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

Question

Charles Arocha · Accepted Answer

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

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.

Elijah Abram · Answer

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] #

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] #

Charles Arocha · Answer

undefined To find the points where the graph of the function r=e^theta has horizontal tangents, we need to determine the values of theta that satisfy the condition.

First, we find the derivative of the function with respect to theta, which is given by dr/dtheta = e^theta.

To find the points where the graph has horizontal tangents, we set the derivative equal to zero and solve for theta: e^theta = 0.

However, since e^theta is always positive, there are no values of theta that satisfy this equation. Therefore, the graph of the function r=e^theta does not have any points with horizontal tangents.