`x=4cos^2theta , y=2sintheta` Find all points (if any) of horizontal and vertical tangency to the curve.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Parametric curve (x(t),y(t)) has a horizontal tangent if its slope `dy/dx` is zero i.e when `dy/dt=0` and `dx/dt!=0`

It has a vertical tangent if its slope approaches infinity i.e it is undefined, which implies that `dx/dt=0` and `dy/dt!=0`  

Given parametric equations are:

`x=4cos^2(theta) ,y=2sin(theta)`

Here the parameter is `theta`

Let's take the derivative of x and y with respect to `theta`

`dx/(d theta)=4(2cos(theta)d/(d theta)cos(theta))`

`dx/(d theta)=4(2cos(theta)(-sin(theta)))`

`dx/(d theta)=-4(2sin(theta)cos(theta))`

Use trigonometric identity: `sin(2theta)=2sin(theta)cos(theta)`

`dx/(d theta)=-4sin(2theta)`

`dy/(d theta)=2cos(theta)`

For Horizontal tangents, set the derivative of y equal to zero

`dy/(d theta)=2cos(theta)=0`



Let's check `dx/(d theta)` for the above angles,

For `theta=pi/2`

`dx/(d theta)=-4sin(2*pi/2)=-4sin(pi)=0`

For `theta=(3pi)/2`

`dx/(d theta)=-4sin(2*(3pi)/2)=-4sin(3pi)=0`

So, there are no horizontal tangents.

Now for vertical tangents, set the derivative of x equal to zero,

`dx/(d theta)=-4sin(2theta)=0`




Let's check for the above angles,

For `theta=0`

`dy/(d theta)=2cos(0)=2` 

For `theta=pi/2`

`dy/(d theta)=2cos(pi/2)=0`

For `theta=pi`

`dy/(d theta)=2cos(pi)=-2`

For `theta=(3pi)/2`

`dy/(d theta)=2cos((3pi)/2)=0`

So, the curve has vertical tangents at `theta=0,pi`

Now let's find the corresponding x and y coordinates by plugging `theta` in the parametric equation,

For `theta=0`



For `theta=pi`



So, the given parametric curve has vertical tangent at (4,0).

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial