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: \int_{0}^{\sqrt{2 \pi}}2t\sqrt{2} \, dt = 2 \pi \sqrt{2}[/itex]<br /> <br /> Where did I go wrong? <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f615.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":confused:" title="Confused :confused:" data-smilie="5"data-shortname=":confused:" />

 

[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&#039;(t)=-2tsin(t^2)\hat{i}+2tcos(t^2)\hat{j}+2t\hat{k}\\
||r&#039;(t)||=\sqrt{4t^2sin^2(t^2)+4t^2cos^2(t^2)+4t^2}
\int_{0}^{\sqrt{2 \pi}}\frac{r&#039;(t)}{||r&#039;(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&#039;(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: \int_{0}^{\sqrt{2 \pi}}2t\sqrt{2} \, dt = 2 \pi \sqrt{2}[/itex]<br /> <br /> Where did I go wrong? <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f615.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":confused:" title="Confused :confused:" data-smilie="5"data-shortname=":confused:" />

 

[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&#039;(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!