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Calculus

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Consider the parametric curve given by the equations
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 ?

  • Calculus -

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

  • Calculus -

    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

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