For #f(t)= ((lnt)^2/t,t^3)# what is the distance between #f(2)# and #f(4)#?

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Answer 1 · 2017-07-19T12:42:18

#56# (2dp)

We have a parametric function giving us cartesian coordinates for the specified parameter:

# f(t) = ( (lnt)^2/t, t^3 ) #

This with #t=2# we get:

# f(2) = ( (ln 2)^2/2, 8 ) #

And with #t=4# we get:

# f(4) = ( (ln 4)^2 /4, 64 ) #
# " " = ( (ln 2^2)^2 /4, 64 ) #
# " " = ( (2ln 2)^2 /4, 64 ) #
# " " = ( (ln 2)^2, 64 ) #

The the distance. #d#, between these two cartesian coordinates is given by pythagoras:

# d^2 = ((ln 2)^2 - (ln 2)^2/2)^2 + (64-8)^2 #
# " " = ((ln 2)^2/2)^2 + (56)^2 #
# " " = (ln 2)^4/4 + 3136 #

Giving:

# d = sqrt((ln 2)^4/4 + 3136) #
# \ \ ~~56.000515 ... #