Asked by Henry
Can someone show how to solve this problem
Find the equation of the tangent line to the curve given by,
x=3e^t
y=5e^-t
at the point where t=0
Thanks!
Find the equation of the tangent line to the curve given by,
x=3e^t
y=5e^-t
at the point where t=0
Thanks!
Answers
Answered by
Damon
one way:
when t = 0
x = 3
y = 5
dy/dx = dy/dt * dt/dx
dy/dt = -5 e^-t
dx/dt = 3 e^t
so
dt/dx = (1/3) e^-t
dy/dx = -5 e-t (1/3)e^-t
dy/dx = -(5/3) e^-2t = -5/3 when t = 0
so
y = -5/3 x + b
5 = -5/3 * 3 + b
b = 10
so
y = -5 x/3 + 10
when t = 0
x = 3
y = 5
dy/dx = dy/dt * dt/dx
dy/dt = -5 e^-t
dx/dt = 3 e^t
so
dt/dx = (1/3) e^-t
dy/dx = -5 e-t (1/3)e^-t
dy/dx = -(5/3) e^-2t = -5/3 when t = 0
so
y = -5/3 x + b
5 = -5/3 * 3 + b
b = 10
so
y = -5 x/3 + 10
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