- I-Learn, Class Pictures, Learning Targets, Text Book Practice
- Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus
This is day 2 of Unit 3.
Brain Gains (Rapid Recall, Jivin' Generation)
- 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.
- A metal plate covers the region in the first quadrant that lies below the parabola $y=9-x^2$. The density (mass per area) at points in the plate is given by the function $\delta = xy$. Set up an iterated double integral that would give the mass of this metal plate.
Solution
The solution is $\ds\iint_Rdm =\iint_R\delta dydx = \int_0^3\int_0^{9-x^2}(xy)dydx$.
Mathematica will compute the integral and draw the region verify our bounds are correct. If the drawn region is NOT what we expected, then we've definitely made a mistake.
outerB = {x, 0, 3};
innerB = {y, 0, 9 - x^2};
density = x y;
mass = Integrate[density, outerB, innerB]
coordinates = {x, y};
ParametricPlot[coordinates, outerB, innerB, Mesh -> {10, 0}]
- Set up iterated integral formulas that would give the center of mass $\bar x$ and $\bar y$ for the region in the previous part.
Solution
The center of mass is the average $x$-value, meaning we use the average value formula with $f(x,y)=x$ being the function we wish to find the average of. We have $\ds\bar x = \frac{\iint_Rxdm}{\iint_Rdm} =\frac{\iint_Rx\delta dydx}{\iint_R\delta dydx} = \frac{\int_0^3\int_0^{9-x^2}x(xy)dydx}{\int_0^3\int_0^{9-x^2}(xy)dydx}$.
The only difference for $\bar y$ is we change the $x$ to a $y$, so that we are finding the average $y$-value, instead of the average $x$-value. This gives $\ds\bar y = \frac{\iint_Rydm}{\iint_Rdm} =\frac{\iint_Ry\delta dydx}{\iint_R\delta dydx} = \frac{\int_0^3\int_0^{9-x^2}y(xy)dydx}{\int_0^3\int_0^{9-x^2}(xy)dydx}$.
We can quickly compute both with Mathematica.
outerB = {x, 0, 3};
innerB = {y, 0, 9 - x^2};
density = x y;
mass = Integrate[density, outerB, innerB]
xbar = Integrate[x*density, outerB, innerB]/mass
ybar = Integrate[y*density, outerB, innerB]/mass
{xbar, ybar} // N
coordinates = {x, y};
ParametricPlot[coordinates, outerB, innerB, Mesh -> {10, 0}]
When a vector appears inside an integral, then we integrate each component separately. This means we could compute both $\bar x$ and $\bar y$ at the same time with $$\ds(\bar x,\bar y) = \frac{\iint_R(x,y)dm}{\iint_Rdm} =\frac{\iint_R(x,y)\delta dydx}{\iint_R\delta dydx} = \frac{\int_0^3\int_0^{9-x^2}(x,y)(xy)dydx}{\int_0^3\int_0^{9-x^2}(xy)dydx} =\left(\frac{\int_0^3\int_0^{9-x^2}x(xy)dydx}{\int_0^3\int_0^{9-x^2}(xy)dydx},\frac{\int_0^3\int_0^{9-x^2}y(xy)dydx}{\int_0^3\int_0^{9-x^2}(xy)dydx}\right) .$$ Continuing with the Mathematica code from above, we can compute both at the same time.
Integrate[{x,y}*density, outerB, innerB]/mass
- Consider the change-of-coordinates given by $x = 2u+v$ and $y=u-2v$. Draw the region in the $xy$-plane given by $1\leq u\leq 2$ and $0\leq v\leq 1$. Suggestion: first complete the table below.
$$\begin{array}{c|c} (u,v)&(x,y)\\\hline (1,0)&(2+0,1-0) = (2,1)\\ (1,1)&(2+1,1-2) = (3,-1)\\ (2,0)&\\ (2,1)& \end{array}$$
Solution
Here's the region, with Mathematica code.
coordinates = {2 u + v, u - 2 v}
R = ParametricRegion[coordinates, {{u, 1, 2}, {v, 0, 1}}];
Region[R, Axes -> True, AxesLabel -> {x, y}, AxesOrigin -> {0, 0}]
- Compute the differential $(dx,dy)$ for the change-of-coordinates given by $x = 2u+v$ and $y=u-2v$. Write your answer as the linear combination $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix}?\\?\end{pmatrix}du + \begin{pmatrix}?\\?\end{pmatrix}dv.$$
Solution
We have $dx = 2du+1dv$ and $dy = du-2dv$. This gives $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix}2\\1\end{pmatrix}du + \begin{pmatrix}1\\-2\end{pmatrix}dv.$$
Mathematica will compute the derivative of the transformation for us (it will be a matrix). To obtain the differential, we just dot the derivative by $(du,dv)$.
coordinates = {2 u + v, u - 2 v}
Dcoordinates = D[coordinates, {{u, v}}]
Dcoordinates // MatrixForm
Dcoordinates . {du, dv}
- State the area of a parallelogram whose edges are the vectors $(a,b)$ and $(c,d)$.
Solution
The prep had you show that the area is $A = |ad-bc|$.
For the 2 by 2 matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$, the determinant of this matrix is $ad-bc$, precisely the quantity inside this area formula.
m = {{a, b}, {c, d}}
m // MatrixForm
Det[{{a, b}, {c, d}}]
% // Abs
- State the area of a parallelogram whose edges are the vectors $(2,1)$ and $(1,-2)$, so the vectors from the differential of the change-of-coordinates $x = 2u+v$, $y=u-2v$. We call this the Jacobian of the change-of-coordinates given by $x = 2u+v$ and $y=u-2v$, written $\ds\frac{\partial(x,y)}{\partial(u,v)}$.
Solution
We have $$\frac{\partial(x,y)}{\partial(u,v)}= A = |ad-bc| = |(2)(-2)-(1)(1)|=|-5|=5.$$
The formula we developed for area is the absolute value of the determinant of the derivative of the change-of-coordinates.
coordinates = {2 u + v, u - 2 v}
Dcoordinates = D[coordinates, {{u, v}}]
Dcoordinates // MatrixForm
Dcoordinates // Det
% // Abs
Group Problems
- Find the area of the parallelogram whose vertices are $(0,0)$, $(2,4)$, $(-3,1)$, and $(-1,5)$. [Check: 14]
- Consider the change of coordinates given by $x = 3u+2v$ and $y=u-4v$.
- Draw the region in the $xy$-plane given by $0\leq u\leq 1$ and $0\leq v\leq 1$.
- Compute the differential of the change-of-coordinates, and write it in the form $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix} ?\\ ?\end{pmatrix}du+\begin{pmatrix}?\\?\end{pmatrix}dv.$$
- Compute the area of the region you drew (the region should be a parallelogram). [You can use the parallelogram area formula.]
- State the Jacobian $\ds\frac{\partial(x,y)}{\partial(u,v)}$ of this transformation.
- 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 $\bar x = \ds\frac{\int_{-2}^{6}\int_1^5 x dydx}{\int_{-2}^{6}\int_1^5 1 dydx}.$ [Check: 64/32=2.]
- Compute the integral formula $\bar y = \ds\frac{\int_{-2}^{6}\int_1^5 y dydx}{\int_{-2}^{6}\int_1^5 1 dydx}.$ [Check: 96/32=3.]
- Find the area of the triangle whose vertices are $(-2,1)$, $(3,4)$, and $(1,7)$. [Check: 21/2]
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