• anonymous
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?
Mathematics
• Stacey Warren - Expert brainly.com
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SOLVED
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