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
- 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.
Solution
The center of mass should be close to 4 than -1, because there is more mass there. There are many ways to do this problem. One way involves thinking of the 2kg object as 2 single kg objects, and the 3 kg object as 3 single kg objects. This means there are 5 equally weighted objects. We then average the $x$ coordinates of the 4 objects to get $$\bar x = \frac{(-1)+(-1)+4+4+4}{5} = \frac{(-1)2+(4)3}{2+3} = \frac{10}{5}=2.$$ Note that $\bar x = 2$ is closer to 4 than -1, as expected.
- The elevation is $z=y+x^2$, along a path $y-2x=5$. Find the $(x,y)$ location of any maxes or mins.
Solution
With $f(x,y) = y+x^2$ and $g(x,y)=y-2x=5$, we have $\vec \nabla f = (2x,1)$ and $\vec \nabla g = (-2,1)$. The system we must solve is $$2x=\lambda(-2), 1=\lambda 1, y-2x=5.$$ The middle equation gives $\lambda =1$, which when plugged into the first equation gives $x=-1$. In the last equation, this give $y=3$.
- The elevation is $z=y+x^2$, along the path $\vec r(t) =(t,2t+5)$. Find the $(x,y)$ location of any maxes or mins.
Solution
Since $\vec r(t) =(t,2t+5)$, we know $x=t$ and $y=2t+5$. Substitution gives $z=(2t+5)+(t)^2$. We compute $\frac{dz}{dt} = 2+2t$. This derivative equals zero when $t=-1$, so we know $(x,y) = \vec r(-1) = ((-1),2(-1)+5) = (-1,3)$.
The only difference between this problem and the previous is that in this problem we have a parametrization of the constraint (the path the rover is on). The solutions to the two problems are identical, but the path we followed to get the solution changed. When we have a parametrization, we can substitute and then differentiate. Without a parametrization, Lagrange multipliers is a powerful tool.
- For the function $f(x,y)=x^2+xy+y^2-2y$, determine the location of any maxes, mins, or saddles, and classify each location appropriately using eigenvalues.
Solution
The gradient is $\vec \nabla f(x,y) = (2x+y,x+2y-2)$. The gradient equals zero when $2x+y=0$ and $x+2y-2=0$. The first equation tells us $y=-2x$. Plugging this into the second equation gives $x+2(-2x)=2$, or $x=-2/3$. Back substitution tells us $y=4/3$. The only critical point is hence $(x,y) = (-2/3,4/3)$.
The second derivative is $D^2f(x,y) = \begin{bmatrix}2&1\\1&2\end{bmatrix}$. The matrix does not change at the critical point, so this is also $D^2f(-2/3,4/3)$. The eigenvalues are the solution to the equation $(2-\lambda)(2-\lambda)-1=0$. Note that $$(2-\lambda)(2-\lambda)-1 = \lambda^2-4\lambda+3 = (\lambda-3)(\lambda-1).$$ The eigenvalues are $\lambda = 3$ or $\lambda = 1$. Since both are positive, the gradient points away from the critical point. This means the critical point corresponds to a minimum.
Group problems
- Two objects lie on the $z$-axis. The first object has a mass of 2 kg and is located at the point $z=3$ (or rather its center of mass is at that point). The second object has a mass of 4 kg, and after being placed on top of the first object, its center-of-mass is located at the point $z=6$. Find the center-of-mass of the combined system.
- A box lies inside the rectangle $ [-2,6]\times [1,5] $ (so $-2\leq x\leq 6$ and $1\leq y \leq 5$ ).
- What is the center-of-mass $(\bar x,\bar y)$ of the rectangle? (Where is the geometric center?)
- Compute the integral formula $\ds\frac{\int_{-2}^{6}\int_1^5 x dydx}{\int_{-2}^{6}\int_1^5 1 dydx}.$ [Check: 64/32=2. This gives $\bar x$. ]
- Compute the integral formula $\ds\frac{\int_{-2}^{6}\int_1^5 y dydx}{\int_{-2}^{6}\int_1^5 1 dydx}.$ [Check: 96/32=3. This gives $\bar y$. ]
- Compute the integral formula $\ds\frac{\int_1^5 \int_{-2}^{6}x dxdy}{\int_1^5 \int_{-2}^{6}1 dxdy}$, to verify that swapping the order of integration still yields $\bar x = 2$.
- Draw the region described by the bounds of each integral.
- $\ds\int_{0}^{2}\int_{2x}^{4}dydx$
- $\ds\int_{0}^{4}\int_{0}^{y/2}dxdy$
- $\ds\int_{0}^{3\pi/2}\int_{0}^{2+2\cos\theta}rdrd\theta$
- $\ds\int_{-3}^{3}\int_{0}^{9-x^2}\int_{0}^{5}dzdydx$
- $\ds\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{\sqrt{1-x^2}}dydxdz$
- Set up an integral formula to compute each of the following:
- The mass of a disc that lies inside the circle $x^2+y^2=9$ and has density function given by $\delta = x+10$
- The $x$-coordinate of the center of mass (so $\bar x$) of the disc above.
- The $z$-coordinate of the center-of-mass (so $\bar z$) of the solid object in the first octant (all variables positive) that lies under the plane $2x+3y+6z=6$.
- The $y$-coordinate of the center-of-mass (so $\bar y$) of the same object.
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