## 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)

1. anonymous

$\frac{ dy }{ dx } = \frac{ dy / dt }{ dx/dt }$ $\frac{ dy }{ dt} = \cos t + 2 \cos(2t)$ $\frac{ dx }{ dt} = -\sin t -2\sin(2t)$ $\frac{ dy }{ dx } = \frac{ \cos t + 2 \cos (2t) }{ -\sin t -2\sin(2t) }$

2. anonymous

Kind of confused on what to do right now. How to simplify all this trig. D:

3. amistre64

dont worry about the simplification, we simply need to determine that t value

4. amistre64

assuming -1,1 is the stated point what is the value of t, when does x=-1, and y=1?

5. 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)

6. anonymous

I didn't even think that you could do that. Wow...

7. anonymous

Is that a double angle identity?

8. amistre64

no, its just x+y, given that x=-1 and y=1

9. 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?

10. amistre64

er, -1 that is

11. amistre64

do you understand what we are doing at this point? if we know t, we can determine the value of our derivatives.

12. anonymous

No, no I don't. Have am I in Calc. 3 and not know this...

13. 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

14. phi

Here is a graph of the problem

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