I'm trying to find the magnitude of a vector dot{vec{r}}(t) by two different methods, but get different results. Let vec{r}(t)=x(t)hat{x}+y(t)hat{y}. The magnitude of this vector is: norm(vec{r}(t)) = sqrt{x(t)^2 + y(t)^2} The time derivative of vec{r}(t) must then be dot{vec{r}}(t) = dot{x}(t)hat{dot{x}}+dot{y}(t)hat{dot{y}} Now, using the same formula for the magnitude as before yields: ||dot{vec{r}}(t)|| = sqrt{dot{x}(t)^2 + dot{y}(t)^2}

Ruby Briggs

Ruby Briggs

Answered question

2022-07-17

I'm trying to find the magnitude of a vector r ˙ ( t ) by two different methods, but get different results.
Let r ( t ) = x ( t ) x ^ + y ( t ) y ^ . The magnitude of this vector is:
| | r ( t ) | | = x ( t ) 2 + y ( t ) 2
The time derivative of r ( t ) must then be
r ˙ ( t ) = x ˙ ( t ) x ˙ ^ + y ˙ ( t ) y ˙ ^
Now, using the same formula for the magnitude as before yields:
| | r ˙ ( t ) | | = x ˙ ( t ) 2 + y ˙ ( t ) 2
But when I try to find the magnitude of r ˙ ( t ) by differentiating | | r ( t ) | | with respect to time, the chain rule gives me:
d d t | | r | | = | | r | | x d x d t + | | r | | y d y d t = 2 x 2 x 2 + y 2 x ˙ + 2 y 2 x 2 + y 2 y ˙ = x ˙ x | | r | | + y ˙ y | | r | | x ˙ 2 + y ˙ 2
Where it is implicit that all the functions are evaluated in t. Have I made an algebraic mistake, or is it not possible to find | | r ˙ ( t ) | | by differentiation? - and if so, why?

Answer & Explanation

Abbigail Vaughn

Abbigail Vaughn

Beginner2022-07-18Added 15 answers

In one equation, you take the magnitude and then differentiate, and in the other you differentiate and then take the magnitude. I don't see any reason why these two operations would be equivalent.
For example, consider f ( x ) = x 2 . Then, f ( x ) = 2 x and | f ( x ) | = 2 | x | . Meanwhile, | f ( x ) | = x 2 and | f ( x ) | = 2 x. These two functions clearly differ on the negative real numbers.

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