Let d^2A/dt^2=6ti-2… - QuestionCove
OpenStudy (anonymous):

Let d^2A/dt^2=6ti-24t^2j+4sintk. Find A given that A=2i+j and dA/dt=-i-3k at t=0 (VECTOR ANALYSIS SUBJECT)

4 years ago
OpenStudy (queelius):

You said "Find A given that A = 2i + j". I found A -- it is 2i + j. :) I assume you meant to say "given that A(0), or some other input, equals 2i + j? Integrating d^2A/dt^2 to get dA/dt, we get: dA/dt = <3t^2 + c1, 8t^3 + c2, -4cost + c3>. At t = 0, dA/dt = <-1,0,-3>, so... 3(0)^2 + c1 = -1 Finish the rest. Then, when we've figured out what dA/dt is, we'll move on to integrating that to get A.

4 years ago
OpenStudy (anonymous):

*but i don't know the solution

4 years ago
OpenStudy (queelius):

First, solve for c1 in my question above. Then, we'll solve c2 and c3. At that point, we'll have dA/dt. Then we can move on to integrating that to get A.

4 years ago
OpenStudy (anonymous):

But that's the question. "Find A given that..." the answer is A= $(t^2−t+2)i+(1−2t^4)j+(t−4sint)k$

4 years ago
OpenStudy (queelius):

3(0)^2 + c1 = -1. We're just trying to figure out what the constant of integration is for the first component in the dA/dt vector.

4 years ago
OpenStudy (queelius):

c1 = -1. What does c2 equal? Remember that dA/dt = <3t^2 + c1, 8t^3 + c2, -4cost + c3>. And, at t = 0, dA/dt = <-1,0,-3>, so... 8t^3 + c2 = what?

4 years ago
OpenStudy (anonymous):

c2=0 ?

4 years ago
OpenStudy (queelius):

Oh, yeah, I meant to say 8(0)^3 + c2 = 0, in which case you are correct, c2 = 0. However, correct, c3 does not equal 3/4. Here's what we're solving: -4cos(0) + c3 = -3. Solve for c3.

4 years ago
OpenStudy (anonymous):

ah yahh, c3=1

4 years ago
OpenStudy (queelius):

Correct. So, what is dA/dt, given that the constants of integration are <-1, 0, 1>?

4 years ago
OpenStudy (anonymous):

$t^3-t-2t^4-4sint+t$

4 years ago
OpenStudy (queelius):

Maybe you're jumping ahead, I'm not sure. But, we know that dA/dt is <3t^2 + c1, 8t^3 + c2, -4cost + c3>, and we solved the constants c1, c2, and c3. So, plugging those in, we get: dA/dt = <3t^2 -1, 8t^3, -4cost + 1>.

4 years ago
OpenStudy (queelius):

Next, we'll repeat the same process of taking the derivative and then solving for the constants of integration to get A as we did to get dA/dt.

4 years ago
OpenStudy (queelius):

First, could you integrate dA/dt for me?

4 years ago
OpenStudy (queelius):

Ok, it looks like that's what you've done. Have you also solved for the constants of integration for A?

4 years ago
OpenStudy (anonymous):

is it correct? t^3−t−2t^4−4sint+t ?

4 years ago
OpenStudy (queelius):

4 years ago
OpenStudy (queelius):

We know that A(0) = <2, 1, 0>, and we know that A(t) = <t^3 - t + c1, t^4/2 + c2, -4sint + t + c3>. So, ...what's next?

4 years ago
OpenStudy (anonymous):

substitute the values of the constants to get the answer?

4 years ago
OpenStudy (queelius):

Yes. What are those constants? t^3 - t + c1 = 2 t^4/2 + c2 = 1 -4sint + t + c3 = 0

4 years ago
OpenStudy (anonymous):

and then?

4 years ago
OpenStudy (anonymous):

thanks i got it :)

4 years ago
OpenStudy (queelius):

Once you've solve for the constants, substitute them into A(t). The constant vector is <2, 1, 0>, so A(t) = <t^3-t+2, t^4/2+1, -4sint+t>.

4 years ago
OpenStudy (queelius):

Good work.

4 years ago
OpenStudy (anonymous):

okay thanks a lot :) good night and God Bless :)

4 years ago