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xEnOnn
 3 years ago
I suddenly got very confused with something. Suppose I have a parametric equation like this: \[\begin{bmatrix}
x\\
y
\end{bmatrix}=\begin{bmatrix}sin(
t)\\
cos(t)
\end{bmatrix}\]
Why is the tangent vector of this function simply just the derivative of the x and y like this: \[tangent \ vector=\begin{bmatrix}
x'\\
y'
\end{bmatrix}=\frac{\partial }{\partial t}
\begin{bmatrix}sin(
t)\\
cos(t)
\end{bmatrix}=
\begin{bmatrix}cos(
t)\\
sin(t)
\end{bmatrix}\]
Taking the derivative just give me the gradient of the equation. It is just the gradient and not the tangent line yet, isn't it?
xEnOnn
 3 years ago
I suddenly got very confused with something. Suppose I have a parametric equation like this: \[\begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix}sin( t)\\ cos(t) \end{bmatrix}\] Why is the tangent vector of this function simply just the derivative of the x and y like this: \[tangent \ vector=\begin{bmatrix} x'\\ y' \end{bmatrix}=\frac{\partial }{\partial t} \begin{bmatrix}sin( t)\\ cos(t) \end{bmatrix}= \begin{bmatrix}cos( t)\\ sin(t) \end{bmatrix}\] Taking the derivative just give me the gradient of the equation. It is just the gradient and not the tangent line yet, isn't it?

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phi
 3 years ago
Best ResponseYou've already chosen the best response.0I would not call it a tangent line i.e. a line that is tangent to the curve at the point. It is the tangent direction vector. It points in the correct direction, but it's not necessarily tangent to the curve.

xEnOnn
 3 years ago
Best ResponseYou've already chosen the best response.0oh yea...you are right. it is the the tangent direction vector. But what is the rationale behind that the gradient is simply the direction? The gradient is just the rate of change but how does it give that direction?
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