Find a parameterization for the curve shown in the figure below. 140 1.0 -1 -1 デ(t) = where くtく (Note: you need to enter answers in all three answer blanks to receive feedback.)

Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter8: Introduction To Functions
Section8.10: Inverse Variation
Problem 5MRE
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How do you set this up the example is confusing 

1le
1,0
-1
-1
7(t) = (1+t)i+ (2+2t)j&,
where 0
(Note: you need to enter answers in all three answer blanks to receive feedback.)
Solution:
SOLUTION
We want the straight line segment from (1, 2) to (2, 4). The position vector of (1, 1) is i + 2j and the
displacement vector from (1, 2) to (2,4) is i + 2j, so the line has equation
6.
7 = i + 23 + t(i + 23),
or
x = 1+t,
y = 2+ 2t.
This
passes (1, 2) when t = 0 and (2, 4) when t = 1, so a possible parameterization is
7 = æ(t) + y(t)j = (1+ t)ỉ + (2+ 2t)j, 0<t< 1.
Transcribed Image Text:1le 1,0 -1 -1 7(t) = (1+t)i+ (2+2t)j&, where 0 (Note: you need to enter answers in all three answer blanks to receive feedback.) Solution: SOLUTION We want the straight line segment from (1, 2) to (2, 4). The position vector of (1, 1) is i + 2j and the displacement vector from (1, 2) to (2,4) is i + 2j, so the line has equation 6. 7 = i + 23 + t(i + 23), or x = 1+t, y = 2+ 2t. This passes (1, 2) when t = 0 and (2, 4) when t = 1, so a possible parameterization is 7 = æ(t) + y(t)j = (1+ t)ỉ + (2+ 2t)j, 0<t< 1.
Find a parameterization for the curve shown in the figure below.
140
4,0
F1
-1
F(t) =
where
<t<
(Note: you need to enter answers in all three answer blanks to receive feedback.)
Transcribed Image Text:Find a parameterization for the curve shown in the figure below. 140 4,0 F1 -1 F(t) = where <t< (Note: you need to enter answers in all three answer blanks to receive feedback.)
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