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

Answer 1

Ezekiel Alexander

The distance is #=27.17#

We start by calculating #f(2)# and #f(5)#

#f(t)=(t-2, -t^2-2t)#

#f(2)=(0,-8)#

#f(5)=(3,-35)#

The distance between #(0,-8)# and #(3,-35)# is

#=sqrt((3-0)^2+(-35+8)^2)#

#=sqrt(9+729)#

#=sqrt738#

#=27.17#

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Answer 2

Charles Anctil

To find the distance between ( f(2) ) and ( f(5) ), we first need to evaluate ( f(t) ) at ( t = 2 ) and ( t = 5 ), then calculate the distance between the resulting points.

( f(2) = (2 - 2, -(2)^2 - 2(2)) = (0, -8) )

( f(5) = (5 - 2, -(5)^2 - 2(5)) = (3, -35) )

Using the distance formula:

( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} )

Substitute the values:

( d = \sqrt{(3 - 0)^2 + (-35 - (-8))^2} )

( d = \sqrt{3^2 + (-35 + 8)^2} )

( d = \sqrt{9 + (-27)^2} )

( d = \sqrt{9 + 729} )

( d = \sqrt{738} )

Therefore, the distance between ( f(2) ) and ( f(5) ) is ( \sqrt{738} ).

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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.