How Can I Prove These Vector Calculus Relations?

[WMDhamnekar]

Hi,
Let f(t) be a differentiable curve such that $f(t)\not= 0$ for all t. How to show that $\frac{d}{dt}\left(\frac{f(t)}{||f(t)||}\right)=\frac{f(t)\times(f'(t)\times f(t))}{||f(t)||^3}\tag{1}$

My attempt:
$\frac{d}{dt}\left(\frac{1}{||f(t)||}\right)*f(t)+\frac{1}{||f(t)||}*\frac{d}{dt}(f(t))$

$\frac{||f(t)||}{f'(t)\cdot f(t)}*f(t) +\frac{1}{||f(t)||}*\frac{d}{dt}(f(t))$

I want to know whether my last step is correct or wrong.If wrong , how and where to go from here to get R.H.S.of (1)? If yes how to proceed further to get R.H.S.of (1)?

 

[GJA]

Hi Dhamnekar,

You have a nice idea in your attempt. There is an error in the first term on your second line. You either need to apply the quotient rule to $1/\|f(t)\|$ or the chain rule to $(\|f(t)\|)^{-1}.$ Whichever method you choose, you will see that you get something similar to, but different from, what you have now.

I know you can do this, so I want to keep my hints to a minimum, at least initially. Of course, if you have more questions, I'm more than happy to help.

 

[WMDhamnekar]

Hi Dhamnekar,

You have a nice idea in your attempt. There is an error in the first term on your second line. You either need to apply the quotient rule to $1/\|f(t)\|$ or the chain rule to $(\|f(t)\|)^{-1}.$ Whichever method you choose, you will see that you get something similar to, but different from, what you have now.

I know you can do this, so I want to keep my hints to a minimum, at least initially. Of course, if you have more questions, I'm more than happy to help.

Hi GJA,

If I have understood your reply correctly, then the last step in the original question is $-\frac{f(t)}{||f(t)||^2}+\frac{f'(t)}{||f(t)||}$. Now where to go from here to get R.H.S of (1)?

 

[WMDhamnekar]

Hi,
I got the answer. :)

 

[WMDhamnekar]

Continuing this exercise, assume that f'(t) and f''(t) are not parallel. Then $T'(t)\not=0$ so we can define unit principal normal vector N by
$$N(t)=\frac{T'(t)}{||T'(t)||}$$

Now how to show that $$N(t)=\frac{f'(t)\times (f''(t)\times f'(t))}{||f'(t)||*(||f''(t)\times f'(t)||)}$$

Continuing this execise we can define unit binormal vector B $$B(t)=T(t)\times N(t)$$ where $$T(t)=\frac{f'(t)}{||f'(t)||}$$. Note: We have already defined T'(t).
Now how to show that $$B(t)=\frac{f'(t)\times f''(t)}{||f'(t)\times f''(t)||}$$
I want to continue this exercise with one more question related to this question. How does the vectors T(t), N(t), B(t)form a right-handed system of mutually perpendicular unit vectors (called orthonormal vectors) at each point on the curve f(t)? In the answer to this question, I want to clear explanation about Osculating plane, Normal plane and Rectifying plane.:)

 

[WMDhamnekar]

Continuing this exercise, assume that f'(t) and f''(t) are not parallel. Then $T'(t)\not=0$ so we can define unit principal normal vector N by
$$N(t)=\frac{T'(t)}{||T'(t)||}$$

Now how to show that $$N(t)=\frac{f'(t)\times (f''(t)\times f'(t))}{||f'(t)||*(||f''(t)\times f'(t)||)}$$

Continuing this execise we can define unit binormal vector B $$B(t)=T(t)\times N(t)$$ where $$T(t)=\frac{f'(t)}{||f'(t)||}$$. Note: We have already defined T'(t).
Now how to show that $$B(t)=\frac{f'(t)\times f''(t)}{||f'(t)\times f''(t)||}$$
I want to continue this exercise with one more question related to this question. How does the vectors T(t), N(t), B(t)form a right-handed system of mutually perpendicular unit vectors (called orthonormal vectors) at each point on the curve f(t)? In the answer to this question, I want to clear explanation about Osculating plane, Normal plane and Rectifying plane.:)

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