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Question
xsiny-ysinx = pi
a) find dy/dx
b) find equation of tangent line at (pi/2,pi)
We need to remember that in deriving implicit functions :
1. d/dx {x} = 1
2. d/dy {y} = y'
3. (fg)'=f'g+fg'
So,
a)d/dx {xsiny-ysinx=pi} -> x'*siny+x(siny)'-y'*sinx-y(sinx)' = 0
-> 1*siny+x*(cosy)y'-y'*sinx-y*(cosx) = 0 ->
siny+xy'*cosy-y'sinx-ycosx=0 -> siny-ycosx=-xy'*cosy+y'sinx=0 ->
siny-ycosx=y'(sinx-xcosy) -> y'=(ycosx-siny)/(xcosy-sinx).
b) The equation of the tangent line is mx+n , where m=y'(π/2,π).
So, m=(sin[π]-πcos[π/2])/([π/2]cos[π]-sin[π/2])=(0-0)/(-π/2-1)=0.
To find n , we substitute (π/2,π) in the equation of the tangent
line & we get : π=m*π/2+n -> n=π. This means that the tangent line
is horizontal at y=π/2.
Alon.
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Alon Mandes
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Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .
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