Question

In Exercises 49-53 sketch the parametric curve and find its length.

$x=1+t^2,y=3+2t^3,t\in [0,1]$

Short answer

The graphical representation of the points $(1,3)(2,5)$is as follows,

The length of the curve is equals to

$\frac{2}{27}\left[{10}^{\frac{3}{2}}-1\right]$

Step-by-step solution

Given information

The parametric curve is $x=1+t^2,y=3+2t^3,t\in [0,1]$

Calculation

Consider the parametric equations $x=1+t^2,y=3+2t^3,t\in [0,1]$.

The objective is to draw the parametric curve and find the arc length of the curve.

The formula to find the arc length of the curve is

Length of the curve $={\int }_a^b\sqrt{{\left(f^{\text{'}}\left(t\right)\right)}^2+{\left(g^{\text{'}}\left(t\right)\right)}^2}dt$

First find the derivative of the parametric equations$x=1+t^2,y=3+2t^3$.

Take $x=1+t^2$

That is $f\left(x\right)=x=1+t^2$

The derivative with respect to t is written as follows,

$f^{\text{'}}\left(x\right)=\frac{d}{dt}\left(1+t^2\right)$

On simplifying further,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dt}1+\frac{d}{dt}{t}^2 \\ {f}^{\text{'}}\left(x\right)=0+2t \\ {f}^{\text{'}}\left(x\right)=2t \end{gathered}$$

Take $y=3+2t^3$

That is $g\left(y\right)=y=3+2t^3$

The derivative with respect to t is written as follows,

$g^{\text{'}}\left(y\right)=\frac{d}{dt}\left(3+2t^3\right)$

On simplifying further,

$$\begin{gathered} g^{\text{'}}\left(y\right)=\frac{d}{dt}3+\frac{d}{dt}2t^3 \\ {g}^{\text{'}}\left(y\right)=0+2\times 3\times t^2 \\ {g}^{\text{'}}\left(y\right)=6t^2 \end{gathered}$$

Now by using the, Length of the curve $={\int }_a^b\sqrt{{\left(f^{\text{'}}\left(t\right)\right)}^2+{\left(g^{\text{'}}\left(t\right)\right)}^2}dt$.

Length of the curve$={\int }_0^1\sqrt{(2t{)}^2+{\left(6t^2\right)}^2}dt\left[\right.$since $\left.f^{\text{'}}\left(t\right)=2t,g^{\text{'}}\left(t\right)=6t^2\right]$

$$\begin{gathered} ={\int }_0^1\sqrt{4t^2+36t^4}dt \\ ={\int }_0^1\sqrt{4t^2\left(1+9t^2\right)}dt \end{gathered}$$

On further simplification,

$$\begin{gathered} g^{\text{'}}\left(t\right)={\int }_0^1\sqrt{4t^2}\sqrt{\left(1+9t^2\right)}dt \\ {g}^{\text{'}}\left(t\right)={\int }_0^{1}2t\sqrt{\left(1+9t^2\right)}dt\dots \left(1\right) \end{gathered}$$

To solve the integral assume,

$1+9t^2=u$

If you differentiate with respect to t, then

$$\begin{gathered} \frac{d}{dt}\left(1+9t^2\right)=u \\ \frac{d}{dt}1+\frac{d}{dt}9t^2=u \\ 18t·dt=du \\ dt=\frac{du}{18t} \end{gathered}$$

To find the Limits take t=0 and substitute in $1+9t^2=u$then,

$$\begin{gathered} 1+9(0{)}^2=u \\ u=1 \end{gathered}$$

When t=1,

$$\begin{gathered} 1+9(1{)}^2=u \\ u=10 \end{gathered}$$

Substitute $dt=\frac{du}{18t}$and the limits in the equation (1), that is in $2{\int }_0^{1}t\sqrt{4+9t^2}dt$.

Length of the curve$=2{\int }_1^{10}/\sqrt{u}\frac{du}{18t}$

$=\frac{2}{18}{\int }_1^{10}\sqrt{u}du$

Then the length of the curve$=\frac{2}{18}{\left[\frac{u^{1+\frac{1}{2}}}{1+\frac{1}{2}}\right]}_1^{10}$

$=\frac{2}{18}{\left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]}_1^{10}$

$=\frac{2}{18}·\frac{2}{3}{\left[u^{\frac{3}{2}}\right]}_1^{10}$

By substituting the limits.

Length of the curve $=\frac{2}{18}\times \frac{2}{3}\left[{10}^{\frac{3}{2}}-1\right]$

Length of the curve$=\frac{2}{27}\left[{10}^{\frac{3}{2}}-1\right]$

To draw the curve for the parametric equations, first find the points when $t\in [0,1]$.

Given that $t\in [0,1]$

When t=0.

$$\begin{gathered} (x,y)=\left(1+t^2,3+2t^3\right) \\ (x,y)=\left(1+0^2,3+2(0{)}^3\right) \\ (x,y)=(1,3) \end{gathered}$$

When t=1.

$$\begin{gathered} (x,y)=\left(1+t^2,3+2t^3\right) \\ (x,y)=\left(1+1^2,3+2(1{)}^3\right) \\ (x,y)=(2,5) \end{gathered}$$

The graphical representation of the points $(1,3)(2,5)$is as follows,

Therefore, the length of the curve is equal to

$\left[\frac{2}{27}\left[{10}^{\frac{3}{2}}-1\right]\right.$