MoonlitFate
  • MoonlitFate
Find an equation of the tangent to the curve at the point x= cos t + cos (2t) , y = sin t + sin (2t), (-1,1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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MoonlitFate
  • MoonlitFate
\[\frac{ dy }{ dx } = \frac{ dy / dt }{ dx/dt }\] \[\frac{ dy }{ dt} = \cos t + 2 \cos(2t)\] \[\frac{ dx }{ dt} = -\sin t -2\sin(2t)\] \[\frac{ dy }{ dx } = \frac{ \cos t + 2 \cos (2t) }{ -\sin t -2\sin(2t) }\]
MoonlitFate
  • MoonlitFate
Kind of confused on what to do right now. How to simplify all this trig. D:
amistre64
  • amistre64
dont worry about the simplification, we simply need to determine that t value

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amistre64
  • amistre64
assuming -1,1 is the stated point what is the value of t, when does x=-1, and y=1?
amistre64
  • amistre64
-1= cos t + cos (2t) 1 = sin t + sin (2t) adding them might prove beneficial 0 = cos(t) + sin(t) + cos(2t) + sin(2t)
MoonlitFate
  • MoonlitFate
I didn't even think that you could do that. Wow...
MoonlitFate
  • MoonlitFate
Is that a double angle identity?
amistre64
  • amistre64
no, its just x+y, given that x=-1 and y=1
amistre64
  • amistre64
we 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
  • amistre64
er, -1 that is
amistre64
  • amistre64
do you understand what we are doing at this point? if we know t, we can determine the value of our derivatives.
MoonlitFate
  • MoonlitFate
No, no I don't. Have am I in Calc. 3 and not know this...
amistre64
  • amistre64
to 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
  • phi
Here is a graph of the problem
1 Attachment

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