Applications of the derivative
Angle between two curves
Angle between two curves, examples
Angle between two curves
Angle between two curves is the angle subtended by tangent lines at the point where the curves intersect.
 If curves  f1(x) and f2(x) intercept at P(x0, y0) then as shows the right figure.
Angle between two curves, examples
Example:  Find the angle between cubic  y = - x3 + 6x2 - 14x + 14 and quadratic  y = - x2 + 6x - 6 polynomial.
Solution:  To find the point where the curves intersect we should solve their equations as the system of two equations in two unknowns simultaneously. Therefore,
- x3 + 6x2 - 1 4x + 14 = - x2 + 6x - 6   or    x3 - 7x2 + 20x - 20 = 0
the root of the cubic equation we calculate using the formula
 where,
Then, we calculate the slopes of the tangents drown to the given cubic and the quadratic polynomial by evaluating their derivatives at x = 2. Thus,
taking    f2(x) = - x3 + 6x2 - 14x + 14  so that  f '2(x) = - 3x2 + 12x - 14   then  f '2(2) = -
and    f1(x) = - x2 + 6x - 6  so that  f '1(x) = - 2x + 6   then  f '1(2) = 2.
Finally we plug the slopes of tangents into the formula to find the angle between given curves, as shows the figure below.
 We sketch the graphs of the quadratic and the cubic by calculating coordinates of translations x0 and y0, as they are at the same time the coordinates of the maximum and the point of inflection respectively, thus (x0)2 = xmax = - a1/(2a2) = 3,  y0 = f1(x0) = 3, (x0)3 = xinfl = - a2/(3a3) = 2,  y0 = f2(x0) = 2.
Example:  At which point of the cubic yx3 - 3x2 + 2x - 2 is its tangent perpendicular to the line  y = x.
Solution:  Since the slopes of perpendicular lines are negative reciprocals of each other then, the slope of the tangent to the cubic has to be  f ' (x) = - to be perpendicular to the given line whose slope  m = 1
Therefore,
f ' (x) = 3x2 - 6x + 2,   we set   f ' (x) = -1   or   3x2 - 6x + 2 = -1  that gives   x = 1
the abscissa of the tangency point. Then, plug  x = 1 into the given cubic to calculate its ordinate,
yx3 - 3x2 + 2x - 2,     y (1) = - 2   so the tangency point  I(1, -2).
 We sketch the graph of the cubic by calculating coordinates of translations x0 and y0, x0 = xinfl = - a2/(3a3) = 1,   y0 = f(x0) = - 2 what coincide with the coordinates of the point of inflection  I(1, -2). By plugging x0 and y0, with changed signs, into the given cubic, y - 2 = (x + 1)3 - 3(x + 1)2 + 2(x + 1) - 2 we get its source form fs (x) = x3 - x    or    fs (x) = a3x3 + a1x

where  a1 = - (a2)2 /(3a3) + a1,   a1 = -  and   a1 =  tan at,  as shows the figure above.
Calculus contents D