A particle is moving on an elliptic trajectory ( x ( t ) , ( y ( t )) described
ID: 2868277 • Letter: A
Question
A particle is moving on an elliptic trajectory (x(t), (y(t)) described by the equation x2 + 2x + y2 = 19. At the point (3, 2) the y-coordinate changes at the rate dy/dx=25m/s. How fast is the x-coordinate changing at that instant? A particle is moving on an elliptic trajectory (x(t), (y(t)) described by the equation x2 + 2x + y2 = 19. At the point (3, 2) the y-coordinate changes at the rate dy/dx=25m/s. How fast is the x-coordinate changing at that instant? A particle is moving on an elliptic trajectory (x(t), (y(t)) described by the equation x2 + 2x + y2 = 19. At the point (3, 2) the y-coordinate changes at the rate dy/dx=25m/s. How fast is the x-coordinate changing at that instant?Explanation / Answer
x^2 + 2x + y^2 = 19
Deriving this implicitly with respect to time, we get :
2xdx/dt + 2dx/dt + 2ydy/dt = 0
Plug in x = 3 , y = 2 and dy/dt = 25 :
2(3)dx/dt + 2(dx/dt) + 2(2)(25) = 0
8(dx/dt) + 100 = 0
8(dx/dt) = -100
dx/dt = -100/8
dx/dt = -25/2
dx/dt = -12.5 meter per second -----> ANSWER
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