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)

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