Rapid Recall (Brain training - 30 seconds per problem - just do your best - no worries if wrong)
- Let $v=u^2$. Suppose that $x=3u+2v$ and $y=u-4v$. Then $\frac{dx}{du} = 3+2(2u)$ and $\frac{dy}{du} = ???$.
- Suppose $x=3u+2u^2$ and $y=u-4u^2$ for $-1\leq u\leq 2$. Set up an integral to give the arc length of this curve in the $xy$ plane.
- Find the area of a parallelogram whose edges are parallel to the vectors $(3,1)$ and $(2,-4)$.
- Compute the integral $\ds\int_{4x}^{x^2}dy$ (assume $x$ is a constant).
Rapid Recall Solutions
- $\frac{dy}{du} = 1-4(2u)$.
- $\ds\int_{-1}^{2}\sqrt{\left(\frac{dx}{du}\right)^2+\left(\frac{dy}{du}\right)^2}du = \ds\int_{-1}^{2}\sqrt{\left(3+4u\right)^2+\left(1-8u\right)^2}du$.
- $|ad-bc|=|3\cdot (-4)-1\cdot 2| = |-14|=14$.
- $\ds\int_{4x}^{x^2}dy= y\bigg|_{4x}^{x^2} = x^2-4x $.
Group Work
- Find the area of a triangle with vertices $(1,0)$, $(0,2)$, and $(5,5)$.
- Draw the region in the plane described by $-1\leq x\leq 2$ and $x\leq y\leq 4-x^2$.
- Compute the integral $\ds\int_{x}^{4-x^2}dy$ (assume $x$ is a constant).
- Compute the double integral $\ds \int_{-1}^{2}\left(\int_{x}^{4-x^2}dy\right)dx$.
- 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 $\ds \int_{\pi/2}^{\pi}\left(\int_{0}^{5}rdr\right)d\theta$.
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