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We are given:

$x=49-t^2$

$y=t^3-16t$

The curve apparently makes a loop which lies along the x-axis. I need help finding total area inside the loop. I don't know where to even start.

If it helps, in previous parts of the question, I found that

(a) the tangent line is horizontal at $t=\sqrt{16/3}$ and $x = 43.6666666666667$

(b) the tangent line is vertical at $t=0$

Thank you!

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I think this is a Green's theorem problem. The loop is traced as $t$ goes from $-4$ to $4$ (but clockwise.) So the area is

$$-\frac{1}{2} \int_{-4}^4 x \; dy - y \; dx $$ $$ = -\frac{1}{2} \int_{-4}^{4} (49-t^2)(3t^2-16)-(t^3-16t)(-2t) \; dt $$ $$= \frac{8192}{15} = 546.13\ldots.$$

The extra minus sign is because of the clockwise orientation.

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Have you plotted it? I don't find a loop. I find an arc below the $x$ axis from $t=0$ to $t=4$. This is from $x=33$ to $x=49$. Maybe you are supposed to find the area between the $x$ axis and the curve in this region. Here is my plot enter image description here
$t=0$ is the right hand end, at $(49,0).\ \ t=4$ is the point where it crosses the axis at $(33,0)$. If you are to find the area between the curve and the $x$ axis you can just solve the $x$ equation for $t$ and plug into the $y$ equation $$x-49-t^2\\t=\sqrt{49-x}\\y=(49-x)^{3/2}-16(49-x)^{1/2}$$
and you can integrate the last from $x=49$ to $x=33$

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  • $\begingroup$ I understand what you've done and I've tried doing it this way. However, I've failed to come up with the correct answer (I got -273.0666). Thank you for trying though! $\endgroup$ – CodingMee Mar 20 at 3:13
  • $\begingroup$ I believe the answer should be positive, but that is the answer Alpha gets. Why do you think it is wrong? $\endgroup$ – Ross Millikan Mar 20 at 3:35
  • $\begingroup$ I tried both negative and positive just to be sure, and they both come up incorrect. I think it is wrong because my assignment is online and automatically checks our answers. $\endgroup$ – CodingMee Mar 20 at 3:40
  • $\begingroup$ The answer key could be wrong, it could be expecting the exact answer of $\frac {4096}{15}$, or I could be interpreting the problem wrong. $\endgroup$ – Ross Millikan Mar 20 at 3:44
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Making a parametric plot, there is effectively a loop which is symmetric with respect to the $x$ axis; the points where the curve intersect the axis correspond to $x=33$ and $x=49$ as @Ross Millikan already answered.

The major issue is to compute $$I=\int \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt=\int\sqrt{4 t^2+\left(3 t^2-16\right)^2}\,dt$$ which would lead to nasty elliptic integrals.

So, the simplest is to do what @Ross Millikan already answered, that is to say $$t=\pm \sqrt{49-x} \implies y=\pm (x-33)\sqrt{49-x} $$ So, the total area enclosed by the loop is $$A=2\int_{33}^{49}(x-33)\sqrt{49-x}\,dx$$ Using $$\int(x-33)\sqrt{49-x}\,dx=-\frac{2}{15} (49-x)^{3/2} (3 x-67)$$ we end with $A=2 \times \frac{4096}{15}$.

Notice that $\frac{4096}{15}\approx 273.067$

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