Q. 16 - Calculus

Question

Let $v_0=\left(a, b\right)$ and let $v=\left(x, y\right)$ Describe the sets of points in $R^2$ satisfying the following properties:

$$\begin{gathered} \left(a\right). \left|v\right|=4 \\ \left(b\right). \left|v\right|\le 4 \\ \left(c\right). \left|v-v_0\right|=4 \end{gathered}$$

Step-by-Step Solution

Answer

(a) The set of points are $(0, 4), (0, -4), (4, 0) \mathrm{and} (-4, 0)$ etc.

(b)  The set of values of $\left(x, y\right)$ , which satisfy the relation $x^2+y^2\le 16$ will be the required set of points.

The set of points are $(0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (0, 4), (0, -4), (2, 3) \mathrm{etc}.$

(c)  The set of values of $(x, y) \mathrm{and} (a, b)$, which satisfy the relation ${(x-a)}^2+{(y-b)}^2=16$ will be the required set of points.

The set of points $(x, y)$ are $(5, 1), (4, 1) \mathrm{and} (1, 5) \mathrm{etc}.$

The set of points $(a, b)$ are  $(1, 1), (0, 1) \mathrm{etc}$.

Step 1: Part (a)

Given that, $v=(x, y)$

$\left|v\right|=\sqrt{x^2+y^2}$

$$\begin{gathered} \sqrt{x^2+y^2}=4 \\ x^2+y^2=16 (\mathrm{Squaring} \mathrm{both} \mathrm{side}) \end{gathered}$$

When $x=0$,

$$\begin{gathered} y=\sqrt{16} \\ y=\pm 4 \end{gathered}$$

When $y=0$,

$$\begin{gathered} x=\sqrt{16} \\ x=\pm 4 \end{gathered}$$

Thus, the set of points are $(0, 4), (0, -4), (4, 0) \mathrm{and} (-4, 0)$ etc.

Step 1: Part (b)

Given that, $v=(x, y)$

$\left|v\right|=\sqrt{x^2+y^2}$

$$\begin{gathered} \sqrt{x^2+y^2}\le 4 \\ x^2+y^2\le 16 (\mathrm{Squaring} \mathrm{both} \mathrm{side}) \end{gathered}$$

All the points that satisfy the above relationship, will be the required sets of points.

The set of points are,

$(0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (0, 4), (0, -4), (2, 3) \mathrm{etc}.$

Step 1: Part (c)

Given that,

$$\begin{gathered} v=(x, y) \mathrm{and} v_0=(a, b) \\ v-v_0=\left\{(x-a), (y-b)\right\} \\ \left|v-v_0\right|=\sqrt{{(x-a)}^2+{(y-b)}^2} \\ \sqrt{{(x-a)}^2+{(y-b)}^2}=4 \left(\mathrm{Given}\right) \\ { (x-a)}^2+{(y-b)}^2=16 \left(\mathrm{Squaring} \mathrm{both} \mathrm{side}\right) \end{gathered}$$

All the values of $(x, y) \mathrm{and} (a, b)$ that satisfy the above relationship will be the required sets of points.

The values of $(x, y)$ are $(5, 1), (4, 1) \mathrm{and} (1, 5) \mathrm{etc}.$

The values of $(a, b)$ are $(1, 1), (0, 1) \mathrm{etc}$.