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myntfalskj4

myntfalskj4

Answered question

2022-07-16

For f ( t ) = ( 1 2 t - 3 , t e t ) what is the distance between f ( 0 ) and f ( 1 ) ?

Answer & Explanation

Feriheashrz

Feriheashrz

Beginner2022-07-17Added 8 answers

Step 1
We have f ( t ) = ( 1 2 t - 3 , t e t ) , so:
f ( 0 ) = ( - 1 3 , 0 )
f ( 1 ) = ( - 1 , e )
So then by Pythagoras, the distance, d, between f ( 0 ) and f ( 1 ) is given by:
d 2 = ( - 1 - ( - 1 3 ) ) 2 + ( e - 0 ) 2
      = ( - 2 3 ) 2 + ( e ) 2
      = 4 9 + e 2
And so:
d = 4 9 + e 2

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