Rapid Recall
- Compute the eigenvalues of the matrix $\begin{bmatrix}4&3\\2&5\end{bmatrix}$.
Solution
We solve $(4-\lambda)(5-\lambda)-6=0$. This becomes $0=\lambda^2-9\lambda+14=(\lambda-7)(\lambda-2)$. The eigenvalues are $\lambda=7$ and $\lambda=2$.
- 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 $y=10/9$).
- 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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