Let c(t)=x(t)i+y(t)j+z(t)k. Show that ||c(t)||=k where k is a constant if

[crystalh]

I don't understand what this question is asking or how to tackle it. Any help would be appreciated. Thank you.

Let c(t)=x(t)i+y(t)j+z(t)k. Show that ||c(t)||=k where k is a constant if and only if c(t) and c'(t) are orthogonal.

(Note: c, i, j, and k above are vectors).

[Hint: use ||c(t)||^2 = c(t) * c(t)]

 

[lineintegral1]

I think this question has been asked like a dozen times in the past two months. I'm surprised you didn't find it doing a quick google (if you did that). My hint for you is to consider the derivative of $||c(t)||^2$. How do you determine the derivative of a dot product? And how do dot products relate to orthogonal vectors?

 

[crystalh]

I did google it and didn't find anything, who knows.
Anyhow, I was able to get a bit more help on this question, but your suggestions were actually the most helpful to me in getting the point of it. Two vectors are orthogonal if and only if their dot product is zero. I think I get it now. Thanks for your help.

 

[Mark44]

I don't understand what this question is asking or how to tackle it. Any help would be appreciated. Thank you.

Let c(t)=x(t)i+y(t)j+z(t)k. Show that ||c(t)||=k where k is a constant if and only if c(t) and c'(t) are orthogonal.

(Note: c, i, j, and k above are vectors).

It looks like you are using k for two different things. In the first use of k, it's a unit vector. In the second use, it seems to be a scalar.

[Hint: use ||c(t)||^2 = c(t) * c(t)]