# How can I pick which formula to use for a curvature?

I used the first formula to see if I would get the same answer as the second formula:

$$\vec{r}'(t) = 2t\hat{i}+\hat{k}$$

$$||\vec{r}'(t)|| = \sqrt{4t^2+1}$$

$$\vec{T}(t) = \frac{\vec{r}'(t)}{||\vec{r}'(t)||} = \frac{2t\hat{i}+\hat{k}}{\sqrt{4t^2+1}}$$

Therefore,

$$\vec{T}'(t) = \frac{2\hat{i}(\sqrt{4t^2+1}) - 4t(\frac{2t\hat{i}+\hat{k}}{\sqrt{4t^2+1}}) }{4t^2+1}$$

Since the unit vector in the direction of a given vector is the same after multiplying by a positive scalar, I can simplify by multiplying $$(4t^2+1)\;\;\; \sqrt{4t^2+1}$$

The first factor gets rid of the denominator and the second gets rid of fractional powers

$$= 2\hat{i}(4t^2+1) - 4t(2t\hat{i}+\hat{k})$$

$$= 8t^2\hat{i}+2\hat{i} - 8t^2\hat{i} - 4t\hat{k}$$

$$= 2\hat{i}-4t\hat{k}$$

$$= \hat{i} - 2t\hat{k}$$

Thus, $$\vec{T}'(t) = \hat{i} - 2t\hat{k}$$

and consequently,

$$||\vec{T}'(t)|| = \sqrt{1 + 4t^2}$$

So,

$$\kappa = \frac{\sqrt{1 + 4t^2}}{\sqrt{4t^2+1}} = 1$$

although this clearly does not make sense as the graph of the vector function is a parabola so there will be some point t that will be steeper than another point t2.

if my math is wrong i'd like to know where I made an error.

Otherwise, can someone please help explain how to decide between the formulas when they produce different results?

This statement:

"Since the unit vector in the direction of a given vector is the same after multiplying by a positive scalar..."


is incorrect. Clearly, multiplying a vector by a scalar changes its length (hence it is called a scalar...it scales the length), and the length of $$\vec T'(t)$$ in this case is essential since it is part of the definition of curvature. If you must scale anything, you have to multiply both $$\vec T'(t)$$ and $$\vec r'(t)$$ by the same scalar to preserve the ratio of their lengths.

Thus, using the first definition, the correct computation is

$$\kappa=\frac{\|\vec T'(t)\|}{\|\vec r'(t)\|}=\frac{\|(4t^2+1)^{3/2}\vec T'(t)\|}{(4t^2+1)^{3/2}\|\vec r'(t)\|}=\frac{\|2(\hat i-2t\hat k)\|}{(4t^2+1)^{3/2}\sqrt{4t^2+1}}=\frac{2}{(4t^2+1)^{3/2}},$$

and all is right with the world.

Tip 1: The first formula may be less tractable when the unit tangent vector $$\vec T(t)$$ is a bit ugly, which happens in this problem since $$\|\vec r'(t)\|$$ is nonconstant, and hence requires using the quotient rule to compute $$\vec T'(t)$$.

Tip 2: The first formula essentially computes $$\|\frac{d\vec T}{ds}\|$$ where $$s$$ is arc length. So if the parameter $$t$$ represents arc length (e.g. $$\frac{ds}{dt}=\|\vec r'(t)\|$$ is constant), the first formula may then be more wieldy.

• ah i see.. i literally got that definition from some class notes a professor posted with examples... Commented Jan 31, 2022 at 1:42