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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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