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

Answer 1

Cameron Alexander

# ~~ 51 \ units #

We have:

# f(t) = (te^(1-3t),2t^2-t) #

When #t=2# we have:

# f(2) = (2e^(1-6),2*4-2) # # \ \ \ \ \ \ \ = (2e^(-5),6) #

When #t=5# we have:

# f(5) = (5e^(1-15),2*25-5) # # \ \ \ \ \ \ \ = (5e^(-14),-45) #

So, using Pythagoras, the distance, #d#, between these coordinates is given by:

# d^2 = (2e^(-5)-5e^(-14))^2+ (6-(-45))^2 # # \ \ \ = (2e^(-5)-5e^(-14))^2+ (6+45)^2 # # \ \ \ = (2e^(-5)-5e^(-14))^2+ 51^2 #

So that:

# d = sqrt((2e^(-5)-5e^(-14))^2+ 51^2) # # \ \ ~~ 51 \ units #

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Answer 2

Aurora Alvarado

To find the distance between two points ((x_1, y_1)) and ((x_2, y_2)) in a Cartesian coordinate system, you can use the distance formula:

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

In this case, (f(2)) corresponds to ((2e^{-5}, 6)) and (f(5)) corresponds to ((5e^{-14}, 50)). Plugging these values into the distance formula:

[ \text{Distance} = \sqrt{(5e^{-14} - 2e^{-5})^2 + (50 - 6)^2} ]

Solving this equation will give you the distance between (f(2)) and (f(5)).

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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.