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

Question

Isabella Ackerman · Accepted Answer

Slope is given by #dy/dx#. However, this is not explicitly defined when we have a polar function. We have to use the identities for #x# and #y#:

And to find #dy/dx#, we need to see that:

#dy/dx=(dy//d theta)/(dx//d theta)=(d/(d theta)(rsintheta))/(d/(d theta)(rcostheta))#

Using the product rule, we can say that:

#dy/dx=((dr)/(d theta)sintheta+rcostheta)/((dr)/(d theta)costheta-rsintheta)#

Since #r=tan^2theta-theta#, we see that:

#(dr)/(d theta)=2tantheta(sec^2theta)-1#

So:

#dy/dx=((2tanthetasec^2theta-1)sintheta+(tan^2theta-theta)costheta)/((2tanthetasec^2theta-1)costheta-(tan^2theta-theta)sin theta)#

Evaluate this at #theta=(3pi)/8# for the slope.

Elijah Abram · Answer

undefined The slope of $ r = \tan^2(\theta) - \theta $ at $ \theta = \frac{3\pi}{8} $ can be found by taking the derivative of $ r $ with respect to $ \theta $ and then evaluating it at $ \theta = \frac{3\pi}{8} $. The derivative of $ r $ with respect to $ \theta $ is:

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

Evaluating this derivative at $ \theta = \frac{3\pi}{8} $, we get:

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

Calculate the tangent and secant values at $ \frac{3\pi}{8} $, and then substitute into the expression. This will give you the slope of the curve at $ \theta = \frac{3\pi}{8} $.