For #f(t)= (e^t/t,4t+1/t)# what is the distance between #f(2)# and #f(5)#?

Question

1499 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-02T05:42:42

The distance is #sqrt(4e^10-20e^7+25e^4+117)/10#

we will substitute first 2 and then 5:

#f(2)=(e^2/2,4*2+1/2)=(e^2/2,17/2)#

#f(5)=(e^5/5,4*5+1/5)=(e^5/5,101/5)#

Then we will apply the distance formula:

#d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

#d=sqrt((e^5/5-e^2/2)^2+(101/5-17/2)^2)#

#=sqrt(((2e^5-5e^2)/10)^2+((101*2-17*5)/10)^2)#

#=sqrt((4e^10-20e^7+25e^4)/100+(202-85)/100)#

#=sqrt(4e^10-20e^7+25e^4+117)/10#