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

1 Answer

Vinícius Ferraz

Jul 19, 2017

#D = (2 sqrt {37})/3#

Explanation:

#f(0) = (1/-3, 0)#

#f(2) = (1/(2-3), 4)#

#D^2 = (-1/3 - (-1))^2 + (0 - 4)^2 = 4/9 + 16#

#D = sqrt {148/9}#

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