Tangent Line in Polar Coordinates

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Let's see how to derive equation of tangent line when we are given equation of curve `r=f(theta)` in polar coordinates.

We will proceed in the same fashion as with tangent lines to parametric curves, because polar coordinates in some sense similar to parametric curves.

We know that `x=rcos(theta)` and `y=rsin(theta)`.

Since we are given that `r=f(theta)` we should treat `r` as a function of `theta`.

To find slope of tangent line, we need to find `(dy)/(dx)` but in terms of `r` and `theta`. This can be easily done with chain rule:

`(dy)/(dx)=(dy)/(dtheta)*(dtheta)/(dx)=(dy)/(dtheta)*1/((dx)/(dtheta))`.

We have that `(dy)/(dtheta)=(dr)/(dtheta)sin(theta)+rcos(theta)` and `(dx)/(dtheta)=(dr)/(dtheta)cos(theta)-rsin(theta)`.

So, we obtained following fact.

Fact. Slope of tangent line in polar coordinates is `(dy)/(dx)=((dr)/(dtheta)sin(theta)+rcos(theta))/((dr)/(dtheta)cos(theta)-rsin(theta))`.

Example. Find equation of tangent line to `r=1+2cos(theta)` at `theta=pi/4`.

tangent line to curve in polar coordinatesWe have that `(dr)/(dtheta)=-2sin(theta)` , so

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

`=(cos(theta)+2cos(2theta))/(-sin(theta)-2sin(2theta))`.

So, slope of tangent line at point `theta=pi/4` is `m=(dy)/(dx)|_(theta=pi/4)=(cos(pi/4)+2cos(pi/2))/(-sin(pi/4)-2sin(pi/2))=1/7(1-2sqrt(2))`.

Now we need corresponding `x` and `y` coordinates when `theta=pi/4`:

`x=rcos(theta)=(1+2cos(theta))cos(theta)=(1+2cos(pi/4))cos(pi/4)=(1+sqrt(2))/sqrt(2)`.

`y=rsin(theta)=(1+2cos(theta))sin(theta)=(1+2cos(pi/4))sin(pi/4)=(1+sqrt(2))/sqrt(2)`.

So, equation of tangent line that passes through point `theta=pi/4` is `y=1/7(1-2sqrt(2))(x-(1+sqrt(2))/sqrt(2))+(1+sqrt(2))/sqrt(2)` or approximately `y=-0.2612x+2.153`.

As with parametric curves there are curves that have several tangent line at one point. For example, in above example, such point is (0,0). The corresponding value(s) of `theta` we can find by solving equation `1+2cos(theta)=0`. This gives two solutions on interval `[0,2pi]`: `(2pi)/3` and `(4pi)/3`, so there will be two tangent lines at (0,0).

Similarly to parametric curves, curve in polar coordinates has

  • horizontal tangent when `(dy)/(dtheta)=0` (provided that `(dx)/(dtheta)!=0`).
  • vertical tangent when `(dx)/(dtheta)=0` (provided that `(dy)/(dtheta)!=0`).