Question

Vector derivative!! Urgent please help!

Answer 1

\[c_1(t)=\langle e^{5t},\sin t,3t^3\rangle\\
c_2(t)=\langle e^{-t},\cos3t,-4t^3\rangle\]
So,
\[c_1\cdot c_2=e^{5t}e^{-t}+\sin t\cos3t+3t^3(-4t^3)\\
c_1\cdot c_2=e^{4t}\sin t\cos 3t-12t^9\]
Notice that it's not a vector, since you take the dot product. This means \(\dfrac{d}{dt}[c_1\cdot c_2]\) is an expression containing \(t\), not a vector.

Answer 2

yeah but the result I get is freaking huge : e^(5t)*-e^-t+sin(t)*-3sin(3t)+3t^3*-12t^2)*(5e^(5t)*e^-t+cos(t)*cos(3t)+9t^2*-4t^3)

Answer 3

well it is the derivative of the vector im taking.

Answer 4

Sorry, should be
\[c_1\cdot c_2=e^{4t}+\sin t\cos 3t-12t^9\]
So the derivative is
\[4e^{4t}+\cos t\cos 3t-3\sin t\sin3t-108t^8\]

Answer 5

but the law state that a(t)*b'(t)+a'(t)*b(t) for any d/dt (a * b) for the dot product assuming a and b are both vectors, didn't you just do something different?

Answer 6

If \(a\) and \(b\) are vectors, then their dot product can't possibly be a vector. You would apply the product rule as you normally would to a function of one variable.

Answer 7

so you apply the product rule after you've simplified the equation ?

Answer 8

Yes.

Answer 9

mhm it is still wrong :/ Ill try to do it again

Answer 10

You probably learned a rule like this one, yes? http://www.proofwiki.org/wiki/Derivative_of_Dot_Product_of_Vector-Valued_Functions

Answer 11

yeah that is the one I learned. The one where d/dt(r(x)*q(x)) = r(x)*q'(x)+r'(x)*q(x)

Answer 12

Right, well I'm not using the rule directly; I stopped at the second line (if you look at the proof section). If you want to apply it directly, you have
\[\begin{align*}\frac{d}{dt}[c_1\cdot c_2]&=c_1'(t)\cdot c_2(t)+c_1(t)\cdot c_2'(t)\\
&=\langle 5e^{5t},\cos t,9t^2\rangle\cdot\langle e^{-t},\cos3t,-4t^3\rangle+\langle e^{5t},\sin t,3t^3\rangle\cdot\langle -e^{-t},-3\sin3t,-12t^2\rangle\\
&=5e^{4t}+\cos t\cos 3t-36t^5-e^{4t}-3\sin t\sin3t-36t^5\\
&=4e^{4t}+\cos t\cos3t-3\sin t\sin3t-72t^5
\end{align*}\]
Turns out I made an algebraic mistake earlier:

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
\[c_1(t)=\langle e^{5t},\sin t,3t^3\rangle\\
c_2(t)=\langle e^{-t},\cos3t,-4t^3\rangle\]
\[c_1\cdot c_2=e^{4t}+\sin t\cos 3t-12t^\color{red}{6}\]
So the derivative is
\[4e^{4t}+\cos t\cos 3t-3\sin t\sin3t-\color{red}{72t^8}\]
\(\color{blue}{\text{End of Quote}}\)

Same as with the rule.

Answer 13

ahh I get it thanks a lot man Im sorry for being so "stupid" ! haha

Answer 14

You got the second part? I think there's a rule for cross product too

Answer 15

Yeah the cross product was easier to do just took much more effort, I found it easier cause I can separate the components.

Answer 16

Alright, and you're welcome!