# Indefinite integral of vector function

• wizard85
In summary: the product rule applies to a cross product (or a dot product) in exactly the same way as to an ordinary product
wizard85

## Homework Statement

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

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

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

Surely this is a mistake.

Mark44 said:
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?

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.

erm

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

tiny-tim said:
erm

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

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

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

wizard85 said:
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?

Pawan Rochwani
tiny-tim said:
well, plus a constant vector!

why? … is something worrying you?

Thanks for the help... ;)

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...

hi selms05!

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

## 1. What is the definition of an indefinite integral of a vector function?

The indefinite integral of a vector function is a mathematical operation that finds a general solution for the antiderivative of a vector function. It is denoted by ∫F(x)dx and is the inverse of the derivative of the same function.

## 2. How is the indefinite integral of a vector function different from a regular indefinite integral?

The indefinite integral of a vector function is similar to a regular indefinite integral in the sense that it also involves finding the antiderivative of a function. However, in the case of a vector function, the result is a vector rather than a scalar value.

## 3. What are the steps involved in finding the indefinite integral of a vector function?

The steps for finding the indefinite integral of a vector function are as follows:1. Determine the components of the vector function.2. Integrate each component separately using the power rule or other integration techniques.3. Combine the individual integrals to form the indefinite integral of the vector function.

## 4. Can the indefinite integral of a vector function be evaluated at specific values?

Yes, the indefinite integral of a vector function can be evaluated at specific values. This is known as evaluating a definite integral and involves substituting the values into the indefinite integral and finding the resulting vector.

## 5. How is the indefinite integral of a vector function used in real life applications?

The indefinite integral of a vector function is used in various fields of science and engineering, such as physics, mechanics, and electromagnetism. It is used to find the displacement, velocity, and acceleration of objects in motion, as well as to solve problems involving work, energy, and force. It also has applications in signal processing and control systems.

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