Did I solve the calculus problem correctly?

[p53ud0 dr34m5]

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.

given $v(t)=13e^{-.02t}sin(t)$:
a) find a(t)
b) find a position equation assuming s(0)=0
c) find the average velocity along 0<t<pi

a) $$a(t)=\frac{dv}{dt}=13e^{-.02t}cos(t)-.26e^{-.02t}sin(t)]$$
a is one of th easy ones. i had trouble on b, but i think i have the correct answer.
b)$$s(t)=\int13e^{-.02t}sin(t)dt$$
i took out the 13 and got:
$$s(t)=13 \int e^{-.02t}sin(t)dt$$
now, i use integration by parts:
$$u=e^{-.02t}~~~du=-.02e^{-.02t}dt~~~dv=sin(t)dt~~~v=-cos(t)$$
$$-e^{-.02t}cos(t)-\frac{1}{50} \int e^{-.02t}cos(t)dt$$
now, i have to integrate by parts again:
$$u=e^{-.02t}~~~du=-.02e^{-.02t}dt~~~dv=cos(t)dt~~~v=sin(t)$$
$$-e^{-.02t}cos(t)-\frac{1}{50}[e^{-.02t}sin(t)+\frac{1}{50} \int e^{-.02t}sin(t)dt]$$
so:
$$-e^{-.02t}cos(t)-\frac{e^{-.02t}sin(t)}{50}+\frac{1}{2500} \int e^{-.02t}sin(t)dt$$
now, i can set the given integral to the one i have and solve for the first one:
$$\inte^{-.02t}sin(t)dt=-e^{-.02t}cos(t)-\frac{e^{-.02t}sin(t)}{50}+\frac{1}{2500} \int e^{-.02t}sin(t)dt$$
so:
$$\frac{2501}{2500} \int e^{-.02t}sin(t)dt = -e^{-.02t}cos(t)-\frac{e^{-.02t}sin(t)}{50}$$
so:
$$\int e^{-.02t}sin(t)dt=\frac{-2500e^{-.02t}cos(t)}{2501}-\frac{2500e^{-.02t}sin(t)}{125050}$$
if you multiply both sides by the 13 that was taken out at the beginning, you get:
$$13\int e^{-.02t}sin(t)dt=-12.998e^{-.02t}cos(t)-.2599e^{-.02t}sin(t)+C$$
now, i can assume that t=0 and s(0)=0 and solve for c:
$$0=-12.998e^{-.02*0}cos(0)-.2599e^{-.02*0}sin(0)+C$$
$$0=-12.998+C$$
$$C=12.998$$
so:
$$s(t)=-12.998e^{-.02t}cos(t)-.2599e^{-.02t}sin(t)+12.998$$?
i think that's right, but I am not sure.
c) $$\overline{v}(t)=\frac{1}{\pi-0}\int_0^\pi v(t)dt=\frac{s(\pi)}{\pi}\approx 8$$

thanks in advance for any verification! :!)

 

[dextercioby]

Use this formula in your case
$$\int e^{ax}\sin bx \ dx=\frac{a\sin bx-b\cos bx}{a^{2}+b^{2}}e^{ax}+C$$

.It can be proven by 2 times part integration.

Daniel.

 

[p53ud0 dr34m5]

we had to use integration by parts, so we couldn't use the formula that you provided. but could you tell me if my work looks right? that's all I am wondering. id appreciate it.

 

[dextercioby]

Your calculations are almost impossible to follow.The purpose of this HW section is not to check arithmetics.The formula that I've provided is the one you should use and,as i said,but you probably didn't read,it can be proven by 2 times integration by parts.

I'm sorry,that's all i can do.I doubt anyone can do more.

Daniel.

 

[p53ud0 dr34m5]

well, when i did the proof, i didnt get what you had. i got:
$$\int e^{ax}\sin bx \ dx=\frac{a\sin bx-b\cos bx}{a^{2}+b^{2}}e^{ax}+C$$

you have $-bcosax$, and i have $-bcosbx$. i probably missed something, but oh well I am ok now. my calculator has the same answer i have.

 

[dextercioby]

Sorry,it was a typo.I misscopied from the paper.It was almost midnight and i was probably tired...

Daniel.

 

[p53ud0 dr34m5]

does this look better?

$$13\int e^{-.02t}sin(t)dt=13(\frac{-.02\sin(t)-cos(t)}{1.0004})e^{-.02}+C$$

$$s(t)=e^{-.02t}(-12.998\cos(t)-.2599\sin(t))+C$$

if we were to assume that s(0)=0, then we could solve for C:

$$0=e^{-.02*0}(-12.998\cos(0)-.2599\sin(0))$$

$$C=12.998$$

$$s(t)=e^{-.02t}(-12.998\cos(t)-.2599\sin(t))+12.998$$

hopefully, you can help me now. if not, i give up.

 

[dextercioby]

Yes,it looks okay.If you applied the correct formula,then it's correct.

The point c) is just a formality.

Daniel.

 

[p53ud0 dr34m5]

thanks, dextercioby, you are of great help! just one more question: to find the average velocity of a function is it
$$\frac{1}{b-a}\int_a^b v(t)dt$$
that shows the displacement and dividing it by the time should give average velocity? i rarely pay attention in class and never look at my book, because i taught myself this stuff 2 years ago, and I am all memory.

 

[dextercioby]

Yes,it is correct.Standard definition for the average of a function on a domain.

Daniel.

 

[p53ud0 dr34m5]

merci tres beaucoup i have another question that I am going to post as a new thread. tackle it if you please.