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anonymous
 4 years ago
A curve is defined parametrically by x=e^(t) and y=2e^(t). An equation of the tangent line to the curve at t=ln2 is
anonymous
 4 years ago
A curve is defined parametrically by x=e^(t) and y=2e^(t). An equation of the tangent line to the curve at t=ln2 is

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TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0vector form of the line:\[\vec r(t)=<e^t,2e^{t}>\]and the point is\[r(\ln2)=<2,1>\]now we need the derivative of the vector function\[\vec r'(t)=<e^t,2e^{t}>\]and the derivative at that point is\[r'( \ln2)=<2,1>\]the parametric formula for a line parallel to a vector u from a point is\[\vec v(t)<x_0,y_0>+t\vec u=<2,1>+t<2,1>=<2+2t,1t>\]with the components written individually we have\[x=2+2t\]\[y=1t\]
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