Answer to Question #181916 in Calculus for Timothy

Question

Find the equation of the tangent line and normal line to the curve -2e^-x + e^y = 3e^x-y at point (0,0)

Answer 1

Solution

Point (0,0) is not belong to given curve -2e^-x+e^y = 3e^x-y because for x=y=0 we’ll get -1 =3 what is impossible.

Let’s solve this task for equation 2e^-x+e^y = 3e^x-y.

Let’s differentiate left and right parts of equation 2e^-x+e^y = 3e^x-y.

-2e^-x+ y’e^y  = 3(1-y’)e^x-y

y’(e^y +3 e^x-y) = 3 e^x-y +2e^-x => y’ = ( 3 e^x-y +2e^-x ) / (e^y +3 e^x-y)

For point (0,0) the equation of the tangent line is y = y’(0)x where y’(0) = 5/4  from previous expression. So the tangent line is y = 5x/4.

Similarly the equation of the normal line at point (0,0) is y = -x/y’(0) = -4x/5.

Answer

For point (0,0) the tangent line is y = 5x/4 and the normal line is y = -4x/5.