2simple4limple

Calculus Level 5

An Archimedean spiral starting at the origin turns counterclockwise and has its first intersection with y = 0 y = 0 at x = − π x = -\pi . The spiral satisfies

x 2 + y 2 = f ( x y ) . x^{2}+y^{2} = f \left(\frac{x}{y} \right).

Find lim ⁡ z → 0 f ( z ) \displaystyle \lim_{z \rightarrow 0} f(z) .


The answer is 2.4674.

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Jake Lai by Jake Lai , 21, Singapore

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2 solutions

Kenny Lau
Oct 16, 2015

The equation of the spiral is r = θ r=\theta .

Therefore, the parametrized equation in cartesian plane is:

  • x = θ cos ⁡ θ x=\theta\cos\theta
  • y = θ sin ⁡ θ y=\theta\sin\theta

Then:

  • x 2 + y 2 = θ 2 ( cos ⁡ 2 θ + sin ⁡ 2 θ ) = θ 2 = ( arctan ⁡ y x ) 2 x^2+y^2=\theta^2(\cos^2\theta+\sin^2\theta)=\theta^2=\left(\arctan\frac yx\right)^2 .

So:

  • f ( z ) = ( arctan ⁡ 1 z ) 2 f(z)=\left(\arctan\frac1z\right)^2 .

Therefore its limit is ( arctan ⁡ ± ∞ ) 2 = ( ± π 2 ) 2 = π 2 4 (\arctan\pm\infty)^2=\left(\pm\frac\pi2\right)^2=\frac{\pi^2}4 .

6 Helpful 0 Interesting 0 Brilliant 0 Confused
Jake Lai
Jan 15, 2015

We recognise our Archimedean spiral as r = θ r = \theta . Using the usual change of coordinates, we can rewrite r = θ r = \theta as

x 2 + y 2 = tan ⁡ − 1 y x \sqrt{x^{2}+y^{2}} = \tan^{-1} \frac{y}{x}

Squaring both sides and equating with f ( x y ) f(\frac{x}{y}) , we see that

x 2 + y 2 = tan ⁡ − 2 y x = f ( x y ) x^{2}+y^{2} = \tan^{-2} \frac{y}{x} = f(\frac{x}{y})

Letting z = x y z = \frac{x}{y} , we can now rewrite the limit:

lim ⁡ z → 0 tan ⁡ − 2 1 z = π 2 4 ≈ 2.467 \lim_{z \rightarrow 0} \tan^{-2} \frac{1}{z} \ = \boxed{\frac{\pi^{2}}{4} \approx 2.467}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Since your equation just generates the spiral until the first intersection, I tried creating a function to infinitely generate the spiral. However, typing it in desmos would cause too much lag, so here is a more complete equation. I just generated until 10 π 10\pi cause it started to lag.

And this is a picture showing it generated until 60 π 60\pi :

hi you are looking at latex! hi you are looking at latex!

Equation of an infinite spiral:

lim ⁡ k → ∞ ∏ n = 0 k x 2 + y 2 − ( cot ⁡ − 1 ( x y ) + π ( n ) ) 2 = 0 \lim _{ k\rightarrow \infty }{ \prod _{ n=0 }^{ k } x^{ 2 }+y^{ 2 }-\left( \cot ^{ -1 } \left( \frac { x }{ y } \right) +\pi \left( n \right) \right) ^{ 2 }=0 }

Julian Poon - 6 years, 4 months ago

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Ah, that's nice :D

Jake Lai - 6 years, 4 months ago

The problem is somewhat vague, in the sense that Archimedian spirals of the form r = a θ + b r=a\theta+b also satisfy the conditions (the parameter b ∈ R b\in\mathbb{R} represents just rotations with respect to the origin) so there should be an extra condition implying b = 0 b=0 .

José Miguel Manzano - 6 years, 4 months ago

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r = a + b θ r = a+b\theta

x 2 + y 2 = a + b ∗ arctan ⁡ x y \sqrt {x^{2} + y^{2}} = a + b * \arctan \frac { x }{ y }

At (- π , 0 ) , π = a + b ∗ n ∗ π \pi, 0), \pi = a + b * n * \pi and then if we take a = 0 and b = n = 1, it then continues as the solution above. a should be given as 0.

Bhaskar Sukulbrahman - 6 years, 4 months ago

Hmm. How should I have phrased it without giving away r = θ r = \theta explicitly then? Thanks in advance.

Jake Lai - 6 years, 4 months ago

However, that is in polar coordinates, this question just states that but in the common x x and y y axis most people are familiar with.

Julian Poon - 6 years, 4 months ago

Who all over the world says that r = θ r=\theta must be fulfilled???

Andreas Wendler - 5 years, 4 months ago

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