- #1

spaghetti3451

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## Homework Statement

In Euclidean three-space, let ##p## be the point with coordinates ##(x,y,z)=(1,0,-1)##. Consider the following curves that pass through ##p##:

##x^{i}(\lambda)=(\lambda , (\lambda -1)^{2}, -\lambda)##

##x^{i}(\mu)=(\text{cos}\ \mu , \text{sin}\ \mu , \mu - 1)##

##x^{i}(\sigma)=(\sigma^{2} , \sigma^{3}+\sigma^{2} , \sigma )##

(a) Calculate the components of the tangent vectors to these curves at ##p## in the coordinate basis ##\{\partial_{x},\partial_{y},\partial_{z}\}##.

(b) Let ##f=x^{2}+y^{2}-yz##. Calculate ##\frac{df}{d\lambda}, \frac{df}{d\mu}## and ##\frac{df}{d\sigma}##.

## Homework Equations

## The Attempt at a Solution

[/B]

My understanding of the concept of tangent vectors and coordinate basis is a little shaky.

I understand that, for a function ##f=f(x^{i}(\lambda))##, the tangent vector is ##\frac{df}{d\lambda}=\frac{dx^{i}}{d\lambda}\frac{df}{dx^{i}}=\frac{dx^{i}}{d\lambda}\partial_{i}f##, where the set of ##\frac{dx^{i}}{d\lambda}## are the components of the tangent vector and the set of ##\partial_{i}## is called the coordinate basis.

I wonder why ##\frac{df}{d\lambda}## is called the tangent vector. Is it because ##\frac{df}{d\lambda}## points along the tangent to the curve ##f(\lambda)##?