What is the equation of the tangent line of #r=cos(theta-pi/2)/sintheta - sin(2theta-pi)# at #theta=(-3pi)/8#?

Question

Isabella Ackerman · Accepted Answer

Equation of tangent line:

#frac{y+1/2*sqrt{{2-sqrt2}/2}}{x-(2-sqrt2)^{3/2}/4}=-sqrt{19+6sqrt{2}}#

The equation can be simplified to

#r = 1 + sin(2theta)#

using trigonometric identities.

The polar plot is shown here

https://tutor.hix.ai

To find the tangent line, you need to know a point it passes though and its gradient, #frac{dy}{dx}#

We know that at #theta=-{3pi}/8#,

#x=rcostheta#

#= (1+sin(2(-{3pi}/8)))cos(-{3pi}/8)#

#= (2-sqrt2)^{3/2}/4#

#~~ 0.112#

#y=rsintheta#

#= (1+sin(2(-{3pi}/8)))sin(-{3pi}/8)#

#= -1/2*sqrt{{2-sqrt2}/2}#

#~~ -0.271#

The tangent line passes through #(0.112,-0.271)#.

We also know that at #theta=-{3pi}/8#,

#frac{dx}{d theta} = frac{d}{d theta}(r sintheta)#

#= frac{d}{d theta}((1+sin(2theta))*cos(theta))#

#= frac{-2sin(theta)+cos(theta)+3cos(3theta)}{2}#

Substituting #theta=-{3pi}/8#, we have

#frac{dx}{d theta}_{theta=-{3pi}/8} = -1/2*sqrt{{2-sqrt2}/2}#

#~~ -0.271#

Similarly,

#frac{dy}{d theta} = frac{d}{d theta}(r sintheta)#

#= frac{d}{d theta}((1+sin(2theta))*sin(theta))#

#= 2sin(theta)cos(2theta) + (sin(2theta)+1)*cos(theta)#

Substituting #theta=-{3pi}/8#, we have

#frac{dy}{d theta}_{theta=-{3pi}/8} = 1/2*sqrt{frac{26-7sqrt{2}}{2}}#

#~~ 1.42#

From the chain rule, we can calculate the gradient as

#frac{dy}{dx} = frac{ frac{dy}{d theta} }{ frac{dx}{d theta} }#

#= -sqrt{19+6sqrt{2}}#

#~~ 5.24#

Equation of tangent line:

#frac{y-(-1/2*sqrt{{2-sqrt2}/2})}{x-(2-sqrt2)^{3/2}/4}=-sqrt{19+6sqrt{2}}#

Emilia Aguilar · Answer

undefined To find the equation of the tangent line of the polar curve $r = \frac{\cos(\theta - \frac{\pi}{2})}{\sin(\theta)} - \sin(2\theta - \pi)$ at the point where $\theta = -\frac{3\pi}{8}$, we first need to find the derivative $\frac{dr}{d\theta}$ and evaluate it at the given value of $\theta$. Then, we find the slope of the tangent line using this derivative. Finally, we use the point-slope form to write the equation of the tangent line.

1. Find $\frac{dr}{d\theta}$:
$$ r = \frac{\cos(\theta - \frac{\pi}{2})}{\sin(\theta)} - \sin(2\theta - \pi) $$
$$ \frac{dr}{d\theta} = \frac{d}{d\theta} \left( \frac{\cos(\theta - \frac{\pi}{2})}{\sin(\theta)} - \sin(2\theta - \pi) \right) $$

2. Evaluate $\frac{dr}{d\theta}$ at $\theta = -\frac{3\pi}{8}$ to find the slope of the tangent line.

3. Once you have the slope of the tangent line, use the point-slope form of the equation of a line to find the equation of the tangent line. The point on the curve is $(r, \theta)$, where $r$ and $\theta$ are given by the equation and the value of $\theta$ provided.

This process will give you the equation of the tangent line at the point where $\theta = -\frac{3\pi}{8}$.