Brain Gains
- A force given by $\vec F = (y,-x+y)$ acts on an object as it moves along the curve $\vec r(t) =(t^2,t)$ for $-1\leq t\leq 2$. In the integral $\int_C Mdx+Ndy$, state $M$, $N$, $dx$, and $dy$ all in terms of $t$.
Solution
We have $$\int_{-1}^{2} \underbrace{[t]}_{M}\underbrace{[2t]}_{dx/dt}+\underbrace{[-(t^2)+(t)]}_{N}\underbrace{[1]}_{dy/dt}dt.$$
- Set up an integral to find the length of the curve with parametrization $\vec r(t) =(t^2,t)$ for $-1\leq t\leq 2$.
Solution
The velocity is $\vec v(t) = (2t,1)$. The speed is $v = \sqrt{(2t)^2+1^2}$. This means the arc length differential is $$\underbrace{ds}_{\text{little distance}} = \underbrace{\sqrt{(2t)^2+1^2}}_{\text{speed}}\underbrace{dt}_{\text{little time}} = \sqrt{4t^2+1}dt.$$ The arc length is hence $$s=\int_C ds = \int_{-1}^2\sqrt{4t^2+1}dt.$$
- A wire lies along the curve $\vec r(t) =(t^2,t)$ for $-1\leq t\leq 2$. The density (mass per length) of the wire at a point $(x,y)$ on the curve is given by $\delta(x,y) = y+2$, which means little bits of mass are equal to $dm = \delta(x,y) ds$. Set up an integral formula that gives the total mass of the wire.
Solution
The mass is is found by adding up little bits of mass. This gives $$m=\int_C dm = \int_C(y+2) ds = \int_{-1}^2(\underbrace{t}_{y}+2)\underbrace{\sqrt{4t^2+1}dt}_{ds}.$$ We used the fact that $y=t$ from $\vec r(t) =(t^2,t)$ to replace $y+2$ with $(t+2)$.
Group problems
Remember to pass the chalk.
- Consider the curve $C$ parametrized by $\vec r(t) = (t^2, t^3)$ for $0\leq t\leq 2$.
- Give a vector equation of the tangent line to the curve at $t=1$.
- Set up an integral that gives the length of this curve. Just set it up.
- A wire lies along the curve $C$. The density (mass per length) of the wire at a point $(x,y)$ on the curve is given by $\delta(x,y) = y+2$, which means little bits of mass along little length $ds$ are given by $dm = \delta ds$. Set up an integral formula that gives the total mass of the wire.
- The wire contains charged particles. The charge density (charge per length) at a point $(x,y)$ on the curve is given by the product $q(x,y)=xy$, which means little bits of charge along little length $ds$ are given by $dQ = q ds$. Set up an integral formula that gives the total charge on the wire.
- Consider the ray from the origin through the point $P=(-2,2)$. What's the angle between this ray and the positive $x$ axis? What the distance from the origin to $P$?
- Plot the polar points with $(r,\theta)$ given by $(2,0)$, $(2,\pi/6)$, $(-2,\pi/6)$, $(4,\pi/2)$, $(-4,\pi/2)$.
- Give a polar equation of the curve $2x+3y=4$. (So substitute $x=r\cos\theta$ and $y=r\sin\theta$, and then solve for $r$.)
- Give a Cartesian equation of the polar curve $r=\tan\theta\sec\theta$. (Use $x=r\cos\theta$ and $y=r\sin\theta$.)
- We know $x=r\cos\theta$ and $y=r\sin\theta$. Compute $dx$ in terms of $r, \theta,dr, d\theta$. (If you need to, assume that everything depends on $t$, compute derivatives, then multiply by $dt$.)
- Plot the curve $r=3-2\sin\theta$.
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