# HW 1 Solutions

Problem 1.  (a) $\mathbb{R}^2 = \left\{(x,y) : x, y \in \mathbb{R} \right\}$.

(b) $\mathbb{R}^3 = \left\{(x,y,z) : x, y, z \in \mathbb{R} \right\}$.

(c) $\mathbb{R}^{2016} = \left\{\left(x_1, x_2, \cdots, x_{2016}\right) : x_i \in \mathbb{R} \right\}$.

Problem 2.  (a) $f:\mathbb{R}^2 \to \mathbb{R}$

(b) $f:\mathbb{R}^3 \to \mathbb{R}$.

(c ) $G:\mathbb{R}^2 \to \mathbb{R}^2$.

(d)  $F:\mathbb{R}^{100} \to \mathbb{R}^{101}$

(e) $c:\mathbb{R} \to \mathbb{R}^3$.

Problem 3.  skipped.

Problem 4.  (a) The plane is given by $z = -12$.

(b) The plane is given by $x = 7$.

(c ) This sphere has as its radius $R = \left|\langle 3, 1, 4 \rangle\right| = \sqrt{26}$ and has as its center $\vec{x}_0 = (2, 1, 3)$.  Its equation is then

$\displaystyle (x-2)^2 + (y-1)^2 + (z-3)^2 = 26.$

(d) To find the equation of this sphere we need to find its center and we need to find its radius.  Its center point is the mid-point or average of the two given points and so

$\displaystyle \vec{x}_0 = \frac{1}{2}\left(-3+7, 1+9, -5+-1\right) = \left(2, 5, -3\right)$.

The radius can be found by computing the distance between the two given points and dividing it by 2:

$\displaystyle R = \frac{\sqrt{(-10)^2+(-8)^2+(-4)^2}}{2} = \frac{\sqrt{180}}{2}$.

We can now use our sphere equation to find an equation for this sphere.

Problem 5.  For the function $h(x,y) = 8 - \sqrt{4-x^2-y^2}$ we cannot plug in any pairs $(x,y)$ that make the expression $4-x^2-y^2$ negative.  This means we need

$\displaystyle 4-x^2-y^2 \geq 0$

$\displaystyle 4 \geq x^2 + y^2$.

This inequality defines a region in the plane that is a filled in circle of radius $2$, centered at the origin.  If we call this region $D \subset \mathbb{R}^2$, then we have that the domain of $h(x,y)$ is $D$.

Problem 6.  (a) $|\vec{v}| = \sqrt{25} = 5.$

(b) Since $|\vec{w}| = \sqrt{18}$ one unit vector is

$\displaystyle \vec{u} = \frac{\vec{w}}{|\vec{w}|} = \frac{1}{\sqrt{18}}\langle3, -3\rangle$.

(c ) $|\vec{v}| = \sqrt{37}$.  The components of $\vec{u}$ are

$\displaystyle \vec{u} = \langle \, \frac{2}{\sqrt{37}}, \frac{3}{\sqrt{37}}, \frac{5}{\sqrt{37}}\, \rangle$

(d)  In general, given $\vec{v} = \langle x, y, z \rangle$ a unit vector parallel to $\vec{v}$ is

$\displaystyle \vec{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{1}{\sqrt{x^2+y^2+z^2}}\langle x, y, z \rangle = \left\langle \frac{x}{\sqrt{x^2+y^2+z^2}}, \frac{y}{\sqrt{x^2+y^2+z^2}}, \frac{z}{\sqrt{x^2+y^2+z^2}}\,\right\rangle$.

Note that this works provided $\vec{v} \neq \vec{0}$ (otherwise we’d be trying to divide by $|\vec{v}| = 0$).

Problem 7.  (skipped)