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

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

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Answer 1 · 2017-10-14T00:47:43

$\frac{π - 4}{π + 4}$ $(pi-4)/(pi+4)$

Explanation:

First, note that we can simplify the equation:

$r = \frac{{sin}^{2} θ}{θ - θ {cos}^{2} θ} = \frac{{sin}^{2} θ}{θ (1 - {cos}^{2} θ)} = \frac{{sin}^{2} θ}{θ {sin}^{2} θ} = \frac{1}{θ}$ $r=sin^2theta/(theta-thetacos^2theta)=sin^2theta/(theta(1-cos^2theta))=sin^2theta/(thetasin^2theta)=1/theta$

The slope of the equation is given by $\frac{d y}{d x}$ $dy/dx$ .

To do this, we need to find the use the relationship between $r, θ$ $r,theta$ and $x, y$ $x,y$ . Recall that $x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$ .

Thus, we can do:

$\frac{d y}{d x} = \frac{d y / d θ}{d x / d θ} = \frac{\frac{d}{d θ} r sin θ}{\frac{d}{d θ} r cos θ}$ $dy/dx=(dy//d theta)/(dx//d theta)=(d/(d theta)rsintheta)/(d/(d theta)rcostheta$

Since we have $r = θ^{- 1}$ $r=theta^-1$ , we get:

$\frac{d y}{d x} = \frac{\frac{d}{d θ} θ^{- 1} sin θ}{\frac{d}{d θ} θ^{- 1} cos θ}$ $dy/dx=(d/(d theta)theta^-1sintheta)/(d/(d theta)theta^-1costheta)$

Using the product rule:

$\frac{d y}{d x} = \frac{- θ^{- 2} sin θ + θ^{- 1} cos θ}{- θ^{- 2} cos θ - θ^{- 1} sin θ}$ $dy/dx=(-theta^-2sintheta+theta^-1costheta)/(-theta^-2costheta-theta^-1sintheta)$

Multiply through by $\frac{θ^{2}}{θ^{2}}$ $theta^2/theta^2$ :

$\frac{d y}{d x} = \frac{- sin θ + θ cos θ}{- cos θ - θ sin θ}$ $dy/dx=(-sintheta+thetacostheta)/(-costheta-thetasintheta)$

Recalling that $sin (\frac{π}{4}) = cos (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ $sin(pi/4)=cos(pi/4)=1/sqrt2$ , we see that the slope at $θ = \frac{π}{4}$ $theta=pi/4$ the slope is:

$\frac{d y}{d x} = \frac{- \frac{1}{\sqrt{2}} + \frac{π}{4 \sqrt{2}}}{- \frac{1}{\sqrt{2}} - \frac{π}{4 \sqrt{2}}}$ $dy/dx=(-1/sqrt2+pi/(4sqrt2))/(-1/sqrt2-pi/(4sqrt2))$

Multiply through by $\frac{4 \sqrt{2}}{4 \sqrt{2}}$ $(4sqrt2)/(4sqrt2)$ :

$\frac{d y}{d x} = \frac{- 4 + π}{- 4 - π} = \frac{π - 4}{π + 4}$ $dy/dx=(-4+pi)/(-4-pi)=(pi-4)/(pi+4)$

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

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What is the slope of the tangent line of r=sin2θθ−θcos2θr=(sin^2theta)/(theta-thetacos^2theta) at θ=π4theta=(pi)/4?

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