What is the slope of the polar curve $f (θ) = 2 θ + {tan}^{2} θ - cos θ$ $f(theta) = 2theta + tan^2theta - costheta$ at $θ = \frac{π}{8}$ $theta = (pi)/8$ ?

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Answer 1 · 2017-01-06T02:10:46

0.48799, nearly.

In polar form. slope

$\frac{d y}{d x}$ $(dy)/(dx)$

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

$=(r'sintheta+rcostheta)/(r'costheta-rsintheta)$ ,

where $r'=(dr)/(d theta)$ .

Here,

$r =2 theta+tan^2theta-costheta=pi/4+tan^2(pi/8)-cosn(pi/8)$

at $theta=pi/8 = 22.5^o$

$=0.20466$ , nearly.

$r'=2+2tantheta sec^2theta+sinthea$

$=2+2tan(pi/8)sec^2(pi/8)+sin(pi/8)$

#=3.335325, nearly.

Now the slope at $theta = pi/8$ is

$=(r'sintheta+rcostheta)/(r'costheta-rsintheta)$ ,

where $r'=(dr)/(d theta) and theta = pi/8$

Upon substitutions and simplification,

slope = 0.48799, nearly.

What is the slope of the polar curve f(theta) = 2theta + tan^2theta - costheta f(θ)=2θ+tan2θ−cosθf(theta) = 2theta + tan^2theta - costheta at theta = (pi)/8θ=π8theta = (pi)/8?