8) Find the unit tangenit vector TilE) at the poin Let ř(t)= <-5t'+3, Be -S sin (E)) 3t -3t -5cos (E)) 4. K-25€ -3t -9e 412 +(-969 + (-Scos (t) -3ti2

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter1: Vectors
Section1.3: Lines And Planes
Problem 33EQ
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I am missing a concept in this type of problem somewhere because I keep getting them wrong. Can someone see what I am missing?
8) Find the wnlit tanqenit vector TiCt) at.thc point to
3t
-3t
5cos (e)>
4.
-3t
K-25¢,-9¢
-3ti
(19:9:(-Scos(E))
+ 25 sin'(t)
T (e) = <-25t",-9c-Scos (t)>
t 8le
-3t
+25 sin?cE)
F(0) = Ko, 9,-5)
Jot
0 +81 +o
-3t
- SCos CE))
u2sE 8le *+25 sin (E)
366)
7(0) = <-256)", -9e
4
-S cos (6) y
<0-9,-5)
-3.0
625 (6)" + 81e
+ 25 Sin C6)
0t 8 1+0
N81
(0) = 0,-1,-5
Transcribed Image Text:8) Find the wnlit tanqenit vector TiCt) at.thc point to 3t -3t 5cos (e)> 4. -3t K-25¢,-9¢ -3ti (19:9:(-Scos(E)) + 25 sin'(t) T (e) = <-25t",-9c-Scos (t)> t 8le -3t +25 sin?cE) F(0) = Ko, 9,-5) Jot 0 +81 +o -3t - SCos CE)) u2sE 8le *+25 sin (E) 366) 7(0) = <-256)", -9e 4 -S cos (6) y <0-9,-5) -3.0 625 (6)" + 81e + 25 Sin C6) 0t 8 1+0 N81 (0) = 0,-1,-5
Expert Solution
Step 1

Given curve rt=<-5t5+3,3e-3t,-5sint>

We need to find tangent vector Tt at t=0.

Tangent vector is given by:

Tt=r'tr't.

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