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

1 Answer
Nathan L.
Aug 3, 2017

#m = 6.757#

Explanation:

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" ")|)#

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