Asked by Abdullah11
Find the equation for the tangent line to the curve at the point defined by the given value of t.also find d²y/dx² at this point x=SECt, y=TANt,where t=π/4.
Answers
Answered by
Steve
x(?/4) = ?2
y(?/4) = 1
dy/dx = (dy/dt)/(dx/dt) = sec^2(t)/(sect tant) = sect/tant = csc t
so at t=?/4, y' = ?2 and the tangent line is
y-1 = ?2 (x-?2)
y = ?2 x - 1
Note that x^2 = 1+y^2. So, to check the graphs, see
http://www.wolframalpha.com/input/?i=plot+x%5E2-y%5E2%3D1,+y%3D%E2%88%9A2+x+-+1
for the 2nd derivative,
d²y/dx² = d/dx (dy/dx)
= d/dt(dy/dx) / dx/dt
= -cot^2(t)/(sect tant)
= -cos^4(t)/sin(t)
y(?/4) = 1
dy/dx = (dy/dt)/(dx/dt) = sec^2(t)/(sect tant) = sect/tant = csc t
so at t=?/4, y' = ?2 and the tangent line is
y-1 = ?2 (x-?2)
y = ?2 x - 1
Note that x^2 = 1+y^2. So, to check the graphs, see
http://www.wolframalpha.com/input/?i=plot+x%5E2-y%5E2%3D1,+y%3D%E2%88%9A2+x+-+1
for the 2nd derivative,
d²y/dx² = d/dx (dy/dx)
= d/dt(dy/dx) / dx/dt
= -cot^2(t)/(sect tant)
= -cos^4(t)/sin(t)
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