Differentiation of vector function

In summary, the homework statement is trying to find a function that is differentiable and has a gradient of 1. The student is trying to find a way to use the gradient to prove the desired result. However, they are not sure how to do it. A hint may help, so thanks.
  • #1
boombaby
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Homework Statement


suppose f is a differentiable mapping of R1 into R3 such that |f(t)|=1 for every t. Prove that [tex]f'(t)\cdot f(t)=0[/tex].
I guess it is more proper to write [tex](\nabla f)(t) \cdot f(t)=0[/tex], where [tex](\nabla f)(t)[/tex] is the gradient of f ant t.


Homework Equations





The Attempt at a Solution


it is then equivalent to prove [tex]\sum\;(Df_{i})(t)\cdot f_{i}(t)=0[/tex], but I've no idea of how to use the |f(t)|=1 to deduce the disired results, although the equation has an easy geometrically interpretation
Any hint may help, thanks a lot.
 
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  • #2
The f'(t) notation is fine. f(t) is a vector. You generally use the gradient notation for scalar functions, not vectors. And |f(t)|=1 is the same as saying f(t).f(t)=1, right?
 
  • #3
just differentiate both side and it is done...Thanks!
and thanks for the explanation about gradient...I misunderstood it...
Here is why I thought it improper..f'(t) is a linear transformation from R^1->R^3, whose basis are different things from the basis of R^3, so I thought it is weird to apply the dot product to f'(t) and f(t)...
 
  • #4
For functions of a single variable it's simpler just to think of f'(t) as an element of R^3 rather than as a linear transformation. If f(t)=(x(t),y(t),z(t)), f'(t)=(x'(t),y'(t),z'(t)).
 
  • #5
it took me years to open and refresh this forums...my connecting to some websites sucks...
Thinking of it as a vector in R3 is okay and easy. okay, I found it more sensible in this way.
take 1 as a basis of R^1. So f'(t)(1) = [tex]\sum^{3}_{1} (Df_{i})(t)e_{i}[/tex], where e_i is the basis of R^3, and dot product makes sense
 
  • #6
Sure, the linear transformation is f'(t)(h)=(x'(t),y'(t),z'(t))*h. That can be identified with a vector in R^3 in a pretty obvious way. I would say you are just throwing the 'h' away altogether.
 
  • #7
Saying |f(t)|= 1 is the same as saying [itex]f(t)\cdot f(t)= 1[/itex]. Use the product rule.
 
  • #8
it's been solved, Thanks anyway:)
 

1. What is the definition of differentiation of a vector function?

Differentiation of a vector function is the process of finding the rate of change of the vector function with respect to its independent variable. It involves finding the derivative of each component of the vector function.

2. What are the applications of differentiation of vector functions?

Differentiation of vector functions is used in various fields such as physics, engineering, and economics, to analyze and model real-world phenomena. It is particularly useful in studying motion, velocity, and acceleration of objects.

3. How is the derivative of a vector function calculated?

The derivative of a vector function is calculated by finding the derivative of each component of the vector function with respect to the independent variable. This can be done using the rules of differentiation, such as the power rule and product rule.

4. Can a vector function have multiple derivatives?

Yes, a vector function can have multiple derivatives. The first derivative represents the rate of change of the vector function, while the second derivative represents the rate of change of the first derivative. Higher-order derivatives can also be calculated to analyze the behavior of the vector function.

5. How is differentiation of vector functions related to integration?

Differentiation and integration are inverse operations. This means that if a vector function is differentiated, its antiderivative can be found by integrating. Similarly, if a vector function is integrated, its derivative can be found by differentiating. Both operations are important in solving various mathematical and real-world problems.

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