71 Find the tan gential. Component aT and the normal componenteN O e accekration a funcrion of t. if +Ct) = (2+2, 7+3> %3D as a T Ct) : %3D a N (t)=

Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Problem Statement:**

Find the tangential component \( a_T \) and the normal component \( a_N \) of acceleration as a function of \( t \) if \( \mathbf{r}(t) = \langle 2t^2, 7t^3 \rangle \).

\( a_T(t) = \, ? \)

\( a_N(t) = \, ? \)

**Explanation:**

This problem involves finding components of acceleration for the vector function \( \mathbf{r}(t) \). The vector \( \mathbf{r}(t) \) describes a position in terms of the parameter \( t \), with components along \( x \) and \( y \) axes given by \( 2t^2 \) and \( 7t^3 \), respectively.

To solve this:

1. **Calculate the velocity vector \( \mathbf{v}(t) \):**
   - **Derivative of position vector:** 
   \[
   \mathbf{v}(t) = \frac{d}{dt}\langle 2t^2, 7t^3 \rangle = \langle 4t, 21t^2 \rangle
   \]

2. **Calculate the acceleration vector \( \mathbf{a}(t) \):**
   - **Derivative of velocity vector:**
   \[
   \mathbf{a}(t) = \frac{d}{dt}\langle 4t, 21t^2 \rangle = \langle 4, 42t \rangle
   \]

3. **Find the magnitudes:**
   - **Speed \( v(t) \):** 
   \[
   v(t) = \|\mathbf{v}(t)\| = \sqrt{(4t)^2 + (21t^2)^2} = \sqrt{16t^2 + 441t^4}
   \]

4. **Calculate the tangential component \( a_T(t) \):**
   - **Using the formula \( a_T = \frac{d}{dt}v(t) \):**

5. **Calculate the normal component \( a_N(t) \):**
   - **Using the formula \( a_N = \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2} \):**

This involves further differentiating and simplifying to arrive
Transcribed Image Text:**Problem Statement:** Find the tangential component \( a_T \) and the normal component \( a_N \) of acceleration as a function of \( t \) if \( \mathbf{r}(t) = \langle 2t^2, 7t^3 \rangle \). \( a_T(t) = \, ? \) \( a_N(t) = \, ? \) **Explanation:** This problem involves finding components of acceleration for the vector function \( \mathbf{r}(t) \). The vector \( \mathbf{r}(t) \) describes a position in terms of the parameter \( t \), with components along \( x \) and \( y \) axes given by \( 2t^2 \) and \( 7t^3 \), respectively. To solve this: 1. **Calculate the velocity vector \( \mathbf{v}(t) \):** - **Derivative of position vector:** \[ \mathbf{v}(t) = \frac{d}{dt}\langle 2t^2, 7t^3 \rangle = \langle 4t, 21t^2 \rangle \] 2. **Calculate the acceleration vector \( \mathbf{a}(t) \):** - **Derivative of velocity vector:** \[ \mathbf{a}(t) = \frac{d}{dt}\langle 4t, 21t^2 \rangle = \langle 4, 42t \rangle \] 3. **Find the magnitudes:** - **Speed \( v(t) \):** \[ v(t) = \|\mathbf{v}(t)\| = \sqrt{(4t)^2 + (21t^2)^2} = \sqrt{16t^2 + 441t^4} \] 4. **Calculate the tangential component \( a_T(t) \):** - **Using the formula \( a_T = \frac{d}{dt}v(t) \):** 5. **Calculate the normal component \( a_N(t) \):** - **Using the formula \( a_N = \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2} \):** This involves further differentiating and simplifying to arrive
Expert Solution
Step 1

Given

r(t)=2t2,7t3r'(t)=4t,21t2r''(t)=4,42t

Tangential component of acceleration vector is

aT=r'(t).r''(t)r'(t)=4t(4)+21t2(42t)(4t)2+(21t2)2=882t3+16tt16+441t2=882t2+1616+441t2

 

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