Finding tangent vector for a curve at a given point

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For a problem like this where I have to find the tangent vector at the point $(0,0,1)$:

$$r(t) = \sin(t)i + (t^2 − t)j+\sqrt{1 + t}k$$

I know I would take the derivative of the problem, but would I substitute $1$ for each $t$ value, or would I substitute $0$ for the $t$ values at $i$ and $j$, and $1$ for the $t$ value at $k$?

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2 Answers

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You need to figure out which value of $t$ corresponds to the point $(0,0,1)$, that is, which number $t$ solves the equation $r(t) = (0,0,1)$?

Then take the derivative of $r$ with respect to $t$, and plug in the value that you found above. The result will be a tangent vector for the curve at the point $(0,0,1)$.

What do you get?

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Neither. You must find the value of $t$ for which your function $r(t)$ is at the point $(0,0,1)$.

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