what is the equation of the line tangent to the curve with parametric equations x=3e^-t, y=6e^t at the point?

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what is the equation of the line tangent to the curve with parametric equations x=3e^-t, y=6e^t at the point where t=0?

2007-12-29 06:53:19 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

2007-12-29 08:00:30 · answer #1 · answered by Anonymous · 0⤊ 3⤋

dx/dt = -3e-t
dy/dt= 6e^t
So dy/dx = 6e^t/-3e^-t = -2e^2t
dy/dx = -2 at t = 0

2007-12-29 16:07:25 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋