Question

What is the slope of the tangent line of `r=2sin(theta/4)*cos(theta/2)` at `theta=(9pi)/2`?

Answer 1

The slope is `~~ -0.1`

Explanation:

Tangents to a polar curve

From the reference

`dy/dx = ((dr(theta))/(d theta)sin(theta) + r(theta)cos(theta))/((dr(theta))/(d theta)cos(theta) - r(theta)sin(theta))`

I used WolframAlpha to compute the derivative

`(dr(theta))/(d theta) = 1/4(3cos((3theta)/4) - cos(theta/4))`

I used an Excel spreadsheet

I entered the following into cell J1:

=PI()*9/2

I used cell J2 to evaluate `(dr((9pi)/2))/(d theta)` by entering the following:

=1/4(3COS(3*J1/4) - COS(J1/4))

I used cell J3 to evaluate `r((9pi)/2)` by entering the following:

=2SIN(J1/4)COS(J1/2)

I used J4 to evaluate the dy/dx and give us the slope, m by entering the following:

=(J2SIN(J1) + J3COS(J1))/(J2COS(J1) - J3SIN(J1))

The value that was returned had 5 decimal places but the second digit was 0 so I rounded to 1 place:

`m ~~ -0.1`