Rapid Recall
- Give an equation of the tangent plane to $f(x,y)=x^2+4xy+3y^2$ at the point $(2,1)$.
Solution
Differentials give $df = (2x+4y)dx+(4x+6y)dy$. Our point is $x=2$ and $y=1$, so we have $dx=x-2$ and $dy=y-1$. Substitution into the original function gives $f(2,1) = 15$ and so $df = z-15$. Substitution into the differential gives $$ \underbrace{z-15}_{dz} = \underbrace{(2(2)+4(1))}_{f_x(2,1)}\underbrace{(x-2)}_{dx}+\underbrace{(4(2)+6(1))}_{f_y(2,1)}\underbrace{(y-1)}_{dy} \quad \text{or}\quad z-15 = 8(x-2)+14(y-1) .$$
- For the function $f(x,y)=x^2+4xy+3y^2$, compute the first derivative $Df(x,y)$ and the second derivative $D^2f(x,y)$.
Solution
Remember that the derivative is a matrix. We can write the differential $df = (2x+4y)dx+(4x+6y)dy$ as the product of a matrix and a vector using $df = \begin{bmatrix}2x+4y&4x+6y\end{bmatrix}\begin{pmatrix}dx\\dy\end{pmatrix}$. This gives $$Df(x,y) = \begin{bmatrix}2x+4y&4x+6y\end{bmatrix}.$$ The differential of the first derivative is $$dDf(x,y) = \begin{pmatrix}2\\4\end{pmatrix}dx+ \begin{pmatrix}4\\6\end{pmatrix}dy =\begin{bmatrix}\begin{pmatrix}2\\4\end{pmatrix}&\begin{pmatrix}4\\6\end{pmatrix}\end{bmatrix}\begin{pmatrix}dx\\dy\end{pmatrix}. $$ This gives $$D^2f(x,y) = \begin{bmatrix}\begin{matrix}2\\4\end{matrix}&\begin{matrix}4\\6\end{matrix}\end{bmatrix}.$$
- A rover travels along the path $2x-4y=12$. The elevation is given by $z=x+y^2-3$. Find the $(x,y)$ location of the lowest elevation reached by the rover along this path.
Solution
- We have $f(x,y) = x+y^2-3$, $g(x,y)=2x-4y$, and $c=12$. Gradients are $\vec \nabla f = (1,2y)$ and $\vec \nabla g = (2,-4)$. The system $\vec \nabla f = \lambda \vec \nabla g$, along with $g=c$ gives the equations $$1=\lambda 2,\quad 2y=\lambda (-4),\quad 2x-4y=12.$$
The solution is $\lambda = 1/2$, $y=-1$, and $x=4$. The requested location is $(x,y)=(4,-1)$.
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 surface at 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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