Let P=(a,b)
P lies on x^{3}-8x
The slope dy/dx at P is equal to the slope of the straight line between (a,b) and (4,0)
I am working with derivatives and the problem is:
The flight of a spaceship is described by the curve y = x^3 - 8x, x> 0. if the engine is shut off at point P, the space ship will fly off along the tangent line at P. If the spaceship is moving left to right, at what point should the engine be shut down in or to reach the point (4,0)
I know dy/dx is 3x^2 - 8
I'm not getting the answer though. not sure what I am missing, I know I should know how to solve this but it is escaping me at the moment. help would be greatly appreciated.
thank you!
In space there is no free fall. If the engin is turned off at a point P(x,y), it continues in a rectilinear motion in the direction where it was at P. This direction is the tangent line to the accelerated trajectory y = x^3 - 8x ar P. The change in y direction is delta(y) = (0-y) and the change in x is delta(x) = (4-x) where x and y are the coordinates of P. Then the tan of the angle opposite to delta(y) is just delta(y)/delta(x). But this is the slope of the tangent line to the accelerated trajectory whch is y'. Equate these slopes and solve for x, the use that value of x in y = x^3 - 8x to locate P.
if I put 4 in and do f(4) and f'(4) - get 32 and 40, 32-40 is -8 which is the point for y and solve for x in f(x) x = 2 but also x = -1 +/1 sqrt(5).
Can you please show me how you would work this out, I think I am overworking it.
I know the answer is (2,-8)