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Mathematics 15 Online
OpenStudy (anonymous):

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?

OpenStudy (amistre64):

derive it :)

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

well i think i got t=2 and 6

OpenStudy (amistre64):

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

OpenStudy (anonymous):

ok, well then what?

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

ok, well that helps. Thank you! :D

OpenStudy (amistre64):

youre welcome :)

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