Regions bounded by a spiral: Let Rₙ be the region bounded by t...

Question

Regions bounded by a spiral: Let Rₙ be the region bounded by the nth turn and the (n+1)st turn of the spiral r = e⁻ᶿ in the first and second quadrants, for θ ≥ 0 (see figure).

c. Evaluate lim(n→∞) Aₙ₊₁/Aₙ.

Spiral graph of r = e^(-θ) showing shaded regions R₁ and R₂ between spiral turns in first and second quadrants.

Accepted Answer

Welcome back, everyone. In this problem for the spiral R equals e to the negative2 theta with theta greater than or equal to 0, let RN be the region between the nth and N + 1 turns in the first and the second quadrants. Find the limit as N approaches infinity of AN + 1 divided by AN. A says it's E to the 8 pi, B E to the -4, C E to the negative 2, and the D says it's 1. Now how are we going to figure out our limit of basically the consecutive areas? Well, first, let's try to figure out the area AN of our region RN. Recall that for N as an integer, then our area AN is going to be equal to a half of the integral between alpha and beta. Of R as a function of theta squared with respect to theta. Now if we think about it, we already have R as a function of theta. It's equal to e to the negative 2 theta, and our angular limits for the 1st and 2nd quadrants alpha, the first one would be 2 pi n. And beta. The second would be 2 pie and plus pie, because our angle theater is sorry, goes from 2 into 2 and plus pie. So now that we have all this information, we can use it to evaluate the integral to find AN, form our ratio, and then take our limit. So let's find A N. So now this means that AN is going to be equal to a half of our integral between 2 pi N and 2N plus pi of E to the negative2 theta R as a function of theta squared with respect to theta. Now, if we think about it here, we could simplify this to become a half of our integral. OK. Of the negative for theater, the theater. And no If we use the exponent exponential integral rule, since the integral of e to the power of a theater with respect to theater equals 1 divided by a of e to the a theater plus C. Then given here a over coefficient of theta equals -4, then this would mean. OK, that our area AN will be equal to a half of 25% of E to the -4 theta between our limits of 2 pi N and the 2N plus pi. Now, if we expand this a bit further, we'll have a 1/2 multiplied by 4th, OK. Which will be equal to 1/8 and if we leave negative e to the 4 theta within our limits and evaluate, that's going to be 1/8 of E to the negative 8 N minus e to the negative 8 N minus 4. I know when we simplify here, we're going to get AN to be equal to 1/8 of E to the negative Api N. Multiplied by 1 minus E to the -4 pi, OK. So this is our expression for A N. Thus we can use this to form our ratio. Because no, this means A N + 1 divided by AN is going to be equal to 1/8 of E to the negative 8 pi multiplied by N + 1. Multiplied by 1 minus e to the -4 pi. All divided by 1/8 of e to the negative 8 by n. Multiplied by 1 minus e to the negative 4 pi. Now if we simplify here 1 minus e to the -4 and 8 and e to the negative 8 n will all cancel and no that would simply leave us with e to the negative 8 pi. So if this is the ratio of AN + 1 to AN. Then our limit as N approaches infinity. Of a n + 1 divided by AN is really equal to the limit as n approaches infinity of e to the negative 8 pi. But e negative 8 pi is a constant. There's no n term in here, and the limit as N approaches infinity of a constant would simply be equal to that constant. Thus, our limit is. Our limit of our ratio of A N + 1 to AN would be equal to E to the negative 8pi. If we look back at our answer choices, that means that A. Is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Regions bounded by a spiral: Let Rₙ be the region bounded by the nth turn and the (n+1)st turn of the spiral r = e⁻ᶿ in the first and second quadrants, for θ ≥ 0 (see figure).
c. Evaluate lim(n→∞) Aₙ₊₁/Aₙ.

Key Concepts

Polar Coordinates and Area Calculation

Behavior of Exponential Functions and Limits

Sequences and Ratio of Areas

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