Indefinite integral of vector function

[wizard85]

Homework Statement

If $${\vec{V}(t)$$ is a vector function of $$t$$, find the indefinite integral:

$$\int (\vec{V}\times \frac{d^2\vec{V}}{dt^2}) \,dt$$

Homework Equations

The Attempt at a Solution

I have solved it by decomposing and integrating each terms of vector $$\vec{V}\times \frac{d^2t}{dt^2}$$, so I have:

$$\vec{V}\times \frac{d^2\vec{V}}{dt^2}= \hat i (V_y * \frac{d^2V_z}{dt^2} - V_z * \frac{d^2V_y}{dt^2}) - \hat j (V_x * \frac{d^2V_z}{dt^2} - V_z * \frac{d^2V_x}{dt^2}) + \hat k (V_x * \frac{d^2V_y}{dt^2} - V_y * \frac{d^2V_x}{dt^2})$$

and the integral will be:

$$<br /> \int (\vec{V}\times \frac{d^2\vec{V}}{dt^2}) \,dt = \hat i (\int (V_y * \frac{d^2V_z}{dt^2})\,dt - \int (V_z * \frac{d^2V_y}{dt^2})\,dt) - \hat j (\int (V_x * \frac{d^2V_z}{dt^2})\,dt - \int (V_z * \frac{d^2V_x}{dt^2}))\,dt + \hat k (\int (V_x * \frac{d^2V_y}{dt^2})\,dt - \int (V_y * \frac{d^2V_x}{dt^2}),dt)<br />$$where $$\hat i$$,$$\hat j$$ and $$\hat k$$ are the three unit vectors with respect to frame of reference $$(x,y,z)$$My questions is: that's the right way or there exist some other way to resolve it in a more faster manner?
Thanks to all who will answer me ;)

 

[Mark44]

Why do you have the cross product with
$$\frac{d^2t}{dt^2}?$$

Surely this is a mistake.

 

[wizard85]

Why do you have the cross product with
$$\frac{d^2t}{dt^2}?$$

Surely this is a mistake.

Yeah, sorry. Really the integral was

$$\int (\vec{V}\times \frac{d^2\vec{V}}{dt^2}) \,dt$$

In fact it includes a vector as a result of the cross product whose components are function of t. Does anybody have any idea?

 

[Mark44]

OK, that makes more sense. For your question, it might be that there is a simpler manner to approach this, but I don't know it. What you're doing is what I would do.

 

[tiny-tim]

erm …

Hint: what is the derivative of V x dV/dt?

 

[wizard85]

erm …

Hint: what is the derivative of V x dV/dt?

$$<br /> \frac{d(\vec{V}\times \frac{d\vec{V}}{dt})}{dt} = \vec{V}\times \frac{d^2\vec{V}}{dt^2} + \frac{d\vec{V}}{dt} \times \frac{d\vec{V}}{dt} = \vec{V}\times \frac{d^2\vec{V}}{dt^2}<br />$$

so result of the integral is $$\vec{V}\times \frac{d\vec{V}}{dt}$$, is it exact? :)

 

[tiny-tim]

so result of the integral is $$\vec{V}\times \frac{d\vec{V}}{dt}$$, is it exact? :)

well, plus a constant vector!

why? … is something worrying you? ​

 

[wizard85]

well, plus a constant vector!

why? … is something worrying you? ​

Thanks for the help... ;)

 

[selms05]

wizard85 is there anyway you can expand on how you took the derivative of the cross product of V and dV/dt ? I am doing the same problem and can not bridge that gap...

 

[tiny-tim]

hi selms05!

the product rule applies to a cross product (or a dot product) in exactly the same way as to an ordinary product