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Consider the Limacon: $\gamma(t)=((1+3cost)cost, (1+3cost)sint)$.

(i) Compute $A(\gamma)=\frac{1}{2}\int_\gamma (x\frac{dy}{dt}-y\frac{dx}{dt})dt$.

(ii) Determine the osculating circle $C$ at $(4,0)$.

(iii) Show that $\gamma$ lies entirely inside $C$.

(iv) Compare $A(C)$ and $A(\gamma)$ and explain what you find.

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For question (i) I've found the derivatives of $y=(1+3cost)sint)$ and $x=((1+3cost)cost$ respectively. Then plugged them into $A(\gamma)=\frac{1}{2}\int_\gamma (x\frac{dy}{dt}-y\frac{dx}{dt})dt$ and calculated the integral. But what makes me doubt is that I assumed $t∈[0,2\pi]$ to sort out the finite integral, which gave a result of $\frac{11}{2}\pi$.

For question (ii), the point $(4,0)$ gives $t=0$. Then I used $\kappa(0)=||\gamma''(0)||=7$ to get the curvature and $R=\frac{1}{\kappa(0)}=\frac{1}{7}$ for the radius of the osculating circle. The unit tangent vector is $T(0)=\frac{\gamma'(0)}{||\gamma'(0||}=(0,1)$ and the unit normal vector is $N(0)=(-1,0)$ by rotating $T$ for $\frac{\pi}{2}$ counter clockwisely. Then the centre of the circle is $\gamma_c(0)=\gamma(0)+RN(0)=(\frac{27}{7},0)$. Therefore the osculating circle is $(x-\frac{27}{7})^2+y^2=\frac{1}{49}$.

However, question (iii) asks to show that the curve $\gamma$ lies entirely inside C. From my results in (i) and (ii), $A(\gamma)=\frac{11}{2}\pi$ while $A(C)=\frac{\pi}{49}$, $A(\gamma)>A(C)$, which makes $\gamma$ less possible to lie entirely inside $C$. Also, I can find nothing from these results to answer question (iv).

Could someone please help me find out where I've gone wrong in (i) and (ii), and thus how to proof the statement in (iii)? Thanks heaps!!

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The graph of the limacon is here

enter image description here

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The question was sloppy not to specify the domain of the parametrization. You are correct to take $t\in [0,2\pi]$. You have the correct value for $A(\gamma)$. On the other hand, you have to be careful about interpreting it, since the curve intersects itself.

Your serious mistake, proceeding to (ii), is that $\gamma$ is not an arclength (speed one) parametrization, so your formula for curvature needs to be corrected by using the chain rule. (You can find lots of examples of that on posts here.) Once you do that, you should get an answer that makes sense.

Note that to check (iii), you're going to have to show that the distance from the center of the osculating circle to points on the curve is always (except for the single point $(4,0)$) less than the radius. The trigonometry (if you consider the square of radius minus the square of the distance) simplifies nicely.

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  • $\begingroup$ Thanks for your answer. I tried $\kappa= \frac{||\gamma'' \times \gamma'||}{||\gamma'||^3}= \frac{7}{16}$, which gave the radius $R=\frac{16}{7}$. This still seems weird with the area.. $\endgroup$ – Evelyn Venne Aug 14 '18 at 23:36
  • $\begingroup$ This is correct. So the area of the osculating circle is just under $21\pi/4$, which is indeed less than $11\pi/2$. Well, have you sketched the curve and interpreted what $A(\gamma)$ actually means? $\endgroup$ – Ted Shifrin Aug 15 '18 at 0:02
  • $\begingroup$ Yea I attached the graph to the question. $A(\gamma)$ = outer loops + 2*inner loops, is it? $\endgroup$ – Evelyn Venne Aug 15 '18 at 0:32
  • $\begingroup$ Yes, good. Because as $t$ increases, you go around both loops counterclockwise. $\endgroup$ – Ted Shifrin Aug 15 '18 at 1:09

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