Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question

3.2.7

**Problem Statement** Find the unit tangent vector to the curve defined by \[ \vec{r}(t) = \langle 5t, -5t, \sqrt{1 - t^2} \rangle \] at \( t = 0 \). **Solution:** To find the unit tangent vector, we need to determine the derivative of the vector function \(\vec{r}(t)\) and then evaluate it at \( t = 0 \). Then, we will normalize this vector to obtain the unit tangent vector. 1. **Differentiate \(\vec{r}(t)\):** \[ \vec{r}'(t) = \left\langle \frac{d}{dt}(5t), \frac{d}{dt}(-5t), \frac{d}{dt}\left(\sqrt{1 - t^2}\right) \right\rangle \] \[ = \langle 5, -5, \frac{-t}{\sqrt{1-t^2}} \rangle \] 2. **Evaluate \(\vec{r}'(t)\) at \( t = 0 \):** \[ \vec{r}'(0) = \langle 5, -5, 0 \rangle \] 3. **Find the magnitude of \(\vec{r}'(0)\):** \[ \|\vec{r}'(0)\| = \sqrt{5^2 + (-5)^2 + 0^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \] 4. **Normalize \(\vec{r}'(0)\) to find the unit tangent vector \(\vec{T}(0)\):** \[ \vec{T}(0) = \frac{\vec{r}'(0)}{\|\vec{r}'(0)\|} = \frac{\langle 5, -5, 0 \rangle}{5\sqrt{2}} = \left\langle \frac{5}{5\sqrt{2}}, \frac{-5}{5\sqrt{2}}, 0 \right\rangle \] \[ = \left\langle \frac{1}{
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Transcribed Image Text:**Problem Statement** Find the unit tangent vector to the curve defined by \[ \vec{r}(t) = \langle 5t, -5t, \sqrt{1 - t^2} \rangle \] at \( t = 0 \). **Solution:** To find the unit tangent vector, we need to determine the derivative of the vector function \(\vec{r}(t)\) and then evaluate it at \( t = 0 \). Then, we will normalize this vector to obtain the unit tangent vector. 1. **Differentiate \(\vec{r}(t)\):** \[ \vec{r}'(t) = \left\langle \frac{d}{dt}(5t), \frac{d}{dt}(-5t), \frac{d}{dt}\left(\sqrt{1 - t^2}\right) \right\rangle \] \[ = \langle 5, -5, \frac{-t}{\sqrt{1-t^2}} \rangle \] 2. **Evaluate \(\vec{r}'(t)\) at \( t = 0 \):** \[ \vec{r}'(0) = \langle 5, -5, 0 \rangle \] 3. **Find the magnitude of \(\vec{r}'(0)\):** \[ \|\vec{r}'(0)\| = \sqrt{5^2 + (-5)^2 + 0^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \] 4. **Normalize \(\vec{r}'(0)\) to find the unit tangent vector \(\vec{T}(0)\):** \[ \vec{T}(0) = \frac{\vec{r}'(0)}{\|\vec{r}'(0)\|} = \frac{\langle 5, -5, 0 \rangle}{5\sqrt{2}} = \left\langle \frac{5}{5\sqrt{2}}, \frac{-5}{5\sqrt{2}}, 0 \right\rangle \] \[ = \left\langle \frac{1}{
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