Question

How do you find the distance travelled from $t=0$ to $t=3$ by a particle whose motion is given by the parametric equations $x=5{{t}^{2}}$,$y={{t}^{3}}$?

Answer 1

For this question we have to find the distance travelled by a particle whose motion is given by a parametric equation.
So, for solving this we have to eliminate t and find relation between x and y. after finding the relation we will use the formula $\int{\sqrt{1+{{\left[ {{f}^{'}}\left( x \right) \right]}^{2}}}}dx$ and solve the integration to find the distance.

Complete step by step solution:
So, firstly we will bring the relation between x and y. So, we can eliminate the t in our solution process.
Given, $x=5{{t}^{2}}$ and $y={{t}^{3}}$
So, here we will make it as a subject and get it in terms of x.
$\Rightarrow x=5{{t}^{2}}$
$\Rightarrow {{t}^{2}}=\dfrac{x}{5}$
$\Rightarrow t=\sqrt{\dfrac{x}{5}}$
Here we got t as $\Rightarrow t=\sqrt{\dfrac{x}{5}}$ in terms of x. So, we will substitute this t in the given y terms and eliminate the t and bring relation between x and y.
$\Rightarrow y={{t}^{3}}$
$\Rightarrow y={{\left( \sqrt{\dfrac{x}{5}} \right)}^{3}}$
$\Rightarrow y={{\left( \dfrac{x}{5} \right)}^{\dfrac{3}{2}}}$
So, let us consider it as $\Rightarrow f\left( x \right)={{\left( \dfrac{x}{5} \right)}^{\dfrac{3}{2}}}$. The derivative of it using the formula \Rightarrow y={{x}^{t}}$$$$\Rightarrow \dfrac{dy}{dx}=t{{x}^{t-1}} will be as follows.
$\Rightarrow f\left( x \right)={{\left( \dfrac{x}{5} \right)}^{\dfrac{3}{2}}}$.
$\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{3}{2}{{\left( \dfrac{x}{5} \right)}^{\dfrac{1}{2}}}$
Here now the limits of the integration will be as follows.
When $t=0,t=3$ the value of x will be $x=0,x=45$ when we substitute the t value in the given $\Rightarrow x=5{{t}^{2}}$.
So, this $x=0,x=45$ will be limits of the integral $\int{\sqrt{1+{{\left[ {{f}^{'}}\left( x \right) \right]}^{2}}}}dx$which gives us required solution.
$\Rightarrow \int{\sqrt{1+{{\left[ {{f}^{'}}\left( x \right) \right]}^{2}}}}dx$
$\Rightarrow \int\limits_{0}^{45}{\sqrt{1+{{\left( \dfrac{3}{2}\sqrt{\dfrac{x}{5}} \right)}^{2}}}}dx$
$\Rightarrow \int\limits_{0}^{45}{\sqrt{1+\left( \dfrac{9}{4}\times \dfrac{x}{5} \right)}}dx$
$\Rightarrow \int\limits_{0}^{45}{\sqrt{1+\left( \dfrac{9x}{20} \right)}}dx$
$\Rightarrow \dfrac{1}{\sqrt{20}}\int\limits_{0}^{45}{\sqrt{20+9x}}dx$
Now, we will consider that $\Rightarrow 20+9x=r$ so that it makes our further simplification much easier.
So, we get,
$\Rightarrow 20+9x=r$
Here we differentiate this equation. After differentiation the equation will be reduced as follows.
$\Rightarrow dx=\dfrac{1}{9}dr$
So, the new limits using the equation $\Rightarrow 20+9x=r$. We get,
When, $x=0,x=45$ then the limits will become $r=20,r=425$ respectively.
So, the integral will become after substituting the new limits and r as $20+9x=r$ as follows.
$\Rightarrow \dfrac{1}{\sqrt{20}}\int\limits_{0}^{45}{\sqrt{20+9x}}dx$
$\Rightarrow \dfrac{1}{\sqrt{20}}\int\limits_{20}^{425}{\dfrac{1}{9}\sqrt{r}dr}$
Here we will use the integration formula $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}$ then the equation will be reduced as follows.
$\Rightarrow \dfrac{1}{9\sqrt{20}}\left[ \dfrac{2}{3}{{r}^{\dfrac{3}{2}}} \right]$ from $425$ to $20$as boundary of limit.
$\Rightarrow \dfrac{2}{27\sqrt{20}}\left[ {{\left( 425 \right)}^{\dfrac{3}{2}}}-{{\left( 25 \right)}^{\dfrac{3}{2}}} \right]$
$\Rightarrow 143.64$
Therefore, the required solution for the given question is $143.64$.

Note:
Students must be very careful in calculations. Students must be having good knowledge in calculus in doing problems of this kind. Here students must be careful that they should know that we are finding the distance and not displacement so we must use formula $\int{\sqrt{1+{{\left[ {{f}^{'}}\left( x \right) \right]}^{2}}}}dx$ and do the further process.