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Vector Derivative and Tangent – A tangent calculation for a curve in a vector presentation – Exercise 3836

Exercise

Calculate the tangent equation for the curve given as an intersection of the plains

z=x^2

x=y^2

At the point

M(1,1,1)

Final Answer

\vec{x}=(2,1,4)t+(1,1,1)

Solution

Coming soon…

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