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Brain Gains (Rapid Recall)
- Consider the function $f(x,y) = 2x^2+3xy+y^2$. Compute $f_x$ and then the second partials $f_{xx}$ and $f_{xy}$. Then compute $f_y$ and then the second partials $f_{yx} $ and $f_{yy} $.
Solution
We have
- $f_x = 4x+3y$
- $f_{xx} = 4$
- $f_{xy} = 3$
- $f_y = 3x+2y$
- $f_{yx} = 3$
- $f_{yy} = 2$
The first derivative of $f$ is the matrix $$Df(x,y) =\begin{bmatrix}f_x&f_y\end{bmatrix} =\begin{bmatrix}4x+3y&3x+2y\end{bmatrix}.$$ The second derivative of $f$, or Hessian of $f$, is the matrix $$D^2f(x,y) = \begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix} = \begin{bmatrix}4&3\\3&2\end{bmatrix} .$$ In general, the second derivative depends on $x$ and $y$. In this particular case, the second derivative is a constant matrix.
- Compute the eigenvalues of the matrix $\begin{bmatrix}4&3\\3&2\end{bmatrix}$. [Note, the eigenvalues are irrational in this example. The quadratic formula or completing the square will help.]
Solution
We solve $(4-\lambda)(2-\lambda)-9=0$. This becomes $0=\lambda^2-6\lambda-1$. The quadratic formula gives $$\lambda = \frac{-(-6)\pm\sqrt{(-6)^2-4(1)(-1)}}{2(1)} = \frac{6\pm\sqrt{40}}{2} = 3\pm\sqrt{10}.$$ Note that $\lambda = 3+\sqrt{10}>0$ and $\lambda = 3-\sqrt{10}<0$, which means the eigenvalues differ in sign. Any critical point for this function will correspond to a saddle point.
- See WolframAlpha for a plot to see the saddle point.
You can also solve for the eigenvalues by completing the square. This gives $$\begin{align} \lambda^2-6\lambda-1 &= \lambda^2-6\lambda +(9-9)-1 \\ &= (\lambda^2-6\lambda +9)-9-1\\ &= (\lambda-3)^2-10 \end{align}.$$ Because $0 = (\lambda-3)^2-10$, we know $(\lambda-3)^2=10$ which means $\lambda-3=\pm\sqrt{10}$ or $\lambda = 3\pm\sqrt{10}$.
- The surface $4x+y^2z=8$ passes through the point $P=(3,2,-1)$. Give an equation of the tangent plane to this surface at $P$.
Solution
Differentials tell us $$4\,dx+2yz\,dy+y^2\,dz=0.$$ We know $x=3$, $y=2$, and $z=-1$. We also know that if $Q=(x,y,z)$ is another point on the plane, then the change from $P$ to $Q$ is $dx = x-3$, $dy=y-2$, and $dz=z+1$. Substitution (plug it in, plug it in) gives the equation of the tangent plane as $$4(\underbrace{x-3}_{dx})+2(\underbrace{2}_{y})(\underbrace{-1}_{z})(\underbrace{y-2}_{dy})+(\underbrace{2}_{y})^2(\underbrace{z+1}_{dz})=0.$$
- Let $f(x,y)=x^2+y$, and $g(x,y)=4x+3y$. Solve the system $\vec \nabla f = \lambda \vec \nabla g$ and $g(x,y)=12$.
Solution
We have $\vec \nabla f = (2x,1)$ and $\vec \nabla g = (4,3)$. The equation $\vec \nabla f = \lambda \vec \nabla g$ gives us $2x=\lambda\cdot 4$ and $1 = \lambda 3$. The second equation tells us $\lambda =1/3$, and the first equation tells us $x=\lambda\cdot 2=2/3$. Substitution into $4x+3y=12$ tells us $y=(12-8/3)/3$.
Group problems
- A rover travels along the line $g(x,y)=2x+3y=6$. The surrounding terrain has elevation $f(x,y)=x^2+4y$. The rover reaches a local minimum along this path, and our job is to find the location of this minimum.
- Compute $\vec \nabla f$ and $\vec \nabla g$.
- Write the system of equations that results from $\vec \nabla f=\lambda\vec \nabla g$ together with $g(x,y) = 6$.
- Solve the system above (you should get $x=4/3$ and $y=10/9$).
- Use LagrangeMultipliers.nb to check your work and visualize the rover's path and how it relates to the elevation contours. Scroll down to the "All Code in One Block" section, and update f, g, and c.
- Consider the function $f(x,y,z) = 3xy+z^2$. We'll be analyzing the level surface that passes through the point $P=(1,-3,2)$.
- Compute the differential $df$, and then evaluate the differential at $P$.
- For a level surface, the output remains constant (so $df=0$). If we let $(x,y,z)$ be a point on the surface really close to $P$, then we have $dx=x-1$, $dy=y-(-3)$ and $dz = z-?$. Plug this information into the differential to obtain the differential at $P$ to obtain an equation of the tangent plane.
- Give an equation of the tangent plane to the level surface of $f$ that passes through $(1,2,-3)$.
- Give an equation of the tangent plane to the level surface of $f$ that passes through $(a,b,c)$.
- What relationship exists between the gradient of $f$ at $P$ and the tangent plane through $P$?
- Suppose a plane passes through the point $(a,b,c)$ and has normal vector $(A,B,C)$. Give an equation of that plane.
- Give an equation of the tangent plane to $xy+z^2=7$ at the point $P=(-3,-2,1)$.
- Give an equation of the tangent plane to $z=f(x,y)=xy^2$ at the point $P=(4,-1,f(4,-1))$.
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