Winston T.

asked • 09/11/19

Find the points on the given curve where the tangent line is horizontal or vertical

Assume (0≤θ≤2π). Enter your answers as a comma-separated list of ordered pairs.)


r=e^(θ)





horizontal tangent     (rθ) = 
 
vertical tangent     (rθ) =


Apparently I keep getting this answer wrong.


Patrick B.

The derivative is never zero
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09/11/19

1 Expert Answer

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Doug C. answered • 09/11/19

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Winston T.

I'm still confused on how you got that answer, considering how the question asked just for 1 tangent specifically for each "(r,θ)" yet you have 2 for each. In my work, I tried having both horizontal dy/dx=0 => x+y=0 => θ =arctan(-1) =n(pi) - (pi)/4 where n is any integer. As 0≤θ≤2π, n=1,2 and for vertical tangent dy/dx=0 and dy/dx=0 => x+y=0 => θ =arctan(-1) =m(pi) - (pi)/4 where m is any integer. As 0≤θ≤2π, m=0,1. However this is consider wrong
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09/11/19

Doug C.

The question is what did you get for dy/dx (in terms of theta)? I got: dy/dx = (costheta + sintheta) / (cos theta - sin theta). In that case dy/dx = 0 when the numerator = 0, or when cos theta = -sin theta). That happens in the interval from 0 to 2pi when theta = 3pi/4 or 7pi/4. The vertical tangents happen when dy/dx has denominator = 0, or when costheta = sintheta. That happens at theta = pi/4 or 5pi/4. Are you entering answers is some online tool? If yes, did you try entering the answers I provided?
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09/11/19

Doug C.

Try attaching the following to the URL for the Desmos online graphing calculator: "calculator/05xnebgnob" (www.desmos.com/). You will see a graph of r = e^theta along with the points where vertical and horizontal tangents are located.
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09/11/19

Doug C.

Or: www.desmos.com/calculator/eufkkbrbld shows the tangent lines
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09/11/19

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