Presentations
5.5 -5.12Pacing Tracker
- The quizzes have included questions for 22 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 4. Have you started your self-directed learning project for each unit?
- The 6th project can be over any topic from the entire semester. Feel free to get started on this one as soon as you have an idea.
- Remember you can submit only one SDL project per week. Plan ahead and don't let yourself get behind.
Brain Gains
- Consider the two vectors $\vec u = (a,b,c)$ and $\vec v = (x,y,z)$. Compute $\vec u\cdot \vec v$. What must this equal if the two vectors are orthogonal?
Solution
We have $\vec u\cdot \vec v = ax+by+cz$. If the vectors are orthogonal, then this dot product equals zero.
- Use the cross product formula $\vec u \times \vec v = \left<u_2 v_3 - u_3 v_2, u_3 v_1 - u_1 v_3, u_1 v_2 - u_2 v_1\right>$ to compute the cross product of $\vec u = (a,b,c)$ and $\vec v = (d,e,f)$.
Solution
The cross product is $(bf-ce, cd-af, ae-bd)$.
There are many pneumonic devices people have invented to remember this. The most common is to use a 3D determinant computation, but I find many introductory students always forget to negate the middle term when using this formula.
My favorite pneumonic device is to write the first vector horizontally twice, underneath these write the second vector twice, ignore the two ends ends, and then compute the area of a 2D parallelogram three times. We denote the area of a 2D parallelogram with edges $(b,c)$ and $(e,f)$ using the symbol $\begin{vmatrix}b&c\\e&f\end{vmatrix}$ (vertical edges, not brackets or paranthesis). Here's my pneumonic device for doing the cross product by hand. $$\begin{array}{ccccccr} (a,& b,& c )&(a,& b,& c)&(\text{write first vector twice})\\ (d,& e,& f )&(d,& e,& f)&(\text{put second vector below})\\ &(\begin{vmatrix}b&c\\e&f\end{vmatrix},&\begin{vmatrix}c&a\\f&d\end{vmatrix},&\begin{vmatrix}a&b\\d&e\end{vmatrix})&&&(\text{ignore ends, find 3 areas})\\ &(bf-ce,&cd-af,&ae-bd)&&&(\text{simplify}) \end{array} $$
In the end, you can always just look it up or have software compute it.
- A wire lies along the curve $\vec r(t) =(t^3, 3t^2)$ for $0\leq t\leq 7$ and has a uniform density of $\delta = 5$, set up an integral formula that gives the mass of this curve.
Solution
Remember that mass is found by adding up little masses, so $m = \int_C dm$. A little mass $dm$ is equal to density $\delta$ multiplied by little length $ds$, which gives $dm = \delta ds$. We found $ds$ earlier in the semester by multiplying speed by a little time. The velocity is $\frac{d\vec r}{dt} = (3t^2,6t)$ which means the speed is $v(t) = \sqrt{(3t^2)^2+(6t)^2}$. The gives $$m = \int_C dm= \int_C \delta ds = \int_C 5 v(t) dt = \int_0^7 5 \sqrt{(3t^2)^2+(6t)^2}dt.$$
- Consider the region $R$ that is bounded by the lines $y=0$, $x=4$, and $y=x/2$. The density (mass per area) is given by $\delta(x,y) = x+y^2$.
- Draw the region.
- Set up a double integral to compute the mass using $\ds\int_{?}^{?}\int_{?}^{?}\delta(x,y) dydx$.
- Set up a double integral to compute the mass using $\ds\int_{?}^{?}\int_{?}^{?}\delta(x,y) dxdy$.
- Set up a double integral to compute the $x$-coordinate of the center of mass.
- Set up a double integral to compute the $y$-coordinate of the center of mass.
Solutions
The graph is a triangle underneath the line $y=x/2$ for $0\leq x\leq 4$. The requested integrals are
- $\ds\int_{0}^{4}\int_{0}^{x/2}x+y^2 dydx$.
- $\ds\int_{0}^{2}\int_{2y}^{4}x+y^2 dxdy$.
For the last two, you can use either of the above sets of bounds. For the $x$ coordinate of the center-of-mass, we have $$\ds\bar x = \frac{\iint_R x dm}{\iint_R dm} = \frac{\iint_R x \delta dA}{\iint_R \delta dA} = \frac{\int_{0}^{4}\int_{0}^{x/2}(x)(x+y^2) dydx}{\int_{0}^{4}\int_{0}^{x/2}x+y^2 dydx} = \frac{\int_{0}^{2}\int_{2y}^{4}(x)(x+y^2) dxdy}{\int_{0}^{2}\int_{2y}^{4}x+y^2 dxdy}.$$ For the $y$ coordinate of the center-of-mass, we have $$\ds\bar y = \frac{\iint_R y dm}{\iint_R dm} = \frac{\iint_R y \delta dA}{\iint_R \delta dA} = \frac{\int_{0}^{4}\int_{0}^{x/2}(y)(x+y^2) dydx}{\int_{0}^{4}\int_{0}^{x/2}x+y^2 dydx} = \frac{\int_{0}^{2}\int_{2y}^{4}(y)(x+y^2) dxdy}{\int_{0}^{2}\int_{2y}^{4}x+y^2 dxdy}.$$
Group Discussion
- Consider the integral $\ds\int_{0}^{3}\int_{0}^{x}dydx$.
- Shade the region whose area is given by this integral.
- Compute the integral.
- Now compute $\ds\int_{0}^{x}\int_{0}^{3}dxdy$. Do you get a single number or an expression involving a variable?
- Adjust the bounds on the previous integral (keeping the order $dxdy$) so that the bounds describe the same region as original integral, but make 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. $$\ds\int_{?}^{?}\int_{?}^{?}dxdy$$
- A parallelogram has edge lengths of $a$ and $b$. The acute angle in the parallelogram is $\theta$. Explain why the area of the parallelogram is $ab\sin\theta$. [Hint: Draw a picture, label the edges, add an extra line to form a right triangle with $\theta$ as one of the angles.]
- Draw the region described by the bounds of each integral. (Use the Mathematica notebook Integration.nb to check your work.)
- $\ds\int_{0}^{3\pi/2}\int_{0}^{2+2\cos\theta}rdrd\theta$
- $\ds\int_{-3}^{3}\int_{0}^{9-x^2}\int_{0}^{5}dzdydx$
- $\ds\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{\sqrt{1-x^2}}dydxdz$
- 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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