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

Answer 1

Christopher Allers

# f(t)=(t-1,sqrt(t)-6) # # t=0 => f(0)=(0-2, sqrt(0)-0) = (-2,0)# # t=1 => f(1)=(1-2, sqrt(1)-1) = (-1,0)#

These two points lie in a straight line do distance between the points = 1

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Answer 2

Charles Anctil

To find the distance between ( f(0) ) and ( f(1) ) for the function ( f(t) = (t-2, \sqrt{t} - t) ):

Calculate ( f(0) ): [ f(0) = (0 - 2, \sqrt{0} - 0) = (-2, 0) ]
Calculate ( f(1) ): [ f(1) = (1 - 2, \sqrt{1} - 1) = (-1, 0) ]
Use the distance formula: [ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]
Plug in the values: [ \text{Distance} = \sqrt{(-1 - (-2))^2 + (0 - 0)^2} = \sqrt{1^2 + 0^2} = \sqrt{1} = 1 ]

So, the distance between ( f(0) ) and ( f(1) ) is 1 unit.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.