Vector Derivative and Tangent – A tangent calculation for a curve in a vector presentation – Exercise 3828

Exercise

Calculate the tangent equation for the curve

r(t)=.(tet+3)i+4+5tj+arctan(2t)k\vec{r}(t)=.(te^{-t}+3)\vec{i}+\sqrt{4+5t}\vec{j}+\arctan(2t)\vec{k}

At a point corresponding to t = 0.

Final Answer

x=(1,54,2)t+(3,2,0)\vec{x}=(1,\frac{5}{4},2)t+(3,2,0)

Solution

Coming soon…

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