Rapid Recall
0. I'll start by graphing for you the curve $r=3-2\cos\theta$ in the $xy$-plane.
1. Given that $\ds \frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta-r\sin\theta$ and $\ds \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta+r\cos\theta$, write a vector equation, in the $xy$-plane, for the line tangent to the curve $r=3-2\cos\theta$ at $\theta=\pi/2$ .
Solution
Note that when $\theta=\pi/2$, we have $r=3$. The $(x,y)$ coordinate is $(0,3)$. Since $\frac{dr}{d\theta}=-2\sin\theta$, we get $\ds \frac{dx}{d\theta} = (2\sin\theta)\cos\theta-(3-2\cos\theta)\sin\theta$ and $\ds \frac{dy}{d\theta} = (2\sin\theta)\sin\theta+(3-2\cos\theta)\cos\theta$. At $\theta=\pi/2$, the above gives $\frac{dx}{d\theta}=(2)(0)-(3)(1) = -3$ and $\frac{dy}{d\theta}=(2)(1)+(3)(0) = 2$. Putting this all together, we have $$ \vec r(t) = (x,y) = (-3,2)t+(0,3).$$
2. Write an integral formula for the arc length of the curve $r=3-2\cos\theta$.
Solution
We get $\int_0^{2\pi}\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta$, using the values computed above for $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$. If you want to actually plug all this in, then we have $$\int_0^{2\pi}\sqrt{\left((2\sin\theta)\cos\theta-(3-2\cos\theta)\sin\theta\right)^2+\left((2\sin\theta)\sin\theta+(3-2\cos\theta)\cos\theta\right)^2}d\theta.$$
3. Find the area of a parallelogram whose edges are the vectors $(a,b)$ and $(c,d)$.
Solution
Group problems
- Consider the polar curve $r=\frac{2\theta}{\pi}$.
- Draw the curve (did you get a spiral) for $0\leq \theta\leq 4\pi$.
- Compute $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$
- Set up an integral formula to calculate the length of the curve for $0\leq \theta\leq 2\pi$.
- Find the slope $dy/dx$ at $\theta=\pi$.
- Give an equation of the tangent line at $\theta = \pi$.
- 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: $(x,y) = (-2,9)t+(0,-10)$.]
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