Rapid Recall
1. Let $v=u^3$. Also let $x=2u+v$ and $y=u-2v$. Compute $dx$ and $dy$ in terms of $u$ and $du$.Solution
We have $dv = 3u^2du$. This gives $dx=2du+dv = 2du+3u^2du$ and $dy=du-2(3u^2du)$.
2. 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 $$(x,y) = (14,-23)t+(12,-14).$$
3. When we add up lots of little changes in $x$, so $\int_C dx$, what do we get?
Solution
Adding up lots of little changes in $x$ gives the total change in $x$. This means $\int_C dx = x_{final}-x_{initial}$.
4. Draw the region whose area is given by the double integral $$\int_{-1}^{1}\int_{x}^{2-x^2}dydx.$$
Solution
The integral gives us the inequalities $-2\leq x\leq 1$ and $x\leq y\leq 2-x^2$. We need to be below the parabola $y=2-x^2$ and above the line $y=x$. We shade the region bounded between these two curves, with $x$ values between $-1$ and $1$. The curves actually touch at $x=-2$ and $x=1$, so we are missing a tiny bit of area between the curves, namely the part from $x=-2$ to $x=-1$.
Group problems
- Find the area of a triangle with vertices $(1,0)$, $(0,2)$, and $(5,5)$.
- 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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