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Polar coordinates FP2 - Perpendicular to the initial line

I'm trying to understand why θ = pi is not a valid solution. I understand that if there was a tangent at θ = pi, it would intersect the curve twice but that is only if the tangent was long enough. That brings me onto my question, are tangents defined to be infinite in length or are they finite?

Also, the other solutions for θ = 2pi/3, -2pi/3. The latter is equivalent to 4pi/3 (anti-clockwise direction) so would I gain the marks if I quoted 4pi/3 rather than -2pi/3 as an answer?



(edited 10 years ago)
Original post by GPODT
I'm trying to understand why θ = pi is not a valid solution. I understand that if there was a tangent at θ = pi, it would intersect the curve twice but that is only if the tangent was long enough. That brings me onto my question, are tangents defined to be infinite in length or are they finite?

Also, the other solutions for θ = 2pi/3, -2pi/3. The latter is equivalent to 4pi/3 (anti-clockwise direction) so would I gain the marks if I quoted 4pi/3 rather than -2pi/3 as an answer?





because theta = pi is not perpendicular to the initial line as the question asks
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Smaug123
15
Original post by GPODT
I'm trying to understand why θ = pi is not a valid solution. I understand that if there was a tangent at θ = pi, it would intersect the curve twice but that is only if the tangent was long enough. That brings me onto my question, are tangents defined to be infinite in length or are they finite?

I didn't read the answer, but "tangent at a point" is a local property. You take the tangent line to be "short enough" - as long as it is some finite length (that is, not zero). For instance, the graph of y=x^2 on [-1,1] and y=|x| otherwise, that's got a tangent at x=3/4 even though the tangent line intersects with the graph later on.
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GPODT
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Original post by Ilovemaths96
because theta = pi is not perpendicular to the initial line as the question asks


Neither is theta = 0 though ? Theta = pi is just a coordinate (think so anyway).
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GPODT
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Original post by Smaug123
I didn't read the answer, but "tangent at a point" is a local property. You take the tangent line to be "short enough" - as long as it is some finite length (that is, not zero). For instance, the graph of y=x^2 on [-1,1] and y=|x| otherwise, that's got a tangent at x=3/4 even though the tangent line intersects with the graph later on.


If a tangent is finite, then I'm unsure why theta = pi is not a solution.
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Smaug123
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Original post by GPODT
If a tangent is finite, then I'm unsure why theta = pi is not a solution.

Theta = pi corresponds to r=0. There, the function isn't differentiable (it looks roughly the same as |x| does, just on its side), so there's no good way to define the tangent. After all, since the gradient goes from positive to negative on either side of that cusp, you'd expect it to be 0 somewhere; by symmetry, it can only be zero at the cusp itself; so the tangent must be y=0. But if we were to walk around the curve, our direction would point due upwards as we walked over the cusp, and that's definitely not y=0. There are two perfectly plausible derivatives here.
Original post by GPODT
Neither is theta = 0 though ? Theta = pi is just a coordinate (think so anyway).


thete = pi is a half line
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Smaug123
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Original post by Ilovemaths96
thete = pi is a half line

Yes, but it yields a coordinate when plugged into the equation.

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