Question

How do you find the points where the graph of the function `r=e^theta` has horizontal tangents and what is the equation?

Answer 1

`theta = - pi/4, (3 pi)/4, .....`

Explanation:

Decompose polar into Cartesian as we are looking for slope wrt horizontal:

`x = r cos theta, qquad dx = dr cos theta - r sin theta \ d theta`

`y = r sin theta, qquad dy = dr sin theta + r cos theta \ d theta`

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

`= (r_theta sin theta + r cos theta)/( r_theta cos theta - r sin theta \ )`

Now: `r = e^theta, qquad r_theta = e^ theta = r`

`implies (dy)/(dx) = (tan theta + 1)/( 1 - tan theta \ )`

`(dy)/(dx) = 0 implies qquad tan theta = -1`

• `implies theta = - pi/4, (3 pi)/4, .....`

• `r = e^(- pi/4), e^((3 pi)/4), .....`

You can take this further, eg: `lim_(theta to pi/2) (tan(theta) + 1)/(1 - tan(theta)) = -1` is maybe worth exploring.

Answer 2

Depending upon the chosen convention for the domain of `theta`

`theta = -pi/4,pi/4 \ \` if `theta in [-pi,pi]`

Or:

`theta = pi/4, (3pi)/4 \ \` if `theta in [0,2pi]`

Explanation:

We have a polar equation:

`r = e^(theta)`

We have a horizontal tangent when `(dy)/(d theta) = 0`

Using the cartesian relationship `y=rsin theta` then:

`y= e^(theta)sin theta`

So differentiating wrt `theta`, and applying the product rule, we have:

`(dy)/(d theta) = e^(theta)((d)/(d theta) sin theta ) + ((d)/(d theta) e^(theta))sin theta`

`\ \ \ \ \ = e^(theta)cos theta + e^(theta)sin theta`

We require `theta` st `(dy)/(d theta) = 0`, thus:

`(dy)/(d theta) = 0 => e^(theta)cos theta + e^(theta)sin theta`

`:. e^(theta)(cos theta + sin theta ) = 0`

`:. cos theta + sin theta = 0 \ \ \` as `e^(theta) gt 0 AA theta in RR`

`:. sin theta =- cos theta`

`:. tan theta =- 1`

And depending upon the chosen convention for the domain of `theta`, we have:

`theta = -pi/4,pi/4 \ \` if `theta in [-pi,pi]`

Or:

`theta = pi/4, (3pi)/4 \ \` if `theta in [0,2pi]`