Find the derivative of the vector function r(t) = In(10 – t) i+ v10+tj+ 8e-8* k r'(t) = (

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Question
**Objective:** Find the derivative of the vector function.

**Given:**

\[ \mathbf{r}(t) = \ln(10 - t^2) \, \mathbf{i} + \sqrt{10 + t} \, \mathbf{j} + 8e^{-8t} \, \mathbf{k} \]

**Solution:** The derivative of the vector function is expressed as \(\mathbf{r}'(t)\) and is calculated as:

\[ \mathbf{r}'(t) = \left\langle \frac{d}{dt}\left(\ln(10 - t^2)\right), \frac{d}{dt}\left(\sqrt{10 + t}\right), \frac{d}{dt}\left(8e^{-8t}\right) \right\rangle \]

**Step-by-Step Calculation:**

1. **First Component:**

   Derivative of \(\ln(10 - t^2)\):

   \[
   \frac{d}{dt}[\ln(10 - t^2)] = \frac{-2t}{10 - t^2}
   \]

2. **Second Component:**

   Derivative of \(\sqrt{10 + t}\):

   \[
   \frac{d}{dt}[\sqrt{10 + t}] = \frac{1}{2\sqrt{10 + t}}
   \]

3. **Third Component:**

   Derivative of \(8e^{-8t}\):

   \[
   \frac{d}{dt}[8e^{-8t}] = -64e^{-8t}
   \]

**Final Derivative:**

\[ \mathbf{r}'(t) = \left\langle \frac{-2t}{10 - t^2}, \frac{1}{2\sqrt{10 + t}}, -64e^{-8t} \right\rangle \]

This expression provides the rate of change of the vector function \(\mathbf{r}(t)\) with respect to \(t\).
Transcribed Image Text:**Objective:** Find the derivative of the vector function. **Given:** \[ \mathbf{r}(t) = \ln(10 - t^2) \, \mathbf{i} + \sqrt{10 + t} \, \mathbf{j} + 8e^{-8t} \, \mathbf{k} \] **Solution:** The derivative of the vector function is expressed as \(\mathbf{r}'(t)\) and is calculated as: \[ \mathbf{r}'(t) = \left\langle \frac{d}{dt}\left(\ln(10 - t^2)\right), \frac{d}{dt}\left(\sqrt{10 + t}\right), \frac{d}{dt}\left(8e^{-8t}\right) \right\rangle \] **Step-by-Step Calculation:** 1. **First Component:** Derivative of \(\ln(10 - t^2)\): \[ \frac{d}{dt}[\ln(10 - t^2)] = \frac{-2t}{10 - t^2} \] 2. **Second Component:** Derivative of \(\sqrt{10 + t}\): \[ \frac{d}{dt}[\sqrt{10 + t}] = \frac{1}{2\sqrt{10 + t}} \] 3. **Third Component:** Derivative of \(8e^{-8t}\): \[ \frac{d}{dt}[8e^{-8t}] = -64e^{-8t} \] **Final Derivative:** \[ \mathbf{r}'(t) = \left\langle \frac{-2t}{10 - t^2}, \frac{1}{2\sqrt{10 + t}}, -64e^{-8t} \right\rangle \] This expression provides the rate of change of the vector function \(\mathbf{r}(t)\) with respect to \(t\).
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