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

1 Answer

#sqrt(25 e^(28) - 20 e^(19) + 4 e^(10) + 1521)#

Explanation:

We have: #f(t) = ((t) / (e^(1 - 3 t)), 2 t^(2) - t)#

First, let's evaluate #f(2)# and #f(5)#:

#=> f(2) = ((2) / (e^(1 - 3 cdot 2)), 2 cdot 2^(2) - 2)#

#=> f(2) = ((2) / (e^(1 - 6)), 2 cdot 4 - 2)#

#=> f(2) = ((2) / (e^(- 5)), 8 - 2)#

#therefore f(2) = (2 e^(5), 6)#

and

#=> f(5) = ((5) / (e^(1 - 3 cdot 5)), 2 cdot 5^(2) - 5)#

#=> f(5) = ((5) / (e^(1 - 15)), 2 cdot 25 - 5)#

#=> f(5) = ((5) / (e^(-14)), 50 - 5)#

#therefore f(5) = (5 e^(14), 45)#

Then, let's evaluate the distance between these two points:

#=> d = sqrt((5 e^(14) - 2 e^(5))^(2) + (45 - 6)^(2))#

#=> d = sqrt((25 e^(28) - 20 e^(19) + 4 e^(10)) + 39^(2))#

#therefore d = sqrt(25 e^(28) - 20 e^(19) + 4 e^(10) + 1521)#