What is the slope of the polar curve #f(theta) = sectheta - csctheta # at #theta = (3pi)/8#?

Question

What is the slope of the polar curve #f(theta) =  sectheta - csctheta # at #theta = (3pi)/8#?

Charles Arocha · Accepted Answer

#m = 6.757#

The slope of the tangent line of a function at a point is equal to the derivative of the function at that point.

With that being said, let's take the derivative

#(df)/(d theta) [f(theta) = sectheta - csctheta]#

The derivative of #sectheta# is #tanthetasectheta#:

#f'(theta) = tanthetasectheta - d/(d theta) [csctheta]#

The derivative of #csctheta# is #-cotthetacsctheta#:

#ul(f'(theta) = tanthetasectheta + cotthetacsctheta#

Or, in terms of #sin# and #cos#:

#ul(f'(theta) = (sintheta)/(cos^2theta) + (costheta)/(sin^2theta)#

Now, to find the slope at the point #(3pi)/8#, we plug it in for #theta#:

#m = (sin((3pi)/8))/(cos^2((3pi)/8)) + (cos((3pi)/8))/(sin^2((3pi)/8)) = color(blue)(2sqrt(10+sqrt2)#

#= color(blue)(ulbar(|stackrel(" ")(" "6.757" ")|)#

Elijah Abram · Answer

undefined The slope of the polar curve $ f(\theta) = \sec(\theta) - \csc(\theta) $ at $ \theta = \frac{3\pi}{8} $ can be found by differentiating the equation with respect to $ \theta $ and then evaluating the derivative at $ \theta = \frac{3\pi}{8} $. The derivative of $ \sec(\theta) $ is $ \sec(\theta) \tan(\theta) $, and the derivative of $ \csc(\theta) $ is $ -\csc(\theta) \cot(\theta) $. Therefore, the derivative of $ f(\theta) $ is $ \sec(\theta) \tan(\theta) + \csc(\theta) \cot(\theta) $. Evaluating this at $ \theta = \frac{3\pi}{8} $, we get the slope of the curve.