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Arclength Problem; stuck on integration

  • Thread starter hsbhsb
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  • #1
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Homework Statement


Find the arclength of $$r(t) = <{t}^2/2,{t}^3/3>$$ from t=-1 to t=1

Homework Equations



I have used this equation for arclength $$\int_{-1}^{1}|{r}'(t)|dt$$

The Attempt at a Solution



After integrating (using u substitution) I have the solution

$$\frac{1}{3}({t^2+1})^\frac{3}{2}$$ evaluated from t = -1 to t = 1. Of course, this yields 0! Which doesn't make any sense for arclength.

I know that the problem is somewhere in my integration step because in mathematica both the ArcLength[ $$\int_{-1}^{1}|{r}'(t)|dt$$] function and the Integrate[$$\sqrt{t^2+t^4}$$] give the same result $$\frac{2}{3}(-1+2\sqrt{2})$$, which seems to be the solution. What am I doing wrong?
 

Answers and Replies

  • #2
SteamKing
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Homework Helper
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Homework Statement


Find the arclength of $$r(t) = <{t}^2/2,{t}^3/3>$$ from t=-1 to t=1

Homework Equations



I have used this equation for arclength $$\int_{-1}^{1}|{r}'(t)|dt$$

The Attempt at a Solution



After integrating (using u substitution) I have the solution

$$\frac{1}{3}({t^2+1})^\frac{3}{2}$$ evaluated from t = -1 to t = 1. Of course, this yields 0! Which doesn't make any sense for arclength.

I know that the problem is somewhere in my integration step because in mathematica both the ArcLength[ $$\int_{-1}^{1}|{r}'(t)|dt$$] function and the Integrate[$$\sqrt{t^2+t^4}$$] give the same result $$\frac{2}{3}(-1+2\sqrt{2})$$, which seems to be the solution. What am I doing wrong?
You've made some kind of mistake in your original calculation, but you don't provide your work, so the details elude us.

I would go back and carefully check the integration, including any u-subs you made.
 
  • #3
ehild
Homework Helper
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$$\frac{1}{3}({t^2+1})^\frac{3}{2}$$ evaluated from t = -1 to t = 1. Of course, this yields 0! Which doesn't make any sense for arclength.
I quess you forgot that $$\sqrt{t^2+t^4}=|t|\sqrt{1+t^2}$$.
 
  • #4
Ray Vickson
Science Advisor
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Homework Statement


Find the arclength of $$r(t) = <{t}^2/2,{t}^3/3>$$ from t=-1 to t=1

Homework Equations



I have used this equation for arclength $$\int_{-1}^{1}|{r}'(t)|dt$$

The Attempt at a Solution



After integrating (using u substitution) I have the solution

$$\frac{1}{3}({t^2+1})^\frac{3}{2}$$ evaluated from t = -1 to t = 1. Of course, this yields 0! Which doesn't make any sense for arclength.

I know that the problem is somewhere in my integration step because in mathematica both the ArcLength[ $$\int_{-1}^{1}|{r}'(t)|dt$$] function and the Integrate[$$\sqrt{t^2+t^4}$$] give the same result $$\frac{2}{3}(-1+2\sqrt{2})$$, which seems to be the solution. What am I doing wrong?
You are making a standard error, but rather than telling you exactly what it is I will attempt to lead you do discover it for yourself---that will server you best in the long run.

Draw a graph of ##f(t) = \sqrt{t^2+t^4}## over ##-1 \leq t \leq 1##. Can you see the meaning of the integral ##\int_{-1}^1 f(t) \, dt##?

Homework Statement


Find the arclength of $$r(t) = <{t}^2/2,{t}^3/3>$$ from t=-1 to t=1

Homework Equations



I have used this equation for arclength $$\int_{-1}^{1}|{r}'(t)|dt$$

The Attempt at a Solution



After integrating (using u substitution) I have the solution

$$\frac{1}{3}({t^2+1})^\frac{3}{2}$$ evaluated from t = -1 to t = 1. Of course, this yields 0! Which doesn't make any sense for arclength.

I know that the problem is somewhere in my integration step because in mathematica both the ArcLength[ $$\int_{-1}^{1}|{r}'(t)|dt$$] function and the Integrate[$$\sqrt{t^2+t^4}$$] give the same result $$\frac{2}{3}(-1+2\sqrt{2})$$, which seems to be the solution. What am I doing wrong?
For an even function ##f(t)## we have ##\int_{-a}^a f(t) \, dt = 2 \int_0^a f(t) \, dt##. Using that is the best way to avoid elementary blunders.
 

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