Question

Find a parametrization $x = x(t)$, $y = y(t)$, $t \in [0, 1]$ for the line segment from $(10, 3)$ to $(-6, -10)$. a) $x(t) = 13t + 10$, $y(t) = 16t + 3$ b) $x(t) = -16t + 10$, $y(t) = -13t + 3$ c) $x(t) = 16t + 10$, $y(t) = 13t + 3$ d) $x(t) = -16t + 10$, $y(t) = 3 - 12t$ e) $x(t) = 10 + 15t$, $y(t) = -13t + 3$

          Find a parametrization $x = x(t)$, $y = y(t)$, $t \in [0, 1]$ for the line segment from $(10, 3)$ to $(-6, -10)$.
a) $x(t) = 13t + 10$, $y(t) = 16t + 3$
b) $x(t) = -16t + 10$, $y(t) = -13t + 3$
c) $x(t) = 16t + 10$, $y(t) = 13t + 3$
d) $x(t) = -16t + 10$, $y(t) = 3 - 12t$
e) $x(t) = 10 + 15t$, $y(t) = -13t + 3$

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Find a parametrization x = x(t), y = y(t), t ∈ [0, 1] for the line segment from (10, 3) to (-6, -10).
a) x(t) = 13t + 10, y(t) = 16t + 3
b) x(t) = -16t + 10, y(t) = -13t + 3
c) x(t) = 16t + 10, y(t) = 13t + 3
d) x(t) = -16t + 10, y(t) = 3 - 12t
e) x(t) = 10 + 15t, y(t) = -13t + 3

Added by Carolina F.

Calculus: Early Transcendentals

James Stewart 8th Edition

Thanks for your feedback!

Find a parametrization x = x(t), y = y(t), t âˆˆ [0,1] for the line segment from (10,3) to (-6,-10). Thank you!!! a) x(t) = 13t + 10, y(t) = 16t + 3 b) x(t) = -16t + 10, y(t) = -13t + 3 c) x(t) = 16t + 10, y(t) = 13t + 3 d) x(t) = -16t + 10, y(t) = 3 - 12t e) x(t) = 10 + 15t, y(t) = -13t + 3 Review Later

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