Consider the given vector equation. R(t) = 5 sin(t)i + 2 cos(t)j (a) find r'(t). R'(t) = incorrect: your answer is incorrect. (b) sketch the plane curve together with position vector r(t) and the tangent vector r'(t) for the given value of t

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solution:

[tex]a)  r(t)=5sint\widehat{i}+2cost\widehat{j}\\

\Rightarrow r'(t)=\frac{d}{dt}(5sint\widehat{i}+2cost\widehat{j})\\

\Rightarrow r'(t)=\frac{d}{dt}(5sint)\widehat{i}+\frac{d}{dt}(2cost)\widehat{j}\\

\Rightarrow r'(t)=5cost\widehat{i}-2sint\widehat{j}\\[/tex][tex]b) at (t)=\frac{\pi }{4},we have\\

r'(\frac{\pi }{4})=\frac{5}{\sqrt{2}}\widehat{i}-\frac{2}{\sqrt{2}}\widehat{j}\\

so, the tangent vector r'(\frac{\pi }{4}) start at(\frac{5}{\sqrt{2}},\frac{-2}{\sqrt{2}})\\

and moves \frac{5}{\sqrt{2}} right and \frac{2}{\sqrt{2}}down[/tex]