Rapid Recall
- Let $f(x,y)=x^2+y$, and $g(x,y)=2x+y$. Solve the system $\vec \nabla f = \lambda \vec \nabla g$ together with $g(x,y)=3$.
Solution
We have $\vec \nabla f = (2x,1)$ and $\vec \nabla g = (2,1)$. The equation $\vec \nabla f = \lambda \vec \nabla g$ gives us $2x=\lambda\cdot 2$ and $1 = \lambda 1$. The second equation tells us $\lambda =1$, and the first equation tells us $x=\lambda=1$. Substitution into $2x+y=3$ tells us $y=1$.
- The surface $x^2+3y^2-4z=-5$ passes through the point $P=(2,1,3)$. Give an equation of the tangent plane to this surface at $P$. Hint: Use differentials.
Solution
Differentials tell us $$2xdx+6ydy-4dz=0.$$ We know $x=2$, $y=1$, and $z=3$. We also know that if $Q=(x,y,z)$ is another point on the plane, then the change from $P$ to $Q$ is $dx = x-2$, $dy=y-1$, and $dz=z-3$. Substitution (plug it in, plug it in) gives the equation of the tangent plane as $$2(2)(x-2)+6(1)(y-1)-4(z-3)=0.$$
- A parallelogram has edges $(5-\lambda, 2)$ and $(3, 4-\lambda)$. Find $\lambda$ so that the area of the parallelogram is zero.
Solution
- The area is $$A=|(5-\lambda)(4-\lambda)-(2)(3)| = |\lambda^2-9\lambda+20-6| = |\lambda^2-9\lambda-14| = |(\lambda - 7)(\lambda - 2)|.$$
This equals zero when $\lambda = 7 $ or $\lambda =2$. We call these eigenvalues.
Group problems
- Find the eigenvalues of the following matrices (pass the chalk after each one).
- $\begin{bmatrix}2&4\\4&2\end{bmatrix}$, $\begin{bmatrix}2&3\\1&4\end{bmatrix}$, $\begin{bmatrix}1&6\\4&3\end{bmatrix}$, $\begin{bmatrix}3&2\\1&2\end{bmatrix}$.
- For each function below, compute $f_x$ and then $f_{xx}$ and $f_{xy}$. Then compute $f_y$ followed by $f_{yx}$ and $f_{yy}$.
- $f(x,y) = 3x^2+5xy+y^2$
- $f(x,y) = xy^2$
- The differential of $f(x,y)=x^2+4y$ is $df = 2xdx+4dy$. At the point $(3,-2)$ this differential is $df = 6dx+4dy$. This differential is sometimes called a "linearization" of the function.
- Consider the level curve of $f$ that passes through $(3,-2)$. Give an equation of the tangent line to this curve at $(3,-2)$. [What are $df,dx,dy$ if we move to a point $(x,y)$ on the level curve that is really close to $(3,-2)$?]
- Consider the surface plot of $f$. Give an equation of the tangent plane to this surface at the point $(3,-2,f(3,-2))$. [What are $df,dx,dy$ if we move to a point $(x,y,z)$ on the surface that is really close to $(3,-2,f(3,-2))$?]
- 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 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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