- What motivates people?
Solution
Rapid Recall
- The points $P=(1,2,3)$ and $Q=(x,y,z)$ lie in a plane. A normal vector to this plane is $\vec n = (4,5,6)$. Give a single number for the dot product $\vec {PQ}\cdot \vec n$, in other words fill in the question mark below. $$4(x-1)+5(y-2)+6(z-3)=?$$
Solution
The vectors $\vec {PQ}$ and $\vec n$ are orthogonal so their dot product is zero. This gives $$4(x-1)+5(y-2)+6(z-3)=0,$$ which is an equation of the plane containing $P$ with normal vector $\vec n$.
- For the change of coordinates $x = 2u+3v$ and $y=4u+5v$ we have $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix}2\\4\end{pmatrix}du+\begin{pmatrix}3\\5\end{pmatrix}dv.$$ What is the Jacobian of this transformation, in other words fill in the blank below: $$\iint_{R_{xy}} dxdy = \iint_{R_{uv}}(?)dudv.$$
Solution
The Jacobian is the area of the parallelogram formed by the partial derivatives $(2,4)$ and $(3,5)$, namely $J=\dfrac{\partial(x,y)}{\partial(u,v)} = |2\cdot 5-3\cdot 4| = 2$. This means $$\iint_{R_{xy}} dxdy = \iint_{R_{uv}}(2)dudv.$$
- For the curve $\vec r(t)=(4t^2, 5t-1,t^3)$, the velocity is $\vec v(t) =(8t,5,3t^2)$ and speed is $v(t) = \sqrt{(8t)^2+(5)^2+(3t^2)^2}$. A wire lies along this curve for $1\leq t\leq 4$, and has density function given by $\delta = xy$. Set up an integral that gives $\bar y$.
Solution
Remember that $ds = vdt = \sqrt{(8t)^2+(5)^2+(3t^2)^2} dt$. We compute $$\bar y =\frac{\int_C y dm}{\int_C dm} =\frac{\int_C y \delta ds}{\int_C \delta ds} =\frac{\int_C y (xy) ds}{\int_C (xy) ds} =\frac{\int_1^4 (5t-1) (4t^2(5t-1)) \sqrt{(8t)^2+(5)^2+(3t^2)^2} dt}{\int_1^4 (4t^2(5t-1)) \sqrt{(8t)^2+(5)^2+(3t^2)^2} dt}. $$
Group problems
- Let $P=(3,0,0)$, $Q=(0,2,0)$, and $R=(0,0,1)$.
- Find a vector that is orthogonal to both $\vec{PQ}$ and $\vec {PR}$. Call it $\vec n$.
- Find the area of the triangle $\Delta PQR$.
- Let $S=(x,y,z)$ be any point on the plane PQR. What is $\vec {PS}\cdot \vec n$ (you should get a number)? Then compute and expand the dot product, to get an equation of the tangent plane.
- A wire lies along the curve $C$ parametrized by $\vec r(t) = (t^2+1, 3t, t^3)$ for $-1\leq t\leq 2$.
- Compute $ds$. (Remember - a little distance equals the product of the speed and a little time.)
- Set up an integral to find $\bar x$, then $\bar y$, then $\bar z$, for the centroid of $C$.
- 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 $2x+3y+6z=6$ (Draw the region as well).
- The $y$-coordinate of the center-of-mass (so $\bar y$) of the same 3D solid.
- Draw the region described the bounds of each integral.
- $\ds\int_{0}^{3}\int_{0}^{\pi}\int_{0}^{5}rdzdrd\theta$
- $\ds\int_{-1}^{1}\int_{0}^{1-y^2}\int_{0}^{x}dzdxdy$
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