FP2 Polar coordinates : tangents at the "cusp"

Find the coordinates of the points on

r=a(1+\cos \theta)

where the tangents are parallel to the initial line

\theta = 0

.

The textbook working says that the tangent at

(0,\pi)

is parallel to the initial line since

\frac{dy}{d\theta} = 0

.

But at this point

\frac{dx}{d\theta}

is also zero. From the graph it seems that the tangent is parallel and not perpendicular to the initial line. But I also watched a video that said this kind of cusp point doesn't have a vertical or horizontal tangent line. So who is right? And is there a way to determine whether the tangent is vertical or horizontal at this kind of point without drawing the graph?

Thanks.

Original post by 0-)

Find the coordinates of the points on

r=a(1+\cos \theta)

where the tangents are parallel to the initial line

\theta = 0

.

The textbook working says that the tangent at

(0,\pi)

is parallel to the initial line since

\frac{dy}{d\theta} = 0

.

But at this point

\frac{dx}{d\theta}

is also zero. From the graph it seems that the tangent is parallel and not perpendicular to the initial line. But I also watched a video that said this kind of cusp point doesn't have a vertical or horizontal tangent line. So who is right? And is there a way to determine whether the tangent is vertical or horizontal at this kind of point without drawing the graph?

Thanks.

At

(0,\pi)

the gradient is undetermined therefore you cannot say that it is parallel to anything, though you can say that the tangent line tends to the gradient of the initial line as the point of tangency approaches the cusp - ie the point

(0,\pi)

Reply 2

0-)

OP

11

Original post by RDKGames

At

(0,\pi)

the gradient is undetermined therefore you cannot say that it is parallel to anything, though you can say that the tangent line tends to the gradient of the initial line as the point of tangency approaches the cusp - ie the point

(0,\pi)

Thank you.

Here's the question and working from the textbook:

And later on there's this question that contains a comment:

Is this all wrong then? I was wondering what they meant by "further work" although if you say the gradient is undetermined then there wouldn't need to be further work.

(edited 6 years ago)

Reply 3

0-)

OP

11

Bump. Can I just get confirmation that the book is wrong here? (See above)

Original post by 0-)

Bump. Can I just get confirmation that the book is wrong here? (See above)

To be honest, I have never come across this situation when working in polar coordinates so I cannot say for certain which is right and which is wrong, however to me it's just not feasible that there is a tangent at

\theta=\pi

due to the gradient being

\frac{0}{0}

. Then again, there are different ways that I think about it and it makes sense from those perspectives; one of which is that

r=p+q\cos(\theta)

will always have a vertical tangent at

\theta=\pi

but as

p \rightarrow q

we notice that at this point we also obtain a horizontal tangent which is what we would be looking for, so at

p=q

we have this tangent, thus it exists.

I cannot explain properly/rigorously why this is the case, but perhaps someone else can, so if the book says there is a tangent, go along with it for now.

By "further work" they mean sketching out the curve and observing whether you'd get a tangent satisfying the requirement at the given point.

Reply 5

Physics Enemy

19

Original post by 0-)

Bump. Can I just get confirmation that the book is wrong here? (See above)

You have to be slightly wary when converting between polar and cartesian, it can bring about problems, in this case at @ = Pi.

Indeterminate is just that, but you can take limits to find the result. I'd expect as @ -> Pi, dy/d@ -> 0 faster than dx/d@, so divide these for dy/dx and result -> 0.

If you know/have a rough sketch and can tell dx/d@ =/= 0 everywhere, just solve for dy/d@.

(edited 6 years ago)

Reply 6

InOrbit

9

A bit of symbol pushing leads to the parametric equation:

f(t) = (x(t),y(t)) = \left(\left(1+\cos t\ \right)\cos t,\left(1+\cos t\ \right)\sin t\right); 0 \leq t \leq 2\pi.

A sufficient condition for a tangent line to have gradient 0 (i.e. horizontal tangent) is for the derivative

\frac{dy}{dx} = \frac{dy}{dt}/\frac{dx}{dt}

to be zero.

In differential geometry, there is this notion of a 'regular parametrisation, where the 'velocity' of a smooth curve is never zero, or

f'(t) \neq (0,0)

for all

t

. (There's also a condition of an open mapping, but whatever). This is nice because a regular parametrisation gives a curve that 'looks' like a smooth line (called a 1d submanifold immersed in the plane), so there are no cusps, intersections, discontinuities or singularities. Consequently, we can just check the derivative for points with a tangent with our desired gradient.

The parametrisation we have is not regular, so there are points with zero derivative or velocity. These are critical points, so in order to determine the tangents with our gradient, we have to check these too.

(edited 6 years ago)

Reply 7

Dr. Green

Since dy/dx is of form 0/0, the values is indeterminate. Further work refers to looking at 1-sided derivatives (tangent slopes). It turns out that both 1-sided derivatives are 0, so the secant slope from the left and the right both approach 0 causing lim_x->0 dy/dx =0. Thus we have a horizontal tangent here too.

FP2 Polar coordinates : tangents at the "cusp"

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