Pacing Tracker
- The quizzes have included questions for 10 objectives. Have you passed at least 80% of them? What are you plans to master those you have't mastered yet?
- We've finished units 1 and 2. Have you started your self-directed learning project for each unit? Ideally you've finished your first SDL, and started (or working on a proposal for) the second.
- Remember you can submit only one SDL project per week, so plan ahead and don't let yourself get behind.
Brain Gains
- Graph the curve $r=3-2\cos\theta$ in the $xy$-plane (in the $r\theta$-plane can be useful).
Solution
We'll do this one together.
- 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 $$\ds \frac{dx}{d\theta} = (2\sin\theta)\cos\theta-(3-2\cos\theta)\sin\theta\text{ and }\ds \frac{dy}{d\theta} = (2\sin\theta)\sin\theta+(3-2\cos\theta)\cos\theta.$$ At $\theta=\pi/2$, the above gives $$\frac{dx}{d\theta}\bigg|_{\theta = \pi/2}=(2)(0)-(3)(1) = -3\text{ and }\frac{dy}{d\theta}=(2)(1)+(3)(0) = 2.$$ Putting this all together, we have $$ (x,y) = (-3,2)t+(0,3)\text{ or }\vec r(t) = (-3,2)t+(0,3).$$
- Write an integral formula for the arc length of the curve $r=3-2\cos\theta$.
Solution
Arc length in general is $\int_\alpha^{\beta}\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta$. From our graph, we discover that we need $0\leq\theta\leq \beta$ for our bounds to trace over the curve once. The values were computed above for $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$. This gives the arc length as $$s=\int_0^{2\pi}\sqrt{\left((2\sin\theta)\cos\theta-(3-2\cos\theta)\sin\theta\right)^2+\left((2\sin\theta)\sin\theta+(3-2\cos\theta)\cos\theta\right)^2}d\theta.$$
Group problems
- Consider the polar curve $r=\frac{2\theta}{\pi}$. Graph is a spiral See Desmos.
- Compute $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$. (Hint: They appear in the integral in the next part.)
- What quantity does the following integral calculate? $$ \int_0^{4\pi}\sqrt{\left(\frac{2}{\pi}\cos\theta-\frac{2\theta}{\pi}\sin\theta\right)^2+\left(\frac{2}{\pi}\sin\theta+\frac{2\theta}{\pi}\cos\theta\right)^2}d\theta$$
- Find the slope $dy/dx$ at $\theta=\pi$. [Check: $\frac{dy}{dx} = \frac{-2}{-2/\pi} = \pi$.]
- 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.
- Compute $dx$ and $dy$ in terms of $u$ and $du$ (we know $dv = 2udu$ since $v=u^2$).
- 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: $(x,y) = (-2,9)t+(0,-10)$.]
- Let $v=u^3$ and use the coordinates $x=2u+v$, $y=u-2v$.
- Draw the curve in both the $uv$-plane, and the $xy$-plane (make a $(u,v)$ and $(x,y)$ table).
- Find $dx$ and $dy$ in terms of $u$ and $du$.
- Find the slope $dy/dx$ at $u=1$.
- Give a vector equation of the tangent line to the curve in the $xy$ plane at $u=1$.
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