Question

Find the angle between the curve 2y=e^-x/2 and y-axis

Answer 1

tan^{-1} (4) is the angle between the curve 2y=e^-x/2 and y-axis.

Given,

2y=e^-x/2

y-axis ⇒ x = 0 should be the equation

Therefore, we have two equations,

2y=e^-x/2

x = 0

substitute the second equation in first equation

2y=e^-(0)/2

2y = e^0

y = 1/2

Therefore, y-axis and e^-x/2  intersect at (0, 1/2)

2y=e^-x/2

differentiating the above equation, we get,

2 dy/dx = -1/2 (e^-x/2 )

dy/dx = -1/4 (e^-x/2 )

dy/dx (0, 1/2) = -1/4 (e^-(0)/2 ) = -1/4

tan ∅ = dy/dx = -1/4

Angle made with y-axis is given by,

= ( π/2 + ∅ )

Therefore,

∅ = tan^{-1} (-1/4)

= π/2 + tan^{-1} (-1/4)

= π/2 - tan^{-1} (1/4)

= cot^{-1} (1/4)

= tan^{-1} 4

Answer 2

The angle between the curve 2y=e−x2 and the y - axis is