/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Find the unit tangent vector at ... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 30

Find the unit tangent vector at the given value of t for the following parameterized curves. $$\mathbf{r}(t)=\left\langle\sqrt{7} e^{t}, 3 e^{t}, 3 e^{t}\right\rangle, \text { for } 0 \leq t \leq 1 ; t=\ln 2$$

Short Answer

Expert verified
Given the parametric equation \(\mathbf{r}(t)=\left\langle\sqrt{7} e^{t}, 3 e^{t}, 3 e^{t}\right\rangle\), find the unit tangent vector at \(t=\ln 2\). The unit tangent vector at \(t = \ln 2\) is: $$\mathbf{T}(\ln 2) = \left\langle\frac{2\sqrt{7}}{10}, \frac{3}{5}, \frac{3}{5}\right\rangle.$$

Step by step solution

01

Find the derivative of the parametric equation

We need the derivative of the parameterized curve with respect to time to find the tangent vector. To find the derivative, take the derivative of each component with respect to t. So: $$\mathbf{r'}(t) = \left\langle\frac{d}{dt}(\sqrt{7} e^{t}), \frac{d}{dt}(3 e^{t}), \frac{d}{dt}(3 e^{t})\right\rangle$$ Computing the derivatives, we get: $$\mathbf{r'}(t) = \left\langle\sqrt{7} e^{t}, 3 e^{t}, 3 e^{t}\right\rangle$$
02

Substitute the given value of t into the derivative

Now, we need to find the tangent vector at the given value of t. Substitute \(t = \ln 2\) into the derivative equation: $$\mathbf{r'}(\ln 2) = \left\langle\sqrt{7} e^{\ln 2}, 3 e^{\ln 2}, 3 e^{\ln 2}\right\rangle$$ Since \(e^{\ln 2} = 2\), we can simplify this expression to: $$\mathbf{r'}(\ln 2) = \left\langle\sqrt{7} \cdot 2, 3 \cdot 2, 3 \cdot 2\right\rangle$$ Thus, the tangent vector at \(t = \ln 2\) is: $$\mathbf{r'}(\ln 2) = \left\langle2\sqrt{7}, 6, 6\right\rangle$$
03

Find the magnitude of the tangent vector

To find the unit tangent vector, we need to normalize the tangent vector from step 2. First, find the magnitude of the tangent vector. The magnitude is given by: $$\|\mathbf{r'}(\ln 2)\| = \sqrt{(2\sqrt{7})^2 + 6^2 + 6^2} = \sqrt{28 + 36 + 36} = \sqrt{100} = 10$$
04

Normalize the tangent vector

Now, we can normalize the tangent vector by dividing it by the magnitude. The unit tangent vector is given by: $$\mathbf{T}(t) = \frac{\mathbf{r'}(t)}{\|\mathbf{r'}(t)\|} = \frac{1}{10}\left\langle2\sqrt{7}, 6, 6\right\rangle$$ So the unit tangent vector at \(t = \ln 2\) is: $$\mathbf{T}(\ln 2) = \left\langle\frac{2\sqrt{7}}{10}, \frac{6}{10}, \frac{6}{10}\right\rangle$$

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Most popular questions from this chapter

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