Fun With Calculus 5

For solving the arc length of a curve in three dimensions, we'll consider the vector notation.

Let $r(t)$ denote the set of points that satisfy the given curves, where $t$ represents the parametric for the curves.

Considering the first curve, if we put $y(t)= 3{t}^{2}$ , we get $x(t)= 3t$ , and putting these two values in the next curve, we get $z(t)=2{t}^{3}$ .

So, the vector $r(t)$ can be written as:

$r(t) = <x(t), y(t), z(t)>$

The arc length for such a curve is given by

${L}_{ab} = \int _{ a }^{ b }{ \sqrt { { [x'(t)] }^{ 2 }+{ [y'(t)] }^{ 2 }{ +[z'(t)] }^{ 2 } } } dt$

For our case, $x'(t) = 3, y'(t) = 6t, z'(t) = 6{t}^{2}$ and comparing the coordinates $(0,0,0)$ and $(3,3,2)$ with $r(t)$ , we get, $a = 0$ and $b = 1$ .

So, our length becomes:

$L =\int _{ 0 }^{ 1 }{ \sqrt { 9+36{ t }^{ 2 }{ +36 }t^{ 4 } } } dt$

which can be written as:

$L = \int _{ 0 }^{ 1 }{ 6{ t }^{ 2 }+3 } dt$

Now solving the integral and inserting limits, we get:

$L = 5 units$

$Let~~ x=t,~~~~y=\dfrac{t^2} 3,~~~~z=\dfrac{2t^3}{27}.\\ \therefore~x'=1,~~~~y'=\dfrac{2t} 3,~~~~z'=\dfrac{2t^2} 9. \\ \therefore~for~point(0,0,0),~t=0.~~~~~~~~~point(3,3,2),~t=3. \\ \displaystyle So~Arc~length=\int_0^3\sqrt{(x')^2+(y')^2+(z')^2}~ dt\\ \displaystyle ~~~~~~~~~~~=\int_0^3\sqrt{1+\dfrac{4t^2} 9+\dfrac{4t^2}{81}}~ dt\\ \displaystyle ~~~~~~=\int_0^3(1+\dfrac{2t^2} 9)~ dt \\ = \displaystyle t+\dfrac{2t^3}{27} |_0^3\\ = 3+2-0-0=\huge ~~\color{#D61F06}{5}.$