Consider the parametric curve given by the equations

Question

x(t)=t^2–14t–4
y(t)=t^2–14t+48
How many units of distance are covered by the point P(t) = (x(t),y(t)) between t=0, and t=11 ?

Answer 1

distance, or displacement? There is a pronounced difference in those.

Answer 2

well to get a distance between points we use the distance formula but first we need the points themselves.
Point 1 is found by subbing t=0 in for t in x(t) and y(t)
Point 2 is found by subbing t=11 in for t
in x(t) and y(t)

I'll do point 1

x(t) = (0)^2 - 14(0) - 4 = -4
y(t) = (0)^2 - 14(0) + 48 = 48
so point 1 is (-4,48)

Hope that helps