Question
find the equation of the tangent line to the curve at the given point by eliminating the parameter:
x=1+sqrt(t) y=e^(t^2) parameter: (2,e)
x=1+sqrt(t) y=e^(t^2) parameter: (2,e)
Answers
x = 1+√t
t = (x-1)^2
y = e^(x-1)^4
y' = e^(x-1)^4 * 4(x-1)^3
y'(2) = 4e
Check. Using the parameters, at t=1,
dy/dt = 2t e^t^2 = 2e
dx/dt = 1/(2√t) = 1/2
dy/dx = 2e/(1/2) = 4e
t = (x-1)^2
y = e^(x-1)^4
y' = e^(x-1)^4 * 4(x-1)^3
y'(2) = 4e
Check. Using the parameters, at t=1,
dy/dt = 2t e^t^2 = 2e
dx/dt = 1/(2√t) = 1/2
dy/dx = 2e/(1/2) = 4e