Find Curvature of r(t)=2ti+2tj+k

[Math10]

Homework Statement

Find the curvature of r(t)=2ti+2tj+k.

Homework Equations

None.

The Attempt at a Solution

The answer is 0 in the book.
I know the formula for curvature is k(t)=abs(r'(t)xr"(t))/abs(r'(t))^3. I know that r'(t)=2i+2j and r"(t)=0, so r'(t)xr"(t)=0? How to find r'(t)xr"(t) and r'(t)^3?

 

[Mark44]

Homework Statement

Find the curvature of r(t)=2ti+2tj+k.

Homework Equations

None.

The Attempt at a Solution

The answer is 0 in the book.
I know the formula for curvature is k(t)=abs(r'(t)xr"(t))/abs(r'(t))^3. I know that r'(t)=2i+2j and r"(t)=0, so r'(t)xr"(t)=0?

Yes.

How to find r'(t)xr"(t) and r'(t)^3?

r'(t)r''(t) = 0, which you already found. What is |r'(t)|? Note that this means the magnitude or length of r'(t).

 

[Math10]

So how do I find abs(r'(t))? That's where I got stuck.

 

[SteamKing]

So how do I find abs(r'(t))? That's where I got stuck.

The notation |r'(t)| does not mean the absolute value of r'(t), it means to find the magnitude of the vector r'(t). Think Pythagoras.

 

[haruspex]

The notation |r'(t)| does not mean the absolute value of r'(t), it means to find the magnitude of the vector r'(t). Think Pythagoras.

The problem is that Math10 quoted the formula using abs() instead of modulus.
It should be $k(t)=\frac{|\dot{\vec r}(t)\times \ddot {\vec r}(t)|}{|\dot{\vec r}(t)|^3}$

 

[Mark44]

So how do I find abs(r'(t))? That's where I got stuck.

Three of us have told you that |r'(t)| does not mean "absolute value". It means "magnitude" or "length" of the vector r'(t).