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Doctors Medicines MCQs Q&A

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I recall that, I have given here the formula, for the polar equation io

the tangent at $P( a, alpha )$, with slope

$m =tanpsi=(r'sin theta + r cos theta)/(r'costheta-rsintheta)$,

evaluated at $(a, alpha)$, as

$rsin(theta-psi)=asin(alpha-psi)$

Here,

$r =5theta+2sin(4theta+2/3pi)$

$alpha=2/3pi=2.0944 radian =120^o$

$a = 10/3pi -sqrt3=8.74$

$r' =5=8cos(4theta+2/3pi) =1,$ at P

$m=0.3400 and psi=20.3^o$.

Now, the equation to the tangent at $P(8.74, 120^o)$ is

$rsin(theta^o-20.3^o)=8.74sin(120^o-20.3^o)=8.74sin(99.7^o)$

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