# Indefinite integral of vector function

1. Feb 2, 2009

### wizard85

1. The problem statement, all variables and given/known data

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

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

2. Relevant equations

3. 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:

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

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 ;)

Last edited: Feb 2, 2009
2. Feb 2, 2009

### Staff: Mentor

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

Surely this is a mistake.

3. Feb 2, 2009

### wizard85

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?

4. Feb 2, 2009

### Staff: Mentor

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.

5. Feb 2, 2009

### tiny-tim

erm

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

6. Feb 2, 2009

### wizard85

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

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

7. Feb 2, 2009

### tiny-tim

well, plus a constant vector!

why? … is something worrying you?

8. Feb 2, 2009

### wizard85

Thanks for the help.... ;)

9. Mar 8, 2011

### 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....

10. Mar 8, 2011

### 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