Presentations
4.31 - 4.36Pacing 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 $\vec F(x,y) = (xe^y,x+y^2)$, compute the derivative $D\vec F(x,y)$.
Solution
We compute the two partial derivatives, namely
- $\vec F_x = (e^y,1)$
- $\vec F_y = (xe^y,2y)$
These vectors are the columns of the derivative of $\vec F$, which means $$D\vec F(x,y) = \begin{bmatrix}e^y&xe^y\\1&2y\end{bmatrix}. $$
- For the function $f(x,y)=x^3+3x^2+y^2+10y$, compute the second derivative.
Solution
Note $\frac{\partial}{\partial x} \vec\nabla f= (6x+6,0)$ and $\frac{\partial}{\partial y}\vec\nabla f = (0,2)$. Placing these vectors into the columns of a matrix gives us $$D^2f(x,y) = \begin{bmatrix}\begin{matrix}6x+6\\0\end{matrix}&\begin{matrix}0\\2\end{matrix}\end{bmatrix}.$$
- Find the critical points of the function $f(x,y)=x^3+3x^2+y^2+10y$.
Solution
There are two critical points, namely $(0,-5)$ and $(-2,-5)$. Note $\vec \nabla f(x,y) = (3x^2+6x,2y+10)$ is the zero matrix precisely when $x=0$ or $x=-2$, and $y=-5$.
- For the function $f(x,y)=x^3+3x^2+y^2+10y$, determine the location of any maxes, mins, or saddles, and classify each location appropriately using eigenvalues.
Solution
At each critical point, we need to (1) evaluate the second derivative, (2) compute the eigenvalues, and (3) classify the point using the eigenvalues.
- At the point $(0,-5)$, the second derivative is $D^2f(0,-5) = \begin{bmatrix}\begin{matrix}6\\0\end{matrix}&\begin{matrix}0\\2\end{matrix}\end{bmatrix}$. The eigenvalues are 6 and 2 (gradient point outwards from the point), which means at $(0,-5)$ we have a minimum.
- At the point $(-2,-5)$, the second derivative is $D^2f(-2,-5) = \begin{bmatrix}\begin{matrix}-6\\0\end{matrix}&\begin{matrix}0\\2\end{matrix}\end{bmatrix}$. The eigenvalues are -6 and 2 (gradient point outwards from the point in one direction, and inwards in another), which means at $(-2,-5)$ we have a saddle point.
- A rover travels along the curve $x^2+y=3$. The elevation near the rover is given by $z = 4 x - 2 y + 1395$. Locate the $(x,y)$ coordinates provided by Lagrange Multipliers.
Solution
We wish to optimize $f(x,y) = 4 x - 2 y + 1395$ subject to the constraint $g(x,y)=x^2+y=3$. We have $\vec\nabla f = (4,-2)$ and $\vec \nabla g = (2x,1)$. The system we must solve is $$4 = \lambda 2x, -2 = \lambda 1, x^2+y=3.$$ The second equation gives $\lambda = -2$. The first equation then provides $x=-1$. This means $y = 3-1 = 2$. The solution is $(x,y)=(-1,2)$.
Group problems
- Consider the function $f(x,y)= x^3+3xy-y^3$. This function has two critical points, namely $(0,0)$ and $(1,-1)$.
- Compute the gradient $\vec \nabla f(x,y)$.
- Compute both $\vec \nabla f(0,0)$ and $\vec \nabla f(1,-1)$. Your work should show that both $(0,0)$ and $(1,-1)$ are critical points. (What value should you obtain, and do you obtain it?)
- Compute the second derivative $D^2f(x,y)$. Then compute both $D^2f(0,0)$ and $D^2f(1,-1)$, the second derivative at these critical points.
- Classify each critical point as a maximum, minimum, or saddle point, by computing the eigenvalues of the second derivative at that point.
- A rover travels along the curve $4x+y=3$. The elevation near the rover is given by $z=y-x^2$. Use Lagrange multipliers to locate the $(x,y)$ coordinates where the rover reaches maximum height. [Check: $(x,y)= (-2,11)$.]
- Two objects lie on the $x$-axis. The first object has a mass of 2 kg and is located at the point $x=-1$ (or rather its center of mass is at that point). The second object has a mass of 3 kg and is located at the point $x=4$. Find the center-of-mass of the combined system.
- Find the directional derivative of $f(x,y)=xy^2$ at $P=(4,-1)$ in the direction $(-3,4)$. [Check: $D_{(-3,4)}f(4,-1) = \vec\nabla f(4,-1)\cdot \frac{(-3,4)}{5}=(-8,16)\cdot \frac{(-3,4)}{5}=88/5$.]
- 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 $(a,b,c)$.
|
Sun |
Mon |
Tue |
Wed |
Thu |
Fri |
Sat |