Question

Find an equation of the line tangent to the curve at the point corresponding to the given value of t. 71 x= cost+t sint, y=sint-tcost;t=4 (Type an equation. Simplify your answer. Type your answer in slope-intercept form. Type an exact answer. Use integers or fractions for any numbers in the equation.)

Answer 1

a = cost + tsint; y = sint -t cost it-Tli at t=1 4 x = cos (2)+ 7TT 4 sin(7) co) x - I VZ + חד st 412 y = sin(TW) - Ycos (2) 4 y=-1 7TT Vz 4 71 y = -1 V2 412 x= cost ttsint. da d (cost etsint) dt at =- sintt d (tsint) dt =-sint + (sinta (1) ute sint) -- sint+ (sint + t cost) =- sinttsint+t cost. da dt t Cost

y=sint -t cost dy (sint -t cost) dt d at d = cost-e (tcost) = cost - cost de (t) ++ eccost)) = cost-(cost-tsint) = tsint. - cost-cost etsint. dy dt dy - dyldt tsint da daldt tcost dy - tant da 7 TT slope at t= sm - 4 dy dalt- הר m- tan (7) m--| mo

m=-1, xoa IT V2 Yo = -1 - TT 4 V2 √2 485 Equation of tangent line is y-Yo = m(x-xo) y - 1 - 1 / 2 חד 71 4022 417 y + 1/2 7 IT 482 I-x+ I - TT 42. aty=f TT -1 V2 2 TTT 412 412 xty = -14 T 4 82 xty=-77 22 y = -x-TT 2/2 y = -x- 782 T 4