Presentations
Pacing Tracker
- The quizzes have included questions for 16 objectives. How many have you passed? What are you plans to master those that you haven't mastered yet?
- We've finished units 1 through 3. Have you started your self-directed learning project for each unit?
- Remember you can submit only one SDL project per week. Plan ahead and don't let yourself get behind.
Brain Gains
- For the vector field $\vec F(x,y) = (5x+6y,xy^2)$, compute the derivative $D\vec F(x,y)$.
Solution
We compute the two partial derivatives, namely
- $\vec F_x = (5,y^2)$
- $\vec F_y = (6,2xy)$
These vectors are the columns of the derivative of $\vec F$, which means $$D\vec F(x,y) = \begin{bmatrix}5&6\\y^2&2xy\end{bmatrix}. $$
- For the function $f(x,y)=x^2+4xy+3y^2-10x-18y$, find the critical points by solving $Df(x,y)=\begin{bmatrix}0&0\end{bmatrix}$.
Solution
The only critical point is $(x,y)=(3,1)$.
The first derivative is $$Df(x,y) = \begin{bmatrix}2x+4y-10&4x+6y-18\end{bmatrix}.$$ To find the critical points, we need this derivative to be the zero matrix, which means we must solve the system of equations $$ 2x+4y-10 = 0,\quad 4x+6y-18=0.$$ This is equivalent to solving $2x+4y=10$ and $4x+6y=18$. Dividing both sides of the second equation by 2 gives the system $2x+4y=10$ and $2x+3y=9$. Subtracting the first equation from the second yields $y=1$, and substitution into either equation yields $x=3$.
- The function $f(x,y) = x^3+3xy+y^3$, the first and second derivatives are $$Df(x,y) = \begin{bmatrix}3x^2+3y&3x+3y^2\end{bmatrix}\quad\text{and}\quad D^2f(x,y) = \begin{bmatrix}\begin{matrix}6x\\3\end{matrix}&\begin{matrix}3\\6y\end{matrix}\end{bmatrix}. $$ A critical point of the function is $(x,y)=(-1,-1)$, which we can verify by computing $$Df(-1,-1) = \begin{bmatrix}3(-1)^2+3(-1)&3(-1)+3(-1)^2\end{bmatrix} = \begin{bmatrix}0&0\end{bmatrix}.$$ Does the function have a maximum or minimum at this critical point? Use the eigenvalues of $D^2f(-1,-1)$ to explain.
Solution
Note that $D^2f(-1,-1) = \begin{bmatrix}\begin{matrix}-6\\3\end{matrix}&\begin{matrix}3\\-6\end{matrix}\end{bmatrix}.$ We find the eigenvalues by solving $(-6-\lambda)(-6-\lambda)-(3)(3) = 0$. This is equivalent to solving $$0=\lambda^2+12\lambda+36-9 = \lambda^2+12\lambda+27 = (\lambda+9)(\lambda+3).$$ The eigenvalues of $D^2f(-1,-1)$ are $\lambda = -9$ and $\lambda = -3$, both of which are negative. We have already seen that negative eigenvalues mean the gradient field points inwards, meaning the critical point $(-1,-1)$ is the location of a local maximum. We can verify all these facts using the following Mathematica notebook.
- A rover travels along the path $2x-4y=12$. The elevation is given by $z=x+y^2-3$. Use Lagrange multipliers to find the $(x,y)$ location of the lowest elevation reached by the rover along this path.
Solution
The solution is $(x,y)=(4,-1)$.
To use Lagrange multipliers, remember you need to identify
- $f(x,y)$ - the thing you wish to optimize - $f(x,y)=x+y^2-3$
- $g(x,y)=c$ - the constraint - $g(x,y)=2x-4y=12$
- Then solve $\vec \nabla f = \lambda \vec \nabla g$ together with $g(x,y)=c$.
The equation $\vec \nabla f = \lambda \vec \nabla g$ gives $(1,2y) = \lambda(2,4)$, which means $1=2\lambda$ and $2y = 4\lambda$. The first equation means $\lambda = \frac{1}{2}$, which means $y=1$ from the second equation. Plugging $y=1$ into $2x-4y=12$ yields $x=4$.
We can check our work, as well as visualize what is happening, with the following Mathematica notebook.
Group problems
- Consider the function $f(x,y)= 2x^2+3xy+4y^2-5x+2y$.
- Find all critical points of $f$. [Check: $(x,y) = (2,-1)$.]
- Determine the eigenvalues of the second derivative at each critical point. [Check: The eigenvalues are $\lambda = 6\pm\sqrt{13}$, so $\lambda \approx 9.6$ or $\lambda \approx 2.4$.]
- At each critical point, do we have a local max, local min, or saddle?
- A rover travels along the curve $x^2y=16$ with $x>0$. The elevation near the rover is given by $z=-x^2-y^2$. Locate the $(x,y)$ coordinates where the rover reaches maximum height. [Check: You should obtain $y=2$ and two different $x$ values, giving two solutions.]
- Give an equation of the tangent plane to $xy+z^2=7$ at the point $P=(-3,-2,1)$. [Check: $(-2)(x-(-3))+(-3)(y-(-2))+2(1)(z-1)=0$. ]
- Give an equation of the tangent plane to $z=f(x,y)=xy^2$ at the point $P=(4,-1,f(4,-1))$. [Check: $z-4 = (-1)^2(x-4)+2(4)(-1)(y-(-1))$.]
- Consider the function $f(x,y,z) = 3xy+z^2$. We'll be analyzing the surface at the point $P=(1,-3,2)$.
- If $dx=0.1$, $dy=0.2$ and $dz=0.3$, then what is $df$ at $P$.
- Find the directional derivative of $f$ at $P$ in the direction $(1,-2,2)$.
- Give an equation of the tangent plane to the level surface of $f$ that passes through $P$.
- Give an equation of the tangent plane to the level surface of $f$ that passes through $(a,b,c)$.
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