For #f(t)= (t-t^3,t^3)# what is the distance between #f(0)# and #f(3)#?

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Answer 1 · 2018-06-08T07:36:40

#3sqrt(145)#

By plugging values, you have

#f(0) = (0-0^3, 0^3) = (0,0)#

#f(3) = (3-3^3, 3^3) = (3-27,27) = (-24,27)#

Both #f(0)# and #f(3)# are points on the plane, so we find the distance between them with the formula

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

In your case, #(x_1,y_1)=(0,0)# and #(x_2,y_2)=(-24,27)#. So, the distance is

#d = \sqrt((0+24)^2+(0-27)^2) = sqrt(24^2+27^2) = sqrt(1305) = sqrt(9*145)=3sqrt(145)#