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anonymous
 one year ago
Find an equation of the tangent to the curve at the point x= cos t + cos (2t) , y = sin t + sin (2t), (1,1)
anonymous
 one year ago
Find an equation of the tangent to the curve at the point x= cos t + cos (2t) , y = sin t + sin (2t), (1,1)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ dy }{ dx } = \frac{ dy / dt }{ dx/dt }\] \[\frac{ dy }{ dt} = \cos t + 2 \cos(2t)\] \[\frac{ dx }{ dt} = \sin t 2\sin(2t)\] \[\frac{ dy }{ dx } = \frac{ \cos t + 2 \cos (2t) }{ \sin t 2\sin(2t) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Kind of confused on what to do right now. How to simplify all this trig. D:

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1dont worry about the simplification, we simply need to determine that t value

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1assuming 1,1 is the stated point what is the value of t, when does x=1, and y=1?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.11= cos t + cos (2t) 1 = sin t + sin (2t) adding them might prove beneficial 0 = cos(t) + sin(t) + cos(2t) + sin(2t)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I didn't even think that you could do that. Wow...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is that a double angle identity?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1no, its just x+y, given that x=1 and y=1

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1we could just work one part of it ... all depends on how simple we can make it for ourselves 1 = cos(t) + cos(2t) what are our solutions for t?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1do you understand what we are doing at this point? if we know t, we can determine the value of our derivatives.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, no I don't. Have am I in Calc. 3 and not know this...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1to form a line, we need a slope and a point, slope = dy/dx \[\frac{ dy }{ dx } = \frac{ \cos t + 2 \cos (2t) }{ \sin t 2\sin(2t) }\] no need to work this out to death since all we really need is a value of t, and not some simplified rewritten formula for dy/dx the point given is (x=1, y=1) so we simply need to assess the value of t for that point from our equations of x and y

phi
 one year ago
Best ResponseYou've already chosen the best response.0Here is a graph of the problem
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