Question

Find the length of the curve. R(t) = 2 i + t2 j + t3 k, 0 ≤ t ≤ 1

Answer 1

curve equation is

$\\ \vec{R}\left ( t \right ) = 2\hat{i}+t^{2}\hat{j} + t^{3}\hat{k}$  ,0≤ t≤ 1

now taking the differentiation

$\\{R}'t = 2t\hat{i} + 3t^{2}\hat{j}$

now taking the modulus

$\left \| {R}'(t) \right \|=$$\sqrt{4t^{2} +9t^{4}}$

                                      = $\sqrt{4 + 9 t^{2} } .t$

now taking the integration

length of the curve =   $\\\int t\sqrt{4 + 9 t^{2}} dt$

now put the value v=  4 + 9t²

                              dv= 18 tdt

now put this value in the above equation

we get

length of the curve =$\\\frac{1}{18}\int \sqrt{v}dv$

now taking integation we get and put the value of the v

we get

= $\frac{1}{18}$× $\frac{2}{3}$×$(4 + 9t^{2} )^{\frac{3}{2} }$

= $\frac{1}{27} ( 4 + 9 t^{2} )^{\frac{3}{2} }$

now find out the length of the curve in the interval from 0 to 1.

length of the curve $= \frac{1}{27} (13^{\frac{3}{2}} -4^{\frac{3}{2}} )\\=\frac{1}{27} (13\sqrt{13} -8)$

Hence proved

Answer 2

Length of a curve is the length of its plot its curve. The length of the given curve for given range of t is: L = 1.44 units approx.

How to find the length of a curve?

If the curve has position vector p(x) for value of x ranging from x = a to x = b,

then, the curve's length is calculated as:

$L = \int_a^b ||p'(x)||dx$ units.

For the given case, we have:

Position vector =  $R(t) = 2\hat i + t^2 \hat j + t^3 \hat k$

Its differentiation gives:

$R'(t) = 2t\hat j + 3t^2\hat k$

Its non negative magnitude is: ||R'(t)|| = $\sqrt{(2t)^2 + (3t^2)^2} = t\sqrt{4+9t^2}$

Thus, as t ranges from a = 0 to b = 1, thus, length of the curve is:

$L = \int_0^1 (t\sqrt{4+9t^2})dt\\\\\text{Let v = 4+9}t^2, \text{then dv = 18tdt}\\and\\t=0\implies v = 4\\t=1 \implies v = 13\\Thus,\\L = \int_4^{13}(\dfrac{\sqrt{v}}{18})dv = \dfrac{1}{18} [\dfrac{2(v)^{3/2}}{3}]^{13}_4 \approx \dfrac{38.87}{27} \approx 1.44 \: \rm units$

Thus,

The length of the given curve for given range of t is: L = 1.44 units approx.

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