Asked by helppls!!!!
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
Answered by
oobleck
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.