Home > Calculus > Determining the Slope and Tangent Lines for a Polar Curve

What is the slope of the polar curve #f(theta) = theta - sectheta+thetasin^3theta # at #theta = (7pi)/12#?

Answer 1

#f'((7pi)/12)=-13.8459#

Given: #f(theta)=theta-sectheta+thetasin^3theta# Differentiating wrt #theta# #f'(theta)=1-secthetatantheta+theta(3sin^2theta)costheta+sin^3theta# #=1-1/costhetasintheta/costheta+sin^3theta+3thetasin^2thetacostheta # #theta=(7pi)/12=1.8326#

#(7pi)/12=pi-(5pi)/12# #costheta=cos(pi-(5pi)/12)=-cos((5pi)/12)=-0.2588# #sintheta=sin(pi-(5pi)/12)=sin((5pi)/12)=0.9659# Substituting the values in #f'(theta)#

#f'((7pi)/12)=1-1/-0.2588xx0.9659/-0.2588+0.9659^3+3xx1.8326xx0.9659^2(-0.2588)# #f'((7pi)/12)=-13.8459#

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Answer 2

Adam Adkins

To find the slope of the polar curve (f(\theta) = \theta - \sec(\theta) + \theta \sin^3(\theta)) at (\theta = \frac{7\pi}{12}), first compute the derivative of (f(\theta)) with respect to (\theta). Then, evaluate the derivative at (\theta = \frac{7\pi}{12}) to find the slope of the curve at that point.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.