Application of Differentiation

1a. y=x e^x, dy/dx=(x+1)e^x
For the tangent to be horizontal, the slope should be 0.
Thus setting dy/dx=0, x=-1.

For x=-1, y = -e^(-1) = -1/e.
Therefore the point is (-1, -1/e).

b. For the tangent to be horizontal,
the normal at that point will be verticle.

Hence, the equation of normal is x=-1.

2a. Solving x=y^2-1/4 and x=1/4-y^2 simultaneously,
we have x=0, y=1/2 or y=-1/2.
Thus the points of intersection are (0, 1/2) and (0, -1/2).

b. For x=y^2-1/4, dx/dy=2y, dy/dx=1/(2y)
For x=1/4-y^2, dx/dy=-2y, dy/dx=-1/(2y)

At (0, 1/2), the product of slope of tangent of both curve
= 1/[2*1/2]*1/[-2*1/2] = -1

At (0, -1/2), the product of slope of tangent of both curve
= 1/[2*-1/2]*1/[-2*-1/2] = -1

Hence, these two curve intersect at right angles.