20201106

Thanks for sharing things in Perusall. Here are the presenters for today. Remember, I start picking presenters at 8am. If you upload something after that time, I may not see it.

• 4.28 -
• 4.29 - Marissa
• 4.30 -
• 4.31 - Santiago
• 4.32 -

Thanks for sharing things in Perusall. Here are the presenters for today. Remember, I start picking presenters at 12 noon. If you upload something after that time, I may not see it.

• 4.27 - Oscar
• 4.28 - Michael
• 4.29 - Trevor
• 4.30 - Alan
• 4.31 -
• 4.32 -

We have $$\begin{align} d(\vec\nabla f) &= d(2xy^4,4x^2y^3)\\ &= ((2dx)y^4+2x(4y^3dy),(8xdx)y^3+4x^2(3y^2dy))\\ &= \begin{pmatrix}(2)y^4\\(8x)y^3\end{pmatrix}dx+\begin{pmatrix}2x(4y^3)\\4x^2(3y^2)\end{pmatrix}dy\\ &= \begin{bmatrix}(2)y^4&2x(4y^3)\\(8x)y^3&4x^2(3y^2)\end{bmatrix}\begin{pmatrix}dx\\dy\end{pmatrix}. \end{align}$$ The last two lines above give the differential of $\vec \nabla f$ as a linear combination of partial derivatives, and then as a matrix product. The last line show the second derivative of $f$ is $$D^2f(x,y) = \begin{bmatrix}(2)y^4&2x(4y^3)\\(8x)y^3&4x^2(3y^2)\end{bmatrix} = \begin{bmatrix}2y^4&8xy^3\\8xy^3&12x^2y^2\end{bmatrix} .$$ We can also obtain this matrix by just directly computing all the second partial derivatives.

• $f_x = 2xy^4$ which means $f_{xx} = 2y^4$ and $f_{xy} = 8xy^3$. These form the first column.
• $f_y = 4x^2y^3$ which means $f_{yx} = 8xy^3$ and $f_{yy} = 12x^2y^2$. These form the second column.

Notice that the direction $\vec{BC}$ is the same for each directional derivative. So all we really need to do is compare the slopes in the "south-east" direction at each of the 5 points.

• Because the contours have labels on them, we see that moving from $A$ to $B$ results in drop, so a negative slope. This is true at $A$, $B$, and $C$.
• Because the distance to level curves increases as we move south east starting at $A$, we can tell that the slope is most negative at $A$.
• At $D$ the slope is zero in the $\vec{BC}$ direction.
• At $E$ the slope is positive.

Ordering them in decreasing order gives $$D_{\vec {BC}}f(E) > D_{\vec {BC}}f(D) > D_{\vec {BC}}f(C) > D_{\vec {BC}}f(B) > D_{\vec {BC}}f(A).$$

Notice that the point $B$ is the exact same in each of these. Also notice that $\vec {AB}$, $\vec {BC}$, $\vec {CD}$, and $\vec{DE}$ all point in the same southeast direction (they have the same unit vector). Since $D_{\vec u}f(P) = \vec \nabla f(P)\cdot\frac{\vec u}{|\vec u|}$, these quantities are all equal. The unit direction is the same, and the point is the same, so the requested slopes are equal. This gives $$D_{\vec {AB}}f(B) = D_{\vec {BC}}f(B) = D_{\vec {CD}}f(B) = D_{\vec {DE}}f(B).$$

The direction $\vec {DC}$ is northwest, while the direction $\vec{DE}$ is southeast. Standing at point $B$, moving southeast results in a drop in height while moving northwest results in an increase in height. As such, we have $$D_{\vec {DC}}f(B) > D_{\vec {DE}}f(B).$$ We can actually say more in this problem Because the point is the same, and the directions are opposite, we know they are equal and opposite, giving us $$D_{\vec {DC}}f(B) = - D_{\vec {DE}}f(B).$$

If $Q = (x,y,z)$ is any other point on the plane, then $\vec {PQ} = (x-a,y-b,z-c)$ is a vector in the plane. Since $\vec n = (A,B,C)$ is normal to the plane, then we know $\vec n$ and $\vec {PQ}$ are orthogonal, which means $\vec n\cdot \vec {PQ}=0$. This gives an equation of the plane as $$(A,B,C)\cdot (x-a,y-b,z-c)=0\quad\text{or}\quad A(x-a)+B(y-b)+C(z-c)=0.$$