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)=
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
8th Edition
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F96c94fd9-7a10-4eda-8759-e5ebee480378%2Ff21de932-aca4-48d2-9d05-24bef14810d4%2F3ivxljq_processed.jpeg&w=3840&q=75)
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
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Tangential component of acceleration vector is
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