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

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

Answer 1

Christopher Agresta

#(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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Isabella Ackerman

To find the slope of ( r = \tan^2(\theta) - \theta^2 - \theta ) at ( \theta = \frac{3\pi}{8} ), we first find the derivative of ( r ) with respect to ( \theta ), ( \frac{dr}{d\theta} ), and then evaluate it at ( \theta = \frac{3\pi}{8} ).

[ r = \tan^2(\theta) - \theta^2 - \theta ]

To find ( \frac{dr}{d\theta} ), we differentiate ( r ) with respect to ( \theta ):

[ \frac{dr}{d\theta} = 2\tan(\theta)\sec^2(\theta) - 2\theta - 1 ]

Now, evaluate ( \frac{dr}{d\theta} ) at ( \theta = \frac{3\pi}{8} ):

[ \frac{dr}{d\theta}\bigg|_{\theta = \frac{3\pi}{8}} = 2\tan\left(\frac{3\pi}{8}\right)\sec^2\left(\frac{3\pi}{8}\right) - 2\left(\frac{3\pi}{8}\right) - 1 ]

Calculate the value to find the slope at ( \theta = \frac{3\pi}{8} ).

By signing up, you agree to our Terms of Service and Privacy Policy

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.