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For #f(t)= (t^3-t^2+1,t^2-t)# what is the distance between #f(2)# and #f(5)#?

Answer 1

Scarlett Allison

#6 sqrt{265}#

#f(2) = (8 - 4 + 1, 4 - 2) = (5,2)#

#f(5) = (125-25+1,25-5) = (101,20)#

dist = #sqrt{96^2 + 18^2} = sqrt 9540 = sqrt{2 ^ 2 * 3 ^ 2 * 5 * 53 }#

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Answer 2

Aurora Alvarado

To find the distance between two points in a Euclidean space, you can use the distance formula. Given two points ( (x_1, y_1) ) and ( (x_2, y_2) ), the distance between them is calculated as follows:

[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]

In this case, the two points are ( f(2) ) and ( f(5) ).

[ f(2) = (2^3 - 2^2 + 1, 2^2 - 2) = (3, 2) ] [ f(5) = (5^3 - 5^2 + 1, 5^2 - 5) = (91, 20) ]

Using the distance formula:

[ \text{Distance} = \sqrt{(91 - 3)^2 + (20 - 2)^2} ] [ = \sqrt{88^2 + 18^2} ] [ = \sqrt{7744 + 324} ] [ = \sqrt{8068} ]

Therefore, the distance between ( f(2) ) and ( f(5) ) is ( \sqrt{8068} ) or approximately ( 89.89 ).

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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.