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Can anyone help me with this calculus problem?

Calculate the length of the path over the given interval:

$c(t)=(5t^2, 10t^3), 1≤t≤3$

I'm not sure how to figure out the equation to graph it.

Any help is appreciated!

Christina
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3 Answers3

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Set $x=5t^2,y=10t^3$. Solve for $x$ in terms of $t$ and substitute it into $y$. To find the length of the path, however, you can use the following formula: $$s=\int^b_a\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}dt$$

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Note: I initially evaluated the integral from $0$ to $3$; it has been amended.

What do we know about $V(D\alpha)$ where $\alpha(t) = (5t^2, 10t^3)$ is a parameterisation of a $1$-manifold in $\mathbb{R^2}$?

$V(D\alpha) = \sqrt{x'(t)^2 + y'(t)^2} = \sqrt{100t^2 + 900t^4} = 10t\sqrt{1 + 9t^2}$.

So we compute $$\int_{t=1}^{3} 10t\sqrt{1+9t^2} \\ = 5/9 \int_{t=1}^{3} 18t\sqrt{1+9t^2} dt \\ = 5/9 \int_{u=10}^{82} \sqrt{u} \\ = 5/9\cdot 2/3\cdot (82^{3/2} - 10^{3/2}) \\ = 10/27(82^{3/2} - 10^{3/2}).$$

Christopher K
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  • This seems like a homework question. Please do not provide the answer itself to a homework question. Rather, provide hints to which the OP may solve the question him/her self. –  Apr 30 '14 at 00:48
  • @SanathDevalapurkar, I thank you for your opinion. If an OP is asking many questions on a given day without showing any initiative, then I will not respond or at most provide hints. However, I usually give the OP the benefit of the doubt and will guide them through a problem as need be. It remains a point of contention in Math.SE over how much help should be given. I tend to find myself in the middle of the spectrum on how much help should be given. I try my best to discern the needs of the students and in this case I felt that an answer was reasonable. – Christopher K Apr 30 '14 at 00:55
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to find $x$ in terms of $y$, try finding $t^6$ in terms of both $x$ and $y$, then set them equal to each other. $$x=5t^2$$ $$x/5=t^2$$ $$x^3/5^3=t^6$$
$$y=10t^3$$ $$y/10=t^3$$ $$y^2/10^2=t^6$$
$$\therefore x^3/5^3=y^2/10^2=t^6$$ $$ y^2=4/5x^3$$ $$ y=\pm 2x\sqrt{x/5}$$

John Joy
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  • +1. How about a hint for the arc-length part of the question? –  Apr 30 '14 at 00:49
  • Why would anyone is their right mind do it this way when everything is parameterised so nicely? – Christopher K Apr 30 '14 at 00:49
  • @ChrisK You have a $\pm$ if you take the square root of $t$. –  Apr 30 '14 at 00:57
  • @SanathDevalapurkar, I was commenting on the overcomplication of the problem not its mathematical accuracy. – Christopher K Apr 30 '14 at 00:58
  • @Chris K - by solving for $y$ in terms of $x$ I can immediately see that the path is somewhere between a line and a parabola (this will aid me in graphing the path), but in actually calculating the length of the line I would keep the parametrization. – John Joy Apr 30 '14 at 15:17
  • @Sanath Devalapurkar - here is my hint $$\int_a^b ds$$. I have nothing further to add to either of your answers. :) – John Joy Apr 30 '14 at 15:24