Question

What is the slope of the polar curve `f(theta) = theta^2 - csctheta` at `theta = (3pi)/4`?

Answer 1

Slope of the curve at
`theta=(3pi)/4`is -1

Explanation:

Given:
`f(theta)=theta^2-csctheta`

Let
`r=f(theta)`

Then,
`r=theta^2-csctheta`

At `theta=(3pi)/4`

`r=((3pi)/4)^2-csc((3pi)/4)`

`((3pi)/4)^2=(9pi^2)/16`

`csc((3pi)/4)=sqrt2`

`r=(9pi^2)/16-sqrt2`

`(dr)/(d(theta))=2theta-(-cscthetacottheta)`

`(dr)/(d(theta))=2theta+cscthetacottheta`

`(dr)/(d(theta))=2((3pi)/4)+csc((3pi)/4)cot((3pi)/4)`

`2((3pi)/4)=(3pi)/2`

`csc((3pi)/4)=sqrt2`

`cot((3pi)/4)=-1`

`(dr)/(d(theta))=(3pi)/2+sqrt2(-1)`

`(dr)/(d(theta))=(3pi)/2-sqrt2`

To find

Slope of the curve at
`theta=(3pi)/4`

`x=rcostheta`

`y=rsintheta`

`dx/(d(theta))=r(-sintheta)+costheta(dr)/(d(theta))`

`dx/(d(theta))=-r(sintheta)+costheta(dr)/(d(theta))`

`r=(9pi^2)/16-1/sqrt2`

`sintheta=sin((3pi)/4)`

`sin((3pi)/4)=1/sqrt2`

`costheta=cos((3pi)/4)`

`cos((3pi)/4)=-1/sqrt2`

`costheta=-1/sqrt2`

`(dr)/(d(theta))=(3pi)/2-sqrt2`

`dx/(d(theta))=-((9pi^2)/16-1/sqrt2)(1/sqrt2)+(-1/sqrt2)((3pi)/2-sqrt2)`

`dx/(d(theta))=1/sqrt2((9pi^2)/16-1/sqrt2)-1/sqrt2((3pi)/2-sqrt2)`

`dx/(d(theta))=1/sqrt2(((9pi^2)/16-1/sqrt2)-((3pi)/2-sqrt2))`

`dy/(d(theta))=rcostheta+sintheta(dr)/(d(theta))`

`r=(9pi^2)/16-1/sqrt2`

`sintheta=sin((3pi)/4)`

`sin((3pi)/4)=1/sqrt2`

`costheta=cos((3pi)/4)`

`cos((3pi)/4)=-1/sqrt2`

`costheta=-1/sqrt2`

`(dr)/(d(theta))=(3pi)/2-sqrt2`

`dy/(d(theta))=((9pi^2)/16-1/sqrt2)(-1/sqrt2)+(1/sqrt2)((3pi)/2-sqrt2)`

`dy/(d(theta))=-1/sqrt2((9pi^2)/16-1/sqrt2)+(1/sqrt2)((3pi)/2-sqrt2)`

`dy/(d(theta))=-1/sqrt2(((9pi^2)/16-1/sqrt2)-((3pi)/2-sqrt2))`

`dy/dx=(dy/(d(theta)))/(dx/(d(theta)))`

`=(-1/sqrt2(((9pi^2)/16-1/sqrt2)-((3pi)/2-sqrt2)))/(1/sqrt2(((9pi^2)/16-1/sqrt2)-((3pi)/2-sqrt2)))`

Cancelling

`dy/dx=-1`

Slope of the curve at
`theta=(3pi)/4`is -1