Question

Problem 3 (12 points) The curve with parametric equations (1 + 2 sin(9) cos(9), y-(1 + 2 sin(θ)) sin(0) is called a limacon and is shown in the figure below. -1 1. Find the point (x,y 2. Find the slope of the line that is tangent to the graph at θ-π/2. 3. Find the slope of the line that is tangent to the graph at (,y)-(1,0) ) that corresponds to θ-π/2.

Answer 1

We are given the parametric curve:

r= (1+2sin(θ)cos(θ), y = (1+2sin(θ))sin(θ)

1.

At $\theta = \frac{\pi}{2}$ we need to find the value of (x,y)

1 2sirn COS

or

r 12(1)(0). y (12(1))(1)

or x = 1 , y = 3

Hence the corresponding (x,y) point is : (1,3)

2.

The slope of the tangent line is nothing but the derivative dy/dx

We could write the derivative as: $\frac{dy}{dx} = \frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}}$

We have:

r= (1+2sin(θ)cos(θ), y = (1+2sin(θ))sin(θ)

or $x =1+sin(2\theta) \textup{, } y = sin(\theta) + 2sin^2(\theta)$

=>

dy d0 cos()+2 * 2sin(0)cos() (2 COS de

or

dy 2cos(20)Cos() d0 cos(0+ 2sin(20) COS de

Hence the slope of the tangent line is:

-------> (1)

dycos(0) +2sin(20) 2cos(20) CAT

Now we need to find the slope at $\theta = \frac{\pi}{2}$

plug $\theta = \frac{\pi}{2}$ in (1)

=>

dy,cos() 2sin(2*5 02 (0) d.x 2(-1) に2 2cos(2*5) 2cos(2 *

(3)

We are to find the slope of the tangent line at (x,y) = (1,0)

We know that $y = (1 + 2sin(\theta))sin(\theta)$

=> plug y = 0 and solve for theta

=> $y = (1 + 2sin(\theta))sin(\theta) = 0$

=> $sin(\theta) = 0 \textup{ and } 1 + 2sin(\theta) = 0$

=> $\theta = 0 \textup{ and } \theta = \frac{11\pi}{6}$ for theta E [0 , 2pi]

Now lets check if we plug $\theta = \frac{11\pi}{6}$ in $x = (1+2sin(\theta)cos(\theta)$ then do we get x = 1 or not

$x = (1+2sin(\frac{11 \pi}{6})cos(\frac{11 \pi}{6}) = 1-\frac{\sqrt{3}}{2}$

=> $\theta = \frac{11\pi}{6}$ is not the value that corresponds to (x,y) = (1,0)

Next we check, if we plug $\theta = 0$ in $x = (1+2sin(\theta)cos(\theta)$ then do we get x = 1 or not

Hence $\theta = 0$ is the parametric point that corresponds to (x,y) = (1,0)

Now from part (2) we know that:

The slope of the tangent line is:

dycos(0) +2sin(20) 2cos(20) CAT

If we plug $\theta = 0$ in the above slope that we would attain the slope of the tangent line at $\theta = 0$ and since $\theta = 0$ corresponds to (1,0)

Hence it would be the slope of the tangent line at (1,0) as well

=> Lets plug $\theta = 0$ in

dycos(0) +2sin(20) 2cos(20) CAT

=> $\frac{dy}{dx}|_{\theta=0}= \frac{cos(0) + 2sin(2*0)}{2cos(2*0)} = \frac{1+2(0)}{2(1)} = \frac{1}{2}$ ---> This is the slope of the tangent line at (1,0)