Vector Derivative and Tangent – A tangent calculation for a curve in a vector presentation – Exercise 3828 Post category:Vector Derivative and Tangent Post comments:0 Comments Exercise Calculate the tangent equation for the curve r⃗(t)=.(te−t+3)i⃗+4+5tj⃗+arctan(2t)k⃗\vec{r}(t)=.(te^{-t}+3)\vec{i}+\sqrt{4+5t}\vec{j}+\arctan(2t)\vec{k}r(t)=.(te−t+3)i+4+5tj+arctan(2t)k At a point corresponding to t = 0. Final Answer Show final answer x⃗=(1,54,2)t+(3,2,0)\vec{x}=(1,\frac{5}{4},2)t+(3,2,0)x=(1,45,2)t+(3,2,0) Solution Coming soon… Share with Friends Read more articles Previous PostVector Derivative and Tangent – A tangent calculation for a curve in a vector presentation – Exercise 3833 Next PostVector Derivative and Tangent – Calculating derivative of a vector function – Exercise 3825 You Might Also Like Vector Derivative and Tangent – Calculating Derivative and a derivative size of a vector function – Exercise 3820 March 3, 2019 Vector Derivative and Tangent – Calculating derivative of a vector function – Exercise 3825 March 3, 2019 Vector Derivative and Tangent – A tangent calculation for a curve in a vector presentation – Exercise 3833 March 4, 2019 Vector Derivative and Tangent – A tangent calculation for a curve in a vector presentation – Exercise 3836 March 4, 2019 Vector Derivative and Tangent – Tangent to the curve in a vector representation parallel to a given plane – Exercise 3839 March 4, 2019 Vector Derivative and Tangent – Unit tangent vector to a curve in a vector presentation – Exercise 3842 March 4, 2019 Leave a Reply Cancel replyCommentEnter your name or username to comment Enter your email address to comment Enter your website URL (optional) Δ
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