What is the slope of $r=tantheta^2-theta^2-theta$ at $theta=(3pi)/8$ ?

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Answer 1 · 2018-04-14T06:29:28

$(dr)/(d theta) = 67.88$

$r=tantheta^2-theta^2-theta$

$(dr)/(d theta) = 2theta(sectheta^2)^2-2theta-1$

At $theta =(3pi)/8$

$(dr)/(d theta ) = 2times(3pi)/8times(sec((3pi)/8)^2)^2-2times(3pi)/8-1$

$(dr)/(d theta) = 67.88$

What is the slope of r=tantheta^2-theta^2-thetar=tantheta^2-theta^2-theta at theta=(3pi)/8theta=(3pi)/8?