf{r}^{\prime}(t)\). $$ \mathbf{r}(t)=4 \sqrt{t} \mathbf{i}+t^{2} \sqrt{t} \mathbf{j}+\ln t^{2} \mathbf{k} $$

Short Answer

Expert verified
Hence, the derivative of given vector function \(\mathbf{r}(t)\) is \(\mathbf{r}'(t) = 2/\sqrt{t}\mathbf{i} + (2t\sqrt{t} + t^{2} /(2 \sqrt{t}))\mathbf{j} + 2/t\mathbf{k}\)

Step by step solution

01

Identify Functions for Differentiation

The first step is to identify the functions to be differentiated in each of the vector components. \(\mathbf{r}(t) = 4\sqrt{t}\mathbf{i} + t^{2}\sqrt{t}\mathbf{j} + \ln t^{2}\mathbf{k}\) is given to us. We identify that \(\mathbf{i}\) component has '4\sqrt{t}', \(\mathbf{j}\) component has 't²\sqrt{t}' and \(\mathbf{k}\) component has '\ln t²'.
02

Differentiate the functions

Differentiating the functions of each component using rules of differentiation, we get the following results. For \(\mathbf{i}\) component: The derivative of \(4\sqrt{t}\) will be \(2/\sqrt{t}\). For \(\mathbf{j}\) component: The derivative of \(t^{2}\sqrt{t}\) can be obtained using the product rule, which gives us \((2t\sqrt{t} + t^{2} /(2 \sqrt{t}))\). For \(\mathbf{k}\) component: The derivative of \(\ln t^{2}\) is \(2/t\).
03

Construct the Derivative Vector

Now, the derivative vector can be assembled by putting these component derivatives to form \(\mathbf{r}^{\prime}(t)\). \(\mathbf{r}^{\prime}(t) = 2/\sqrt{t} \mathbf{i} + (2t\sqrt{t} + t^{2} /(2 \sqrt{t}))\mathbf{j} + 2/t\mathbf{k}\)

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Most popular questions from this chapter

Find the angle \(\theta\) between \(r(t)\) and \(r^{\prime}(t)\) as a function of \(t .\) Use a graphing utility to graph \(\theta(t) .\) Use the graph to find any extrema of the function. Find any values of \(t\) at which the vectors are orthogonal. $$ \mathbf{r}(t)=t^{2} \mathbf{i}+t \mathbf{j} $$

Use the model for projectile motion, assuming there is no air resistance. Use a graphing utility to graph the paths of a projectile for the given values of \(\theta\) and \(v_{0} .\) For each case, use the graph to approximate the maximum height and range of the projectile. (Assume that the projectile is launched from ground level.) (a) \(\theta=10^{\circ}, v_{0}=66 \mathrm{ft} / \mathrm{sec}\) (b) \(\theta=10^{\circ}, v_{0}=146 \mathrm{ft} / \mathrm{sec}\) (c) \(\theta=45^{\circ}, v_{0}=66 \mathrm{ft} / \mathrm{sec}\) (d) \(\theta=45^{\circ}, v_{0}=146 \mathrm{ft} / \mathrm{sec}\) (e) \(\theta=60^{\circ}, v_{0}=66 \mathrm{ft} / \mathrm{sec}\) (f) \(\theta=60^{\circ}, v_{0}=146 \mathrm{ft} / \mathrm{sec}\)

Find \(\mathbf{r}(t)\) for the given conditions. $$ \mathbf{r}^{\prime \prime}(t)=-4 \cos t \mathbf{j}-3 \sin t \mathbf{k}, \quad \mathbf{r}^{\prime}(0)=3 \mathbf{k}, \quad \mathbf{r}(0)=4 \mathbf{j} $$

The position vector \(r\) describes the path of an object moving in space. Find the velocity, speed, and acceleration of the object. $$ \mathbf{r}(t)=\left\langle e^{t} \cos t, e^{t} \sin t, e^{t}\right\rangle $$

Consider the motion of a point (or particle) on the circumference of a rolling circle. As the circle rolls, it generates the cycloid \(\mathbf{r}(t)=b(\omega t-\sin \omega t) \mathbf{i}+b(1-\cos \omega t) \mathbf{j}\) where \(\omega\) is the constant angular velocity of the circle and \(b\) is the radius of the circle. Find the maximum speed of a point on the circumference of an automobile tire of radius 1 foot when the automobile is traveling at 55 miles per hour. Compare this speed with the speed of the automobile.

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