Chapter 10.7, Problem 17AYU

To determine

The graph of the curve of a parametric equations $x\left(t\right)=\sqrt{t},y\left(t\right)=t^{\frac{3}{2}};t\ge 0$, show its orientation. Find the rectangular equation of curve of the parametric equations.

Answer to Problem 17AYU

Solution:

The graph of the parametric equations $x\left(t\right)=\sqrt{t},y\left(t\right)=t^{\frac{3}{2}};t\ge 0$ is

The rectangular equation of the parametric equation $x\left(t\right)=\sqrt{t},y\left(t\right)=t^{\frac{3}{2}};t\ge 0$ is $y=x^3$.

Explanation of Solution

Given information:

The parametric equation is $x\left(t\right)=\sqrt{t},y\left(t\right)=t^{\frac{3}{2}};t\ge 0$.

Explanation:

To draw the given parametric equation, plug some values of $t$ in the given equation and find few points on the curve.

The interval for $t$ is $t\ge 0$.

$\begin{array}{cccc}t& x\left(t\right)=\sqrt{t}& y\left(t\right)=t^{\frac{3}{2}}& \left(x,y\right)\\ 0& 0& 0& \left(0,0\right)\\ 1& \text{1}& \text{1}& \left(1,1\right)\\ 2& \text{1}\text{.414}& \text{2}\text{.828}& \left(\text{1}\text{.414},\text{2}\text{.828}\right)\\ 3& \text{1}\text{.732}& \text{5}\text{.196}& \left(\text{1}\text{.732},\text{5}\text{.196}\right)\\ 4& \text{2}& \text{8}& \left(2,8\right)\end{array}$

The orientation is the curve traced out in a certain direction by the corresponding succession of points $\left(x,y\right)$.

The arrow shows the orientation along the curve as $t$ varies from $t\ge 0$.

As $t$ takes values from $t\ge 0$, the corresponding orientation is from $\left(0,0\right)$ to $\left(2,8\right)$.

Use points in the table to sketch the curve.

The graph of the parametric equation $x\left(t\right)=\sqrt{t},y\left(t\right)=t^{\frac{3}{2}};t\ge 0$ is

Let $x\left(t\right)=\sqrt{t}$ and $y\left(t\right)=t^{\frac{3}{2}}$

By simplifying,

$\Rightarrow x^2=t$ and $y^2=t^3$

By substituting the value of $t=x^2$ in the equation $y^2=t^3$

$\Rightarrow y^2={\left(x^2\right)}^3$

$\Rightarrow y^2=x^6$

$\Rightarrow y^2-x^6=0$

$\Rightarrow y^2=x^6$

$\Rightarrow y=x^3$

The rectangular equation of the parametric equation $x\left(t\right)=\sqrt{t},y\left(t\right)=t^{\frac{3}{2}};t\ge 0$ is $y=x^3$