Brain Gains
- When we add up lots of little areas, so $\int_R dA$, what do we get?
Solution
Total area.
- Finish the following statement: "Adding up lots of little changes in $x$ along a curve $C$, so $\int_C dx$, gives __________."
Solution
The solution is, "Adding up lots of little "changes in $x$" along a curve $C$, so $\int_C dx$, gives "the total change in $x$", or $x_{final}-x_{initial}$."
- Adding up lots of little masses gives the total mass. $m=\int_C dm$
- Adding up lots of little areas gives the total area. $A=\int_C dA$
- Adding up lots of little length gives the total length. $s=\int_C ds$
- Adding up lots of little charges gives the total charge. $Q=\int_C dQ$
- Adding up lots of little forces gives the total force. $F=\int_C dF$
- Adding up lots of little work gives the total work. $W=\int_C dW$
- Adding up lots of little widths gives the total width. $width=\int_C dx$
- Adding up lots of little heights gives the total height. $height=\int_C dy$
- Adding up lots of little "changes in $x$" gives the total "change in $x$." The words and the concepts generalize perfectly. Unfortunately, the notation does not generalize perfectly in this instance (as we think of $x$ as both a number and a vector in the same phrase).
- Adding up lots of little "changes in $y$" gives the total "change in $y$." $\text{total change in y}=\int_C dy$
- Adding up lots of little "changes in time" gives the total "change in time." $\text{total change in time}=\int_C dt$
- 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 work integral $$W=\int_C dW = \int \vec F\cdot d\vec r = \int_C Mdx+Ndy,$$ state $M$, $N$, $dx$, and $dy$ all in terms of $t$.
Solution
We have
- $x = t^2$,
- $y=t$,
- $M = y = t$,
- $N = -x+y = -t^2+t$,
- $dx = \frac{dx}{dt}dt = (2t)dt$, and
- $dy = \frac{dy}{dt}dt = (1)dt$.
The work done by $F$ along $C$ is $$W = \int_{-1}^{2} \underbrace{[t]}_{M}\underbrace{[2t]}_{dx/dt}+\underbrace{[-(t^2)+(t)]}_{N}\underbrace{[1]}_{dy/dt}dt.$$
- 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 $(t+2)$.
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