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?
derive it :)
dx = 2t dy = .(3/4)t^2 -4t +5 ; im guessing that a fraction for the first term
err... left off the cos(t).. dx = 2t+cos(t)
right right, so then i get dy/dx, but then where do i plug in the y = -1 part?
(t^3)/4 - 2t^2 + 5t -3 = -1 (t^3)/4 - 2t^2 + 5t -2 = 0
t^3 -8t^2 +20t -8 = 0
can we factor this? or should we just go straight to wolframalpha and get the zero :)
well i think i got t=2 and 6
well if I did that right, its nasty looking....
ok, well then what?
then we know what t = and plug it into the dy/dx spots ...
but i have two different t values, so is one of them wrong?
dy (3/4)t^2 -4t +5 --- = ------------- dx 2t+cos(t)
it quite possible that there are just simply two places that it matches; but yes, checking would be prudent
ok, well that helps. Thank you! :D
youre welcome :)
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