Rapid Recall 
1. Find the area of a parallelogram whose edges are the vectors $(-1,3)$ and $(4,2)$.
Solution
Remember the formula $A=|ad-bc|$. Here we have $(a,b) = (-1,3)$ and $(c,d)=(4,2)$, so we get $$A = |(-1)(2)-(3)(4)|=14.$$
2. Fill in the missing pieces in $$\int_{-1}^{2}e^{5x}dx = \int_{?}^{?}e^{u}(?)du.$$
Solution
Note that $u=5x$ so we have $du=5x$ or $dx=\frac15 dx$. In addition, when $x=-1$ we have $u=-5$, and when $x=2$ we have $u=10$. This gives $$\int_{-1}^{2}e^{5x}dx = \int_{-5}^{10}e^{u}(\frac{1}{5})du.$$
Solution
We have $dv = 3u^2du$. This gives $dx=2du+dv = 2du+3u^2du$ and $dy=du-2(3u^2du)$.
4. Continuing from above, give a vector equation of the tangent line to the curve $v=u^3$ in the $xy$-plane at $u=2$.
Solution
Note that $u=2$ means $v=8$, so we have $x=2u+v=12$ and $y=u-2v=-14$. We also have $\frac{dx}{du} = 2+3u^2 =14$ and $\frac{dy}{du}=1-6u^2 = -23$. The line passes through $(12,-14)$ and is parallel to $(14,-23)$. The requested equation is $$(x,y) = (14,-23)t+(12,-14).$$
Group problems
- Find the area of a triangle with vertices $(1,0)$, $(0,2)$, and $(5,5)$. (Hint: how does a parallelogram relate to a triangle?)
- Consider the curve $v=u^2$ and use the change-of-coordinates $x=2u+v$, $y=u-2v$.
- Draw the curve in both the $uv$-plane and the $xy$-plane.
- Find the slopes $\frac{dv}{du}$ and $\frac{dy}{dx}$ at $u=-2$.
- Give a vector equation of the tangent line to the curve in the $uv$-plane at $u=-2$. [Check: $(u,v) = (1,-4)t+(-2,4)$.]
- Give a vector equation of the tangent line to the curve in the $xy$-plane at $u=-2$. [Check: $(x,y) = (-2,9)t+(0,-10)$.]
- Draw the region in the $xy$-plane described by $-2\leq x\leq 1$ and $x\leq y\leq 2-x^2$.
- Compute the integral $\ds\int_{x}^{2-x^2}dy$ (assume $x$ is a constant).
- Compute the double integral $\ds \int_{-2}^{1}\left(\int_{x}^{2-x^2}dy\right)dx$.
- Draw the region in the $xy$ plane described by $\pi/2\leq \theta \leq \pi$ and $0\leq r\leq 5$.
- Compute the integral $\ds\int_{0}^{5}rdr$.
- Compute the double integral $\ds \int_{\pi/2}^{\pi}\left(\int_{0}^{5}rdr\right)d\theta$.
- Draw the region in the plane described by $-3\leq y\leq 2$ and $y\leq x\leq 6-y^2$.
- Compute the integral $\ds\int_{y}^{6-y^2}dx$ (assume $y$ is a constant).
- Compute the double integral $\ds \int_{-3}^{2}\left(\int_{y}^{6-y^2}dx\right)dy$.
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi$ and $2\leq r\leq 5$.
- Compute the double integral $\ds \int_{0}^{\pi}\left(\int_{2}^{5}rdr\right)d\theta$.
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi/3$ and $0\leq r\leq 2\sin3\theta$.
- Compute the double integral $\ds \int_{0}^{\pi/3}\left(\int_{0}^{2\sin 3\theta}rdr\right)d\theta$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside the cardiod $r=2+2\cos\theta$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside one petal of the rose $r=3\cos2\theta$.
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