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MoonlitFate

  • 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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  1. moonlitfate
    • one year ago
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    \[\frac{ dy }{ dx } = \frac{ dy / dt }{ dx/dt }\] \[\frac{ dy }{ dt} = \cos t + 2 \cos(2t)\] \[\frac{ dx }{ dt} = -\sin t -2\sin(2t)\] \[\frac{ dy }{ dx } = \frac{ \cos t + 2 \cos (2t) }{ -\sin t -2\sin(2t) }\]

  2. moonlitfate
    • one year ago
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    Kind of confused on what to do right now. How to simplify all this trig. D:

  3. amistre64
    • one year ago
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    dont worry about the simplification, we simply need to determine that t value

  4. amistre64
    • one year ago
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    assuming -1,1 is the stated point what is the value of t, when does x=-1, and y=1?

  5. amistre64
    • one year ago
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    -1= cos t + cos (2t) 1 = sin t + sin (2t) adding them might prove beneficial 0 = cos(t) + sin(t) + cos(2t) + sin(2t)

  6. moonlitfate
    • one year ago
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    I didn't even think that you could do that. Wow...

  7. moonlitfate
    • one year ago
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    Is that a double angle identity?

  8. amistre64
    • one year ago
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    no, its just x+y, given that x=-1 and y=1

  9. amistre64
    • one year ago
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    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?

  10. amistre64
    • one year ago
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    er, -1 that is

  11. amistre64
    • one year ago
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    do you understand what we are doing at this point? if we know t, we can determine the value of our derivatives.

  12. moonlitfate
    • one year ago
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    No, no I don't. Have am I in Calc. 3 and not know this...

  13. amistre64
    • one year ago
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    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

  14. phi
    • one year ago
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    Here is a graph of the problem

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