What is the slope of $r = tan θ - θ$ $r=tantheta-theta$ at $θ = \frac{π}{8}$ $theta=pi/8$ ?

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What is the slope of $r = tan θ - θ$ $r=tantheta-theta$ at $θ = \frac{π}{8}$ $theta=pi/8$ ?

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Answer 1 · 2018-03-19T18:59:33

Derivative with Polar Coordinates is $\frac{\partial x}{\partial y} = \frac{\frac{\partial r}{\partial θ} sin (θ) + r cos (θ)}{\frac{\partial r}{\partial θ} cos (θ) - r sin (θ)}$ $(\partial x) / (\partial y) = ((\partial r) / (\partial theta) sin(theta)+ rcos(theta))/((\partial r) / (\partial theta) cos(theta)- rsin(theta)$

Explanation:

$\frac{\partial r}{\partial θ} = \frac{1}{{cos}^{2} (θ)} + 1$ $(\partial r) / (\partial theta) = 1/cos^2(theta)+1$
$r (θ = \frac{π}{8}) = 0.0 22$ $r(theta = pi/8)=0.0 22$
$\frac{\partial x}{\partial y} = \frac{\frac{tan (θ)}{cos (θ)} + sin (θ) + r cos (θ)}{\frac{1}{cos (θ)} + cos (θ) - r sin (θ)}$ $(\partial x) / (\partial y) = (tan(theta)/cos(theta)+sin(theta)+rcos(theta))/(1/cos(theta)+cos(theta)-rsin(theta))$
$\frac{\partial x}{\partial y} (θ = \frac{π}{8}) = 1.57$ $(\partial x) / (\partial y)(theta=pi/8) =1.57$

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What is the slope of $r = tan θ - θ$ $r=tantheta-theta$ at $θ = \frac{π}{8}$ $theta=pi/8$ ?

1 Answer

Explanation:

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What is the slope of $r = tan θ - θ$ $r=tantheta-theta$ at $θ = \frac{π}{8}$ $theta=pi/8$ ?