What is the slope of the tangent line of $r = \frac{{sin}^{2} θ}{2 θ - θ cos 2 θ}$ $r=(sin^2theta)/(2theta-thetacos2theta)$ at $θ = \frac{π}{4}$ $theta=(pi)/4$ ?

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What is the slope of the tangent line of $r = \frac{{sin}^{2} θ}{2 θ - θ cos 2 θ}$ $r=(sin^2theta)/(2theta-thetacos2theta)$ at $θ = \frac{π}{4}$ $theta=(pi)/4$ ?

1 Answer

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Answer 1 · 2018-01-15T17:21:24

slope is the derivative value at a given value of $θ$ $theta$

Explanation:

dr/d $θ$ $theta$
by quotient rule:
it is
$\frac{2 sin θ cos θ (2 θ - θ cos 2 θ) - ({sin}^{2} θ) (2 - (cos 2 θ + θ (- sin 2 θ) \cdot 2))}{{(2 θ - θ cos 2 θ)}^{2}}$ $(2sinthetacostheta(2theta -thetacos2theta) - (sin^2theta)(2 - (cos2theta + theta(-sin2theta)*2)))/(2theta - thetacos2theta)^2$
now placing the value of $θ = \frac{π}{4}$ $theta = pi/4$
you can find slope
hope u find it helpful :)

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What is the slope of the tangent line of $r = \frac{{sin}^{2} θ}{2 θ - θ cos 2 θ}$ $r=(sin^2theta)/(2theta-thetacos2theta)$ at $θ = \frac{π}{4}$ $theta=(pi)/4$ ?

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What is the slope of the tangent line of r=sin2θ2θ−θcos2θr=(sin^2theta)/(2theta-thetacos2theta) at θ=π4theta=(pi)/4?

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Explanation:

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What is the slope of the tangent line of $r = \frac{{sin}^{2} θ}{2 θ - θ cos 2 θ}$ $r=(sin^2theta)/(2theta-thetacos2theta)$ at $θ = \frac{π}{4}$ $theta=(pi)/4$ ?