# Graph the polar function and its tangents that are horizontal or vertical for r=`sqrt(theta), 0<=theta<=2pi`

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embizze | Certified Educator

We are asked to graph the polar function `r=sqrt(theta), 0<= theta <= 2pi ` , along with its vertical and horizontal tangents.

We can graph by plotting points:

`theta: ` r:

0 0

pi/6 .7236

pi/3 1.0233

pi/2 1.2533

etc... yielding:

We can find the horizontal and vertical tangents by using:

`(dy)/(dx)=((dy)/(d theta))/((dx)/(d theta))=(1/(2sqrt(theta)) sin theta + sqrt(theta) cos theta)/(1/(2sqrt(theta))cos theta-sqrt(theta) sin theta) `

The horizontal tangents occur when the numerator (`(dy)/(d theta) ` ) is zero, while the vertical tangents occur when the denominator is zero.

Solving numerically we get the horizontal tangents when `x ~~ +- .653 `

and the vertical tangents when `x ~~ -1.83 `

Further Reading: