i had this problem on my calculus test the other day that was actually challenging. the whole class tried to get it taken off of the test, because barely any of them did it. :sad: i was wanting to know if i approached it right and yielded the correct answer.(adsbygoogle = window.adsbygoogle || []).push({});

given [itex]v(t)=13e^{-.02t}sin(t)[/itex]:

a) find a(t)

b) find a position equation assuming s(0)=0

c) find the average velocity along 0<t<pi

a) [tex]a(t)=\frac{dv}{dt}=13e^{-.02t}cos(t)-.26e^{-.02t}sin(t)][/tex]

a is one of th easy ones. i had trouble on b, but i think i have the correct answer.

b)[tex]s(t)=\int13e^{-.02t}sin(t)dt[/tex]

i took out the 13 and got:

[tex]s(t)=13 \int e^{-.02t}sin(t)dt[/tex]

now, i use integration by parts:

[tex]u=e^{-.02t}~~~du=-.02e^{-.02t}dt~~~dv=sin(t)dt~~~v=-cos(t)[/tex]

[tex]-e^{-.02t}cos(t)-\frac{1}{50} \int e^{-.02t}cos(t)dt[/tex]

now, i have to integrate by parts again:

[tex]u=e^{-.02t}~~~du=-.02e^{-.02t}dt~~~dv=cos(t)dt~~~v=sin(t)[/tex]

[tex]-e^{-.02t}cos(t)-\frac{1}{50}[e^{-.02t}sin(t)+\frac{1}{50} \int e^{-.02t}sin(t)dt][/tex]

so:

[tex]-e^{-.02t}cos(t)-\frac{e^{-.02t}sin(t)}{50}+\frac{1}{2500} \int e^{-.02t}sin(t)dt[/tex]

now, i can set the given integral to the one i have and solve for the first one:

[tex]\inte^{-.02t}sin(t)dt=-e^{-.02t}cos(t)-\frac{e^{-.02t}sin(t)}{50}+\frac{1}{2500} \int e^{-.02t}sin(t)dt[/tex]

so:

[tex]\frac{2501}{2500} \int e^{-.02t}sin(t)dt = -e^{-.02t}cos(t)-\frac{e^{-.02t}sin(t)}{50}[/tex]

so:

[tex]\int e^{-.02t}sin(t)dt=\frac{-2500e^{-.02t}cos(t)}{2501}-\frac{2500e^{-.02t}sin(t)}{125050}[/tex]

if you multiply both sides by the 13 that was taken out at the beginning, you get:

[tex]13\int e^{-.02t}sin(t)dt=-12.998e^{-.02t}cos(t)-.2599e^{-.02t}sin(t)+C[/tex]

now, i can assume that t=0 and s(0)=0 and solve for c:

[tex]0=-12.998e^{-.02*0}cos(0)-.2599e^{-.02*0}sin(0)+C[/tex]

[tex]0=-12.998+C[/tex]

[tex]C=12.998[/tex]

so:

[tex]s(t)=-12.998e^{-.02t}cos(t)-.2599e^{-.02t}sin(t)+12.998[/tex]?

i think thats right, but im not sure.

c) [tex]\overline{v}(t)=\frac{1}{\pi-0}\int_0^\pi v(t)dt=\frac{s(\pi)}{\pi}\approx 8[/tex]

thanks in advance for any verification!! :!!)

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