Question

Find the length of the curve: (posting via equation editor). i, j, k are unit vectors in the x,y,z direction.

Answer 1

\[\vec{r(t)} = t^2 \vec{i}+t^2\vec{j}+t^3\vec{k}\] from t=0 to t=1

Answer 2

I derived the function but after taking the absolute value of the vector I find it really quite difficult to integrate. \[\left| \frac{\vec{dr(t)}}{dt} \right|=\sqrt{8t^2+9t^4}\]

Answer 3

There's thus probably a simpler solution...

Answer 4

Very cute and quite wisening question . Understanding the 3D nature of this curve gives clarity.
Consider t as time, and the location of the point drawing the curve is parametrical function of time t.
One can immediately see that the i and the j components are equal. That means that the moving point projection on the (x,y) plane is ALL the time ON the diagonal straight line at 45 degrees with the x-axis. And the actual 3-D location of the point - above that straight line at HEIGHT z = t^3
Now let us in introduce a reparametrization
\[u = t^2 ...which...means...that....t^3 = u^{\frac{3}{2}}\]

Answer 5

Here we used the fact that since \[u=t^2\] it follows that\[du =\frac{1}{2} dx *x ===> dx = \frac{2du}{x} = \frac{2du}{\sqrt{u}}\]

Answer 6

Ahh and small correction to the result above - it must be\[dt = \frac{2u}{\sqrt{u}}\]

Answer 7

Introduce a new, more convenient coordinates axis, 1-st w is rotated by 45 deg coordinate axis W : along the straight line x=y, z= 0 or the line (x,x,0).
2-nd axis along the line orthogonal to W: (x,-x, 0) . And the z axis is kept as it is.

As pointed out from the outset, the curve is exactly above that line.