Chapter 10.7, Problem 18AYU

To determine

The graph of curve of a parametric equations $x\left(t\right)=t^{\frac{3}{2}}+1,y\left(t\right)=\sqrt{t};t\ge 0$, show its orientation. Find the rectangular equation of curve of the parametric equations.

Answer to Problem 18AYU

Solution:

The graph of the parametric equations $x\left(t\right)=t^{\frac{3}{2}}+1,y\left(t\right)=\sqrt{t};t\ge 0$ is

The rectangular equation of the parametric equation $x\left(t\right)=t^{\frac{3}{2}}+1,y\left(t\right)=\sqrt{t};t\ge 0$ is $y^2-x^6=0$.

Explanation of Solution

Given information:

The parametric equation is $x\left(t\right)=t^{\frac{3}{2}}+1,y\left(t\right)=\sqrt{t};t\ge 0$.

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 $t\ge 0$.

$\begin{array}{cccc}t& x\left(t\right)=t^{\frac{3}{2}}+1& y\left(t\right)=\sqrt{t}& \left(x,y\right)\\ 0& 1& 0& \left(1,0\right)\\ 1& \text{2}& \text{1}& \left(2,1\right)\\ 2& \text{3}\text{.828}& \text{1}\text{.414}& \left(\text{3}\text{.828},\text{1}\text{.414}\right)\\ 3& \text{6}\text{.196}& \text{1}\text{.732}& \left(\text{6}\text{.196,1}\text{.732}\right)\\ 4& 9& \text{2}& \left(9,2\right)\end{array}$

The orientation is 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 $\left(1,0\right)$ to $\left(9,2\right)$.

Use points in the table to sketch the curve.

The graph of the parametric equation $x\left(t\right)=t^{\frac{3}{2}}+1,y\left(t\right)=\sqrt{t};t\ge 0$ is

Let $x\left(t\right)=t^{\frac{3}{2}}+1$ and $y\left(t\right)=\sqrt{t}$

By simplifying,

$\Rightarrow y^2=t$

By substituting the value of $t=y^2$ in the equation $x\left(t\right)=t^{\frac{3}{2}}+1$,

$\Rightarrow x\left(t\right)={\left(y^2\right)}^{\frac{3}{2}}+1$

$\Rightarrow x=y^3+1$

$\Rightarrow y^3=x-1$

The rectangular equation of the parametric equation $x\left(t\right)=t^{\frac{3}{2}}+1,y\left(t\right)=\sqrt{t};t\ge 0$ is $y^3=x-1$.