Assignment 1

Hannah Galang 30104279 Math 367

Q1.

solution. The domain is given by $Dom(f)=\{(x,y)|y\ge 1+x\}$. The boundary point for this function is $y=1+x$, and the interior points are given by $y>1+x$. In this case, we have that the domain is closed, as it includes all of its boundary points.

Q2.

solution, a). All directional derivatives exist for $\alpha \ge 2$. The proof is as shown below. proof. Let $\hat{u}=<u_1,u_2>$ denote an arbitrary unit vector at the point (0,0) on the function, that is, a vector with length 1. Then, the expression for the directional derivative at that point is given by D_\hat{u}\vec{f}(0,0) = \lim_{t \to 0} \frac{\vec{f}(0+tu_{1}, 0+tu_{2})-\vec{f}(0,0)}{t} Evaluating the limit, we take the following steps:

Noticing that $\vec{f}(0,0)=0$, this then becomes D_\hat{u}\vec{f}(0,0) = \lim_{t \to 0} \frac{\vec{f}(0+tu_{1}, 0+tu_{2})-0}{t} = \lim_{t \to 0} \frac{\vec{f}(0+tu_{1}, 0+tu_{2})}{t} we can then rewrite this above as ${\lim }_{t\to 0}\frac{\vec{f}(0+t{u}_1,0+t{u}_2)}{t}\to {\lim }_{t\to 0}\frac{\vec{f}(t{u}_1,t{u}_2)}{t}$as the zeros are redundant. Then, we write ${\lim }_{t\to 0}\frac{\vec{f}(t{u}_1,t{u}_2)}{t}={\lim }_{t\to 0}\frac{(t{u}_{1}t{u}_2{)}^{\alpha }}{(t{u}_1{)}^2+(t{u}_2{)}^2}\ast \frac{1}{t}$ and we can then write this as ${\lim }_{t\to 0}\frac{(t{u}_{1}t{u}_2{)}^{\alpha }}{(t{u}_1{)}^2+(t{u}_2{)}^2}\ast \frac{1}{t}={\lim }_{t\to 0}\frac{1}{t}\ast \frac{t^{2\alpha }{u}_1^{\alpha }{u}_2^{\alpha }}{t^2{u}_1^2+t^2{u}_2^2}$ $\to {\lim }_{t\to 0}\frac{1}{t}\ast \frac{t^{2\alpha }{u}_1^{\alpha }{u}_2^{\alpha }}{t^2{u}_1^2+t^2{u}_2^2}={\lim }_{t\to 0}\frac{t^{2\alpha }{u}_1^{\alpha }{u}_2^{\alpha }}{t^3(u_1^2+u_2^2)}$ From here, we note that $(u_1^2+u_2^2)=1$, as $\hat{u}$ is a unit vector and $||\hat{u}||=1=\sqrt{u_1^2+u_2^2}=u_1^2+u_2^2$. We then have ${\lim }_{t\to 0}\frac{t^{2\alpha }{u}_1^{\alpha }{u}_2^{\alpha }}{t^3}$ From here, we note that if $\alpha =1$, this results in the limit ${\lim }_{t\to 0}\frac{u_1{u}_2}{t}=\infty$ and the limit is not well defined for the directional derivative at (0,0). However, if $\alpha \ge 2$, this results in the limit ${\lim }_{t\to 0}{t}^{2\alpha -3}{u}_1^{\alpha }{u}_2^{\alpha }$ where $2\alpha -3$ is a positive integer and we have D_\hat{u}\vec{f}(0,0) = \lim_{t \to 0}{t^{2\alpha-3}u_1^{\alpha}u_{2}^{\alpha}}=0 resulting in a well defined limit. Thus, the directional derivative exists in all directions at (0,0) for any $\alpha \ge 2$.

solution, b). The function is totally differentiable at the point (0,0) for any $\alpha \ge 2$. The proof is as shown below. proof. Since we have shown that all directional derivatives exist at the point $(0,0)$, it suffices to show that $D\vec{f}(0,0)=0$. First, note that $f_x(0,0)=0$, and $f_y(0,0)=0$, so we have that $J\vec{f}(0,0)=\left[\begin{array}{cc}0& 0\end{array}\right]$. Plugging this into the definition of $D\vec{f}$, we have: $D\vec{f}(0,0)={\lim }_{(x,y)\to (0,0)}\frac{|f(x,y)-f(0,0)-\left[\begin{array}{cc}0& 0\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]|}{||(x,y)||}\to {\lim }_{(x,y)\to (0,0)}\frac{|f(x,y)|}{||x,y||}$ We then get: ${\lim }_{(x,y)\to (0,0)}\frac{|f(x,y)|}{||x,y||}={\lim }_{(x,y)\to (0,0)}\frac{(xy{)}^{\alpha }}{(x^2+y^2{)}^{\frac{3}{2}}}$ We can convert this to polar coordinates:

${\lim }_{(x,y)\to (0,0)}\frac{(xy{)}^{\alpha }}{(x^2+y^2{)}^{\frac{3}{2}}}={\lim }_{r\to 0}\frac{r^{2\alpha }co{s}^{\alpha }(\theta )si{n}^{\alpha }(\theta )}{r^3}$ Like in the previous case, we have that if $\alpha =1$, then the limit doesn’t exist. However, when $\alpha \ge 2$, we have that $2\alpha$ is a positive integer, and the limit is 0. Since we have then that $D\vec{f}(0,0)=\vec{f}(0,0)$, the function is differentiable at the point $(0,0)$.

Q3.

solution, a. To find the set of points for which the function given by the question is satisfied, we will first find the jacobian of the matrix. The jacobian is given by

$$\left[\begin{array}{cc}\frac{\delta F_1}{\delta x}& \frac{\delta F_1}{\delta y}\\ \frac{\delta F_2}{\delta x}& \frac{\delta F_2}{\delta y}\end{array}\right]$$

We have that $F_1=3x-y$ and $F_2=x^2{y}^2$. From here, we can evaluate the jacobian under these functions:

$$\left[\begin{array}{cc}\frac{\delta F_1}{\delta x}& \frac{\delta F_1}{\delta y}\\ \frac{\delta F_2}{\delta x}& \frac{\delta F_2}{\delta y}\end{array}\right]\to \left[\begin{array}{cc}3& -1\\ y^{2}2x& x^{2}2y\end{array}\right]$$

We know from linear 1 that this matrix is invertible if and only if its determinant is not equal to 0. Since we know that the determinant of a 2x2 matrix is $ad-bc$, we can just take the determinant and solve for which x and y values will yield a 0 determinant:

$$det\left[\begin{array}{cc}3& -1\\ y^{2}2x& x^{2}2y\end{array}\right]\to 6x^{2}y-(-y^{2}2x)\to 6x^{2}y+y^{2}2x$$

Setting this equal to 0, we have: $6x^{2}y+y^{2}2x=0\to x=0$ or $y=0$. So, we have that for any set of points in $R^2$ such that $x\ne 0$ and $y\ne 0$, the inverse function theorem is satisfied, and the function is invertible.

solution, b. We first note that the inverse function theorem applies at the point (1,1), since $x\ne 0$ and $y\ne 0$.

$$\left[\begin{array}{cc}3& -1\\ y^{2}2x& x^{2}2y\end{array}\right]$$

We can calculate $D\vec{f}(1,1{)}^{-1}$ by the following:

$${\left[\begin{array}{cc}3& -1\\ 2& 2\end{array}\right]}^{-1}\to \frac{1}{8}\left[\begin{array}{cc}2& 1\\ -2& 3\end{array}\right]$$

By the inverse function theorem, we have that $D\vec{f}(1,1{)}^{-1}=D\vec{g}(\vec{f}(1,1))$. We double check that $f(1,1)=(2,1)$: $\vec{f}(1,1)=(3(1)-1,1^2\ast 1^2)=(2,1)$ Since this is the case, have that $D\vec{f}(1,1{)}^{-1}=D\vec{g}(\vec{f}(1,1))=D\vec{g}(2,1)$. So, we’ve shown what we wanted to show.

Q4.

solution. First, we set up our matrix of partial deirvatives:

$$\left[\begin{array}{ccccc}3& 2y& 0& v& u\\ yz& xz& xy& v^2& 2uv\end{array}\right]$$

Then we have that:

\begin{bmatrix} 2y & u & 3 & v & 0 \\ xz & 2uv & yz & v^2 & xy \end{bmatrix} \right|_{(1,1,1,1,1)} \rightarrow \begin{bmatrix} 2 & 1 & 3 & 1 & 0 \\ 1 & 2 & 1 & 1 & 1 \end{bmatrix}

We have here that $A=\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]$. Since $detA=(2\ast 2)-(1\ast 1)=3\ne 0$, we note that $A$ is invertible, which allows us to apply the implicit function theorem. Finding $D\vec{g}(1,1,1)$, we have:

$$\left[\begin{array}{cc}-2& 1\\ 1& -2\end{array}\right]\left[\begin{array}{ccc}3& 0& 1\\ 1& 1& 1\end{array}\right]=\left[\begin{array}{ccc}-1& \frac{1}{5}& -\frac{1}{5}\\ \frac{1}{5}& -\frac{2}{5}& -\frac{1}{5}\end{array}\right]$$

So, we have found $D\vec{g}(1,1,1)$, and we’ve shown what we needed to show.