By finding the slopes of the tangent lines to the curve of y=(1/3)x^3+5x at the points where x=3 and x=6, find the acute angle between these lines at the point where they cross.

Question

I don't have any idea where to start. An explanation would be appreciated.

Please and thanks.

Answer 1

slope = dy/dx = x^2 + 5

if x = 3
y = 9+15 = 24 so through (3,24)
here slope = 9+5 = 14
y = mx+b = 24 = 14(3) +b so b = -18
so
y = 14 x - 18

if x = 6
slope = 36+5 = 41
y = 72 + 30 = 102
102 = 41*6 + b
b = -144
y = 41 x - 144

so find intersection of lines
y = 41 x -144 and
y = 14 x - 18

Answer 2

Then you know how to find the slope of each line at that point and find the angle between
tan^-1(slope 1 - slope 2)

Answer 3

Not tan^-1(m1 - m2)
but rather
tan^-1 m1 - tan^-1 m2

Answer 4

whoops, yes