How do you find the range given $x=3-2t$ and $y=2+3t$ for $-2\le t\le 3$?
Answer
582k+ views
Hint: First we need to know the meaning of the range of a parametric equation. Now, we will calculate the range of the values of x and y separately. To do this, first we will select the variable x and then we will write t in terms of x. Now, using the given condition $-2\le t\le 3$ we will find the range of x. Similarly we will find the range of values of y.
Complete step by step answer:
Here we have been provided with the parametric equation $x=3-2t$ and $y=2+3t$ with the condition $-2\le t\le 3$. We have to determine the range.
Since the given equation is a parametric equation so here the definitions of the domain and the range is slightly different from what we generally use for the function $y=f\left( x \right)$. In parametric equations the domain is the set of values of t and the ranges are the sets of values of x and y. So, we need to determine the sets of values of x and y separately.
(1) Let us consider $x=3-2t$.
Now, we can write t in terms of x as:
$\Rightarrow t=\dfrac{3-x}{2}$
$\because -2\le t\le 3$
Substituting the value of t in terms of x we get,
$\Rightarrow -2\le \dfrac{3-x}{2}\le 3$
Multiplying all the terms with 2 we get,
$\begin{align}
& \Rightarrow -4\le 3-x\le 6 \\
& \Rightarrow -7\le -x\le 3 \\
\end{align}$
Multiplying all the terms with -1 and changing the direction of inequality because we are multiplying them with a negative number, we get,
$\begin{align}
& \Rightarrow -3\le x\le 7 \\
\end{align}$
Hence, the range of values of x is $\left[ -3,7 \right]$.
(2) Now, let us consider $y=2+3t$.
We can write t in terms of y as:
$\Rightarrow t=\dfrac{y-2}{3}$
$\because -2\le t\le 3$
Substituting the value of t in terms of y we get,
$\Rightarrow -2\le \dfrac{y-2}{3}\le 3$
Multiplying all the terms with 3 we get,\[\]
$\begin{align}
& \Rightarrow -6\le y-2\le 9 \\
& \because -4\le y\le 11 \\
\end{align}$
Hence, the range of values of y is $\left[ -4,11 \right]$.
Note: You must remember the definition of domain and range for a parametric equation. In the above you cannot write the set of x values as the domain and the set of y values as the range because here we don’t have the function $y=f\left( x \right)$ but the values of x and y depend on a third variable t. So, the domain of this given function will be the set of values of t.
Complete step by step answer:
Here we have been provided with the parametric equation $x=3-2t$ and $y=2+3t$ with the condition $-2\le t\le 3$. We have to determine the range.
Since the given equation is a parametric equation so here the definitions of the domain and the range is slightly different from what we generally use for the function $y=f\left( x \right)$. In parametric equations the domain is the set of values of t and the ranges are the sets of values of x and y. So, we need to determine the sets of values of x and y separately.
(1) Let us consider $x=3-2t$.
Now, we can write t in terms of x as:
$\Rightarrow t=\dfrac{3-x}{2}$
$\because -2\le t\le 3$
Substituting the value of t in terms of x we get,
$\Rightarrow -2\le \dfrac{3-x}{2}\le 3$
Multiplying all the terms with 2 we get,
$\begin{align}
& \Rightarrow -4\le 3-x\le 6 \\
& \Rightarrow -7\le -x\le 3 \\
\end{align}$
Multiplying all the terms with -1 and changing the direction of inequality because we are multiplying them with a negative number, we get,
$\begin{align}
& \Rightarrow -3\le x\le 7 \\
\end{align}$
Hence, the range of values of x is $\left[ -3,7 \right]$.
(2) Now, let us consider $y=2+3t$.
We can write t in terms of y as:
$\Rightarrow t=\dfrac{y-2}{3}$
$\because -2\le t\le 3$
Substituting the value of t in terms of y we get,
$\Rightarrow -2\le \dfrac{y-2}{3}\le 3$
Multiplying all the terms with 3 we get,\[\]
$\begin{align}
& \Rightarrow -6\le y-2\le 9 \\
& \because -4\le y\le 11 \\
\end{align}$
Hence, the range of values of y is $\left[ -4,11 \right]$.
Note: You must remember the definition of domain and range for a parametric equation. In the above you cannot write the set of x values as the domain and the set of y values as the range because here we don’t have the function $y=f\left( x \right)$ but the values of x and y depend on a third variable t. So, the domain of this given function will be the set of values of t.
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