Question

How do you find the length of the curve $x = 1 + 3{t^2}$, $y = 4 + 2{t^3}$, where $0 \leqslant t \leqslant 1$?

Answer 1

Length of a curve is given by $\int\limits_a^b {\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} } dx$. So, we will find $\dfrac{{dy}}{{dx}}$ by finding $\dfrac{{dy}}{{dt}}$ and $\dfrac{{dx}}{{dt}}$ and then dividing them. Then we will substitute $dx = 6tdt$. Then we will put all these values in the formula to find the length of the curve and we will integrate by making proper substitution. Then we will simplify it to find the result.

Complete step by step answer:
We have $x = 1 + 3{t^2}$ and $y = 4 + 2{t^3}$.
Differentiating $x$ with respect to $t$, we get
$\Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {1 + 3{t^2}} \right)$
$\Rightarrow \dfrac{{dx}}{{dt}} = 6t - - - (1)$
Differentiating $y$ with respect to $t$, we get
$\Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{d}{{dx}}\left( {4 + 2{t^3}} \right)$
$\Rightarrow \dfrac{{dy}}{{dt}} = 6{t^2} - - - (2)$
Dividing $(2)$ and $(1)$, we get
$\Rightarrow \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}} = \dfrac{{6{t^2}}}{{6t}}$

On simplification, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = t - - - (3)$
Also, from $(1)$, we can write
$\Rightarrow dx = 6tdt - - - (4)$
As we know that, ${\text{Arc Length}} = \int\limits_a^b {\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} } dx - - - (5)$
Given, $0 \leqslant t \leqslant 1$. Putting the values from $(3)$ and $(4)$ in $(5)$, we get
$\Rightarrow {\text{Arc Length}} = \int\limits_0^1 {\left( {\sqrt {1 + {{\left( t \right)}^2}} } \right)} \times 6tdt$
On rewriting, we get
$\Rightarrow {\text{Arc Length}} = 3 \times \int\limits_0^1 {2t\left( {\sqrt {1 + {t^2}} } \right)} dt - - - (6)$

Now, let $I = \int {2t\left( {\sqrt {1 + {t^2}} } \right)dt}$
Let $1 + {t^2} = u$
Differentiating w.r.t $t$, we get
$\Rightarrow 2t = \dfrac{{du}}{{dt}}$
$\Rightarrow dt = \dfrac{{du}}{{2t}}$
Therefore, we get
$I = \int {2t \times \sqrt u \times \dfrac{{du}}{{2t}}}$
Cancelling the common terms and on rewriting, we get
$I = \int {{{\left( u \right)}^{\dfrac{1}{2}}}du}$
On integrating, we get
$\Rightarrow I = \int {\dfrac{{{{\left( u \right)}^{\dfrac{1}{2} + 1}}}}{{\left( {\dfrac{1}{2} + 1} \right)}}du}$

On simplification, we get
$\Rightarrow I = \dfrac{2}{3}{\left( u \right)^{\dfrac{3}{2}}}$
Substituting back $1 + {t^2} = u$, we get
$\Rightarrow I = \dfrac{2}{3}{\left( {1 + {t^2}} \right)^{\dfrac{3}{2}}}$
So, $\int\limits_0^1 {2t\left( {\sqrt {1 + {t^2}} } \right)} dt = \left[ {\dfrac{2}{3}{{\left( {1 + {t^2}} \right)}^{\dfrac{3}{2}}}} \right]_0^1$
Putting the upper and lower limits, we get
$\Rightarrow \int\limits_0^1 {2t\left( {\sqrt {1 + {t^2}} } \right)} dt = \left[ {\dfrac{2}{3}{{\left( {1 + {{\left( 1 \right)}^2}} \right)}^{\dfrac{3}{2}}} - \dfrac{2}{3}{{\left( {1 + {{\left( 0 \right)}^2}} \right)}^{\dfrac{3}{2}}}} \right]$

On simplification, we get
$\Rightarrow \int\limits_0^1 {2t\left( {\sqrt {1 + {t^2}} } \right)} dt = \left[ {\dfrac{2}{3}{{\left( 2 \right)}^{\dfrac{3}{2}}} - \dfrac{2}{3}{{\left( 1 \right)}^{\dfrac{3}{2}}}} \right]$
On further simplification, we get
$\Rightarrow \int\limits_0^1 {2t\left( {\sqrt {1 + {t^2}} } \right)} dt = \dfrac{2}{3}\left( {{2^{\dfrac{3}{2}}} - 1} \right) - - - (7)$
Putting $(7)$ in $(6)$, we get
$\Rightarrow {\text{Arc Length}} = 3 \times \dfrac{2}{3}\left( {{2^{\dfrac{3}{2}}} - 1} \right)$
Cancelling the common terms, we get
$\therefore {\text{Arc Length}} = 2\left( {{2^{\dfrac{3}{2}}} - 1} \right)$

Therefore, the length of the curve $x = 1 + 3{t^2}$, $y = 4 + 2{t^3}$ is $2\left( {{2^{\dfrac{3}{2}}} - 1} \right)$.

Note: If we want to calculate the length of any curve given by $f(x)$ in an interval $a$ to $b$, then $f(x)$ must be continuous in the interval $a$ to $b$. Also, we require $f(x)$ to be differentiable and its derivative is required as we are integrating an expression involving $f'(x)$.