- 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 4 of Unit 3.
Brain Gains (Rapid Recall, Jivin' Generation)
- Using the change-of-coordinates $x=u-v$ and $y=u+v$, write the differential as a (linear combination) $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix} ?\\ ?\end{pmatrix}du+\begin{pmatrix}?\\?\end{pmatrix}dv.$$ This is part of Task 6.
- A parallelogram has vertices $(0,0)$, $(-2,5)$, $(3,4)$, and $(1,9)$. Find its area. This is part of Task 7.
- Set up an iterated integral formula that would give the $x$-coordinate of the centroid (center of mass, assuming constant density) of the triangular region in the first quadrant that lies below the line $\frac{x}{5}+\frac{y}{7}=1$. This is part of Task 8.
- For the integral $\ds\int_0^5\int_2^{x+2}\cos(xy^2) dydx$, swap the order of integration. In other words, rewrite the integral in the form $\ds\int_?^?\int_?^{?}\cos(xy^2) dxdy$. Drawing the region first will help tremendously. This is part of Task 9.
- Using the change of coordinates $x=2u$, $y=3v$, compute the differential $\begin{pmatrix}dx\\dy\end{pmatrix}$. Use this information to compute the area inside the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$? This is part of Task 10, and leads to the definition of the Jacobian.
Group Problems
- Consider the change-of-coordinates $x=r\cos\theta$ and $y=r\sin\theta$.
- Compute the differential $(dx,dy)$ and write it as a linear combination.
- Compute the Jacobian of this change-of-coordinates, so compute the area of the parallelogram whose edges are given by the vectors $(\cos\theta, \sin\theta)$ and $(-r\sin\theta, r\cos\theta)$ (the partial derivatives of the change-of-coordinates). Simplify your answer as much as possible, as we'll use this result quite often.
- Consider the triangle in the first quadrant satisfying $0\leq y\leq 1$ and $0\leq x\leq y$. We'll assume constant density $\delta =1$.
- Draw the triangle, and make a guess where the center of mass is located.
- Compute the integral formula $\bar x = \ds\frac{\int_{0}^{1}\int_0^y x dxdy}{\int_{0}^{1}\int_0^y 1 dxdy}.$ [Check: (1/6)/(1/2)=1/3.]
- Compute the integral formula $\bar y = \ds\frac{\int_{0}^{1}\int_0^y y dxdy}{\int_{0}^{1}\int_0^y 1 dxdy}.$ [Check: (1/3)/(1/2)=2/3.]
- Consider the integral $\ds\int_{0}^{3}\int_{0}^{x}dydx$.
- Shade the region whose area is given by this integral. (You should obtain a triangle whose area is 9/2, the value of the integral above).
- Someone naively tries to swap the order of integration thinking it's as simple as $\ds\int_{0}^{x}\int_{0}^{3}dxdy$. Compute the integral and show that you do not get a number for the area. How can you recognize there will be a problem without even computing any integrals?
- Correctly adjust the bounds on the previous integral (using the order $dxdy$) so that the bounds describe the same region as original integral, making sure the bounds on $y$ are between two constants. In other words, fill in the question marks below so that the integral's bounds describe the same region as the first part of this problem. $$\ds\int_{?}^{?}\int_{?}^{?}dxdy.$$
- 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.
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