Calculus

posted by Mike

A rocket follows a path given by y=x-1/91x^3 where distances are in miles. If the horizontal velocity is given by vx=1.4x, find the direction of the velocity when the rocket hits the ground (assume level terrain) if time is in minutes.

So far I've only taken the derivative for y and don't know what to do afterwards

Vx=1.4x
Vy=1-(3/91)x^2

  1. Damon

    Maybe you mean
    y=x - (1/91) x^3
    if so

    y = 0 at ground, start and finish
    0 = x - (1/91) x^3
    x = 0 works of course
    x^2 = 91
    x = 9.54

    Vx = 1.4 x = 13.4 miles/min at ground
    what is Vy ?
    dy/dt = dy/dx * dx/dt = [1-(3/91)x^2 ] Vx = [ 1-(3/91)9.54^2 ] 13.4 miles/min

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