Brain Gains
- For the vector field $\vec F(x,y) = (x+y, x^2)$ state $M$ and $N$.
Solution
The notation we'll be using is $\vec F(x,y)=(M,N)$ or $\vec F(x,y,z) = (M,N,P)$. This gives
- $M=x+y$
- $N=x^2$.
- For the curve $\vec r(t) = (t^2, t^3)$, state $x$, $y$, $dx$, $dy$, and $ds$ in terms of $t$.
Solution
We have
- $x = t^2$
- $y = t^3$
- $dx=2t dt$ (the differential of $x$ equals $2t$ multiplied by the differential of $t$. )
- $dy=3t^2 dt$ (the $dt$ here matters. We have $\frac{dy}{dt} = 3t^2$, and we have $dy=3t^2 dt$.)
- $ds = \sqrt{(2t)^2+(3t^2)^2}dt$ - This is the quantity we use for arc length.
Note that $dx$ is the differential of $x$, whereas $\frac{dx}{dt}$ is the derivative of $x$ with respect to $t$.
- Given $\vec F(x,y) = (x+y, x^2)$ and $\vec r(t) = (t^2, t^3)$ state $\vec F(\vec r(t))$.
Solution
Recall that $\vec r(t) = (t^2, t^3)$ means $x=t^2$ and $y=t^3$, so $$\vec F(\vec r(t)) = \vec F(x=t^2, y=t^3) = (t^2+t^3, (t^2)^2).$$
- Set up the work integral $\int_C Mdx+Ndy$ for $\vec F(x,y) = (x+y, x^2)$ and $\vec r(t) = (t^2, t^3)$ for $t\in [-1,3] $ (so replace each term with what it equals in terms of $t$).
Solution
We have the following:
- $M=x+y = t^2+t^3$,
- $N=x^2 = (t^2)^2$,
- $dx = 2tdt$,
- $dy = 3t^2dt$.
This gives $$\int_C Mdx+Ndy =\int_{-1}^3 \underbrace{(t^2+t^3)}_{M}\underbrace{(2tdt)}_{dx}+\underbrace{(t^2)^2}_{N}\underbrace{(3t^2dt)}_{dy} .$$
- Set up the work integral $\int_C \vec F\cdot d\vec r$ for $\vec F(x,y) = (3xy, x+2y)$ and $\vec r(t) = (t^2+1, 2t)$ for $t\in [0,2] $.
Solution
We have the following:
- $F(\vec r(t)) = (3(t^2+1)(2t), (t^2+1)+2(2t))$,
- $\frac{d\vec r}{dt}=(2t,2)$,
- $d\vec r=(2t,2)dt$.
This gives $$\int_C \vec F\cdot d\vec r =\int_{0}^2 \left[\underbrace{(3(t^2+1)(2t))}_{M}\underbrace{(2t)}_{dx/dt}+\underbrace{((t^2+1)+2(2t))}_{N}\underbrace{(2)}_{dy/dt}\right]dt .$$
Group problems
- Use the arc length formula and the parameterization $\vec r(t)=(a\cos t,a\sin t)$ of a circle to verify that the circumference of a circle of radius $a$ is $2\pi a$. $$s=\int_?^? \sqrt{(?)^2+(?)^2}dt.$$
- A force given by $\vec F = (y,-x+y)$ acts on an object as it moves along the curve $\vec r(t) =(t,t^2)$ for $0\leq t\leq 2$. Note that this means $x=t$ and $y=t^2$, so $\vec F = (t^2,-t+t^2)$. Compute $\ds \int_C\vec F\cdot \frac{d\vec r}{dt}dt$ (the work done by the force along the curve).
- Consider the curve $C$ parametrized by $\vec r(t) = (t, t^2)$ for $0\leq t\leq 2$.
- Give a vector equation of the tangent line to the curve at $t=1$ (fill in the blanks below): $$\begin{pmatrix}?\\?\end{pmatrix} = \begin{pmatrix}?\\?\end{pmatrix}t+\begin{pmatrix}?\\?\end{pmatrix}.$$
- Set up an integral that gives the length of this curve. Just set it up (fill in the blanks below). $$s=\int_?^? \sqrt{(?)^2+(?)^2}dt.$$
- 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 $\delta(x,y) = y+2$. This means a little bits of mass along a small length $ds$ is given by $$dm = \delta ds = (y+2)ds = (y(t)+2)\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt.$$ Set up an integral formula that gives the total mass of the wire (put everything in terms of $t$.
- The wire contains charged particles. The charge density (charge per length) at a point $(x,y)$ on the curve is given by the product $q(x,y)=xy$. This means the charge along a little length $ds$ is given by $dQ = (q) ds = (xy)ds$. Set up an integral formula that gives the total charge on the wire.
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