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Find the equation of the line tangent to the r(t)=(t^2+t)*i+sqrt(t)*j when t=3

Find the equation of the line tangent to the r(t)=(t^2+t)*i+sqrt(t)*j when t=3

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Answer #1

(3)=12i+ 13; direction of tangent lien is, tangent line is, (0) –12+ 45j + (9)i zadov)

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Answer #2
First we take derivative of above eq we get r'(t) =(2t+1)i +(1/2√t)j .......(1) Put the value of t=3 in eq 1 we get r'(3) =3i +1/2 j As we know the formula r(t) =r'(t) +t r'(t) Putting the values in it r(t) =r'(3) +3i+1/2 j
answered by: Jabbar
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Find the equation of the line tangent to the r(t)=(t^2+t)*i+sqrt(t)*j when t=3
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