Calculate Arclength of Parametric Curve

[mill]

Homework Statement

Calculate the arclength of the curve given parametrically by

$x=2t^2, y=\frac 8 5 \sqrt 3t^ \frac 5 2, z=2t^3$
for 0≤t≤2

Homework Equations

$S=∫ \sqrt(dx^2 + dy^2 + dz^2)$

The Attempt at a Solution

1. Found derivative of each and input into equation.
$S=∫\sqrt((4t)^2 + (4\sqrt 3 t^ \frac 3 2 )^2 + (6t^2)^2) dt$
2. $S=∫\sqrt(16t^2 + 48t^3 + 36t^4) dt$
3. $S=∫2t \sqrt(4 + 12t + 9t^2) dt$
4. Used complete the square

$=∫2t (t+ \frac 3 2 )dt$

5. Integrated.
$[\frac 2 3 t^3 + \frac 3 2 t^2]$

Which evaluated becomes 34/3. The right answer is 24. I think I have the steps correct. Where did I go wrong?

 

[gopher_p]

Homework Statement

Calculate the arclength of the curve given parametrically by

$x=2t^2, y=\frac 8 5 \sqrt 3t^ \frac 5 2, z=2t^3$
for 0≤t≤2

Homework Equations

$S=∫ \sqrt(x'^2 + y'^2 + z'^2)$

The Attempt at a Solution

1. Found derivative of each and input into equation.
2. $S=∫\sqrt(16t^2 + 48t^3 + 36t^4) dt$
3. $S=∫2t \sqrt(4 + 12t + 9t^2) dt$
4. Used complete the square

=∫2t (t+ (3/2))dt

5. Integrated.
$[\frac 2 3 t^3 + \frac 3 2 t^2]$

Which evaluated becomes 34/3. The right answer is 24. I think I have the steps correct. Where did I go wrong?

You don't need to complete the square; $4+12t+9t^2=(2+3t)^2$ is already a perfect square.

Edit: Technically, $16t^2 + 48t^3 + 36t^4=(4t+6t^2)^2$ is also a perfect square. So you may have been able to omit a few steps if you had been on the lookout for the to happen.

 

[mill]

You don't need to complete the square; $4+12t+9t^2=(2+3t)^2$ is already a perfect square.

Edit: Technically, $16t^2 + 48t^3 + 36t^4=(4t+6t^2)^2$ is also a perfect square. So you may have been able to omit a few steps if you had been on the lookout for the to happen.

Oh, I see. Thanks. I will keep the squares in mind. Although would ^ apply in cases where

$x^4 + x^3 + x$?

Also, if you have any insight into the original problem's answer, that would be awesome.

 

[gopher_p]

Oh, I see. Thanks. I will keep the squares in mind.

It's a good thing to be aware of in these "math class" problems (as opposed to real-world problems) involving arc lengths and surface areas. After all, the problem has to be doable for the math students. Unfortunately the real world isn't so nice and tidy and prearranged for doability, and you get things like Elliptic Integrals (https://en.wikipedia.org/wiki/Elliptic_integral) which aren't doable by elementary means.

Basically what I'm trying to say is that being on the lookout for perfect squares falls under the banner of "strategies for math students trying to solve standard problems in a math class" and isn't useful so much otherwise.

Although would ^ apply in cases where

$x^4 + x^3 + x$?

$x^4 + x^3 + x$ is not a perfect square. One easy way to see that is to (1) note that squares of real numbers are always nonnegative, which means that "perfect-square polynomials" must be non-negative, and (2) recognize that $(-1)^4+(-1)^3+(-1)=-1<0$. So that can't be a perfect square.

Another less clever but more algorithmic way to see that $x^4 + x^3 + x$ is not a perfect square is to see that, if it were, it would have to look something like $x^4 + x^3 + x=(x^2+bx+c)^2$ because it is a degree $4$ polynomial. If you were to FOIL out the RHS, the constant term would be $c^2$. Since $x^4 + x^3 + x$ has no constant term (or rather it's constant term is $0$), $c=0$. Then you're left with (possibly) $x^4 + x^3 + x=(x^2+bx)^2$, but there is no way to get an $x$ term from the RHS.

Also, if you have any insight into the original problem's answer, that would be awesome.

I'm not sure what you're asking me to do here. You seem to have access to an answers sheet which says that the arc length is $24$. Given that the curve starts at $(0,0,0)$ and ends near $(8,16,16)$, and seeing as how the straight line between $(0,0,0)$ and $(8,16,16)$ has length $24$, I'd say $24$ is a good approximation to a lower bound at least.

 

[mill]

I'm not sure what you're asking me to do here. You seem to have access to an answers sheet which says that the arc length is $24$. Given that the curve starts at $(0,0,0)$ and ends near $(8,16,16)$, and seeing as how the straight line between $(0,0,0)$ and $(8,16,16)$ has length $24$, I'd say $24$ is a good approximation to a lower bound at least.

Sorry. I was asking if there is anything obviously wrong about my method of solving this. Evaluating, I am not getting 24. I am getting 34/3. Yes, I have access to the answers sheet but no solutions.

 

[scurty]

Sorry. I was asking if there is anything obviously wrong about my method of solving this. Evaluating, I am not getting 24. I am getting 34/3. Yes, I have access to the answers sheet but no solutions.

You completed the square wrong. You need to factor out a 9 before completing the square (x^2 coefficient has to be 1). Also, the fraction should have been 2/3, not 3/2.

 

[mill]

You completed the square wrong. You need to factor out a 9 before completing the square (x^2 coefficient has to be 1). Also, the fraction should have been 2/3, not 3/2.

I did factor out the 9, but I flipped the 2/3 by accident.

Still

$=∫2t (t+ \frac 2 3 )dt$
$=∫2t^2 + \frac 4 3 t)dt$
$=\frac 2 3 t^3 + \frac 2 3 t^2$

Evaluated at 2 is 8, not 24. Not sure at which step I went wrong.

 

[mill]

Thanks. Got it. Using the (4t+6t^2) got to 24, so something in the complete the square must have went wrong.