Question

What is the slope of the tangent line of `r=theta-sin((10theta)/3-(pi)/8)` at `theta=(pi)/4`?

Answer 1

Slope of the tangent line of
`r=theta-sin((10theta)/3-pi/8)`
at `theta=pi/4` is
`(dy)/(dx)=0.995`

Explanation:

Given:
`r=theta-sin((10theta)/3-pi/8)`
`(dr)/(d(theta))=1-(10/3)cos((10theta)/3-pi/8)`

wkt
`x=rcostheta`
`(dx)/(d(theta))=r(-sintheta)+costheta((dr)/(d(theta)))`
at `theta = pi/4`
`r=pi/4-sin((10pi/4)/3-pi/8)`
`=-0.008`
`(dr)/(d(theta))=1-(10/3)cos((10pi/4)/3-pi/8)`
`=3.029`
`(dx)/(d(theta))=-0.008(-sinpi/4)+cos(pi/4)(3.029)`
`=0.008(0.707)+0.707(3.029)`
`(dx)/(d(theta))=2.147`

`y=rsintheta`
`(dy)/(d(theta))=r(costheta)+sintheta((dr)/(d(theta)))`
At `theta=pi/4`
`(dy)/(d(theta))=-0.008(cospi/4)+sin(pi/4)(3.029)`
`(dy)/(d(theta))=2.136`
Slope of the tangent line of
`r=theta-sin((10theta)/3-pi/8)`
at `theta=pi/4` is
`(dy)/(dx)=((dy)/(d(theta)))/((dx)/(d(theta)))=2.136/2.147`
Thus,
`(dy)/(dx)=0.995`