Arc Length Int. of r(t)=cos(t^2)+sin(t^2)+t^2 from 0 to sqrt(2pi)

[cdotter]

Homework Statement

$r(t)=cos(t^2)\hat{i}+sin(t^2)\hat{j}+t^2\hat{k}$

Compute the arc length integral from t=0 to $t=\sqrt{2 \pi}$

Homework Equations

Arclength = $\int_{a}^{b}||v(t)||\, dt$

The Attempt at a Solution

I did the following:

$$\\r'(t)=-2tsin(t^2)\hat{i}+2tcos(t^2)\hat{j}+2t\hat{k}\\$$
$$||r'(t)||=\sqrt{4t^2sin^2(t^2)+4t^2cos^2(t^2)+4t^2}$$
$$\int_{0}^{\sqrt{2 \pi}}\frac{r'(t)}{||r'(t)||} \, dt$$

I put the integral in Maple and it gave me 1.9 something if I remember correctly?

My professor has the answer as: [tex]\int_{0}^{\sqrt{2 \pi}}2t\sqrt{2} \, dt = 2 \pi \sqrt{2}[/itex]

Where did I go wrong?

 

[Mark44]

Homework Statement

$r(t)=cos(t^2)\hat{i}+sin(t^2)\hat{j}+t^2\hat{k}$

Compute the arc length integral from t=0 to $t=\sqrt{2 \pi}$

Homework Equations

Arclength = $\int_{a}^{b}||v(t)||\, dt$

The Attempt at a Solution

I did the following:

$$\\r'(t)=-2tsin(t^2)\hat{i}+2tcos(t^2)\hat{j}+2t\hat{k}\\$$
$$||r'(t)||=\sqrt{4t^2sin^2(t^2)+4t^2cos^2(t^2)+4t^2}$$
$$\int_{0}^{\sqrt{2 \pi}}\frac{r'(t)}{||r'(t)||} \, dt$$

The integral above doesn't make sense, geometrically. r'(t)/|r'(t)| is a unit tangent vector. When you integrate, you get a vector, not the scalar that you should get when you compute arc length.

Use the formula you showed in Relevant Equations - that's what you should be using.

Also, be sure to simplify $$||r'(t)||=\sqrt{4t^2sin^2(t^2)+4t^2cos^2(t^2)+4t^2}$$

I put the integral in Maple and it gave me 1.9 something if I remember correctly?

My professor has the answer as: [tex]\int_{0}^{\sqrt{2 \pi}}2t\sqrt{2} \, dt = 2 \pi \sqrt{2}[/itex]

Where did I go wrong?

 

[cdotter]

The integral above doesn't make sense, geometrically. r'(t)/|r'(t)| is a unit tangent vector. When you integrate, you get a vector, not the scalar that you should get when you compute arc length.

Use the formula you showed in Relevant Equations - that's what you should be using.

Also, be sure to simplify $$||r'(t)||=\sqrt{4t^2sin^2(t^2)+4t^2cos^2(t^2)+4t^2}$$

Wow I'm mixed up today.

 

[cdotter]

Got the answer. I'm not sure what the heck I was thinking earlier. Thank you for your help!