Asked by Anonymous
Find an equation of the tangent line at the point (π/4,π/4)
sin(π₯βπ¦)=π₯*cos(π¦+π/4)
sin(π₯βπ¦)=π₯*cos(π¦+π/4)
Answers
Answered by
oobleck
sin(x-y) = x cos(y + Ο/4)
cos(x-y) - cos(x-y)y' = cos(y + Ο/4) - x sin(y + Ο/4) y'
y' = [cos(y+ Ο/4) - cos(x-y)] / [x sin(y + Ο/4) - cos(x-y)]
at (Ο/4,Ο/4), that means
y' = [0 - 1] / [Ο/4 - 1] = 1/(1 - Ο/4)
Now you have a point and a slope, so the line is
y - Ο/4 = 1/(1 - Ο/4) (x - Ο/4)
see the graphs at
https://www.wolframalpha.com/input/?i=plot+sin%28x-y%29+%3D+x+cos%28y+%2B+%CF%80%2F4%29%2C+y+-+%CF%80%2F4+%3D+1%2F%281+-+%CF%80%2F4%29+%28x+-+%CF%80%2F4%29+for+0+%3C%3D+x+%3C%3D++%CF%80%2F3%2C+0+%3C%3D+y+%3C%3D++%CF%80%2F3
cos(x-y) - cos(x-y)y' = cos(y + Ο/4) - x sin(y + Ο/4) y'
y' = [cos(y+ Ο/4) - cos(x-y)] / [x sin(y + Ο/4) - cos(x-y)]
at (Ο/4,Ο/4), that means
y' = [0 - 1] / [Ο/4 - 1] = 1/(1 - Ο/4)
Now you have a point and a slope, so the line is
y - Ο/4 = 1/(1 - Ο/4) (x - Ο/4)
see the graphs at
https://www.wolframalpha.com/input/?i=plot+sin%28x-y%29+%3D+x+cos%28y+%2B+%CF%80%2F4%29%2C+y+-+%CF%80%2F4+%3D+1%2F%281+-+%CF%80%2F4%29+%28x+-+%CF%80%2F4%29+for+0+%3C%3D+x+%3C%3D++%CF%80%2F3%2C+0+%3C%3D+y+%3C%3D++%CF%80%2F3
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