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Let $f(x,y)=2x-y$, and consider the path $x=t^4, y=t^4,-1 \leq t \leq 1$.

(a) Compute the integral of $f$ along the path. What does this mean geometrically?

My attempt:

Let $\textbf c(t) = (t^4,t^4)$. We wish to integrate $f(x,y)=2x-y$ over this path:

$$\int_{\textbf c}f\space ds = \int_a^bf\big(x(t),y(t)\big)\|\textbf c'(t)\| \space dt.$$

$$\int_{-1}^1 (2t^4-t^4)\sqrt{32t^6} \space dt = \sqrt{32}\int_{-1}^1 t^7 \space dt = 0.$$

But this is not the correct answer. I think I'm misinterpreting what $x=y=t^4$ means.

(b) Evaluate the arc-length function $s(t)$ and redo part (a) in terms of $s$.

What does the question mean by "redo (a) in terms of $s$"?

Integral
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Bobby Lee
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  • for (a), I think you made a mistake when evaluating |c'|=sqrt(32t^6) |c'|= sqrt(32)t^3 if t>0 and |c'|= -sqrt(32)t^3 if t<0 right? for (b), I have trouble understanding it too –  Apr 09 '14 at 03:42

1 Answers1

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Fisrt you write $$\int_{-1}^1 (2t^4-t^4)\sqrt{32t^6} \ dt = \sqrt{32}\int_{-1}^1 t^4\sqrt{t^6} \ dt = \sqrt{32}\Big(\int_{-1}^0 t^4\sqrt{t^6} \ dt + \int_0^1 t^4\sqrt{t^6} \ dt\Big) $$

Now you need to realize that $\sqrt{t^6} = -t^3$ when $t\in[-1,0]$ and $\sqrt{t^6} = t^3$ when $t\in[0,1]$. Therefore $$\sqrt{32}\Big(\int_{-1}^0 t^4\sqrt{t^6} \ dt + \int_0^1 t^4\sqrt{t^6} \ dt\Big) = \sqrt{32}\Big(\int_{-1}^0 -t^7 \ dt + \int_0^1 t^7 \ dt\Big) = \sqrt{32}\Big( \frac{-t^8}{8}\Big|_{-1}^0 + \frac{t^8}{8}\Big|_0^1 \Big) = \sqrt{32}\Big(\frac{1}{8}+\frac{1}{8}\Big) = \sqrt{2}$$

For part b, probably they are talking about $s(t) = \int_{t_0}^t\|c'(t)\| \ dt$.

Integral
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