A particle moves along a curve given by the parametric equations x(t) = t^2 + sint and y(t) = (t^3/4) - 2t^2 + 5t -3 for time t 0. Find the slope of the path at the time when the particle's y-coordinate is -1. What is the speed of the particle at this point?

Question

A particle moves along a curve given by the parametric equations x(t) = t^2 + sint and y(t) = (t^3/4) - 2t^2 + 5t -3 for time t > 0. Find the slope of the path at the time when the particle's y-coordinate is -1. What is the speed of the particle at this point?

amistre64 · Answer

derive it :)

amistre64 · Answer

dx = 2t

dy = .(3/4)t^2 -4t +5 ; im guessing that a fraction for the first term

amistre64 · Answer

err... left off the cos(t)..

dx = 2t+cos(t)

anonymous · Answer

right right, so then i get dy/dx, but then where do i plug in the y = -1 part?

amistre64 · Answer

(t^3)/4 - 2t^2 + 5t -3 = -1

(t^3)/4 - 2t^2 + 5t -2 = 0

amistre64 · Answer

t^3 -8t^2 +20t -8 = 0

amistre64 · Answer

can we factor this?  or should we just go straight to wolframalpha and get the zero :)

anonymous · Answer

well i think i got t=2 and 6

amistre64 · Answer

well if I did that right, its nasty looking....

anonymous · Answer

ok, well then what?

amistre64 · Answer

then we know what t = and plug it into the dy/dx spots ...

anonymous · Answer

but i have two different t values, so is one of them wrong?

amistre64 · Answer

dy      (3/4)t^2 -4t +5
--- = -------------
dx          2t+cos(t)

amistre64 · Answer

it quite possible that there are just simply two places that it matches; but yes, checking would be prudent

anonymous · Answer

ok, well that helps. Thank you! :D

amistre64 · Answer

youre welcome :)