Question

A point is moving on the graph of 5x³+6y³=xy. When the point is at P=( 1/11 , 1/11 ), its y-coordinate is increasing at a speed of 6 units per second.

What is the speed of the x-coordinate at that time and in which direction is the x - coordinate moving?
(A) 8 units/sec, increasing x
(B) 17/2 units/sec, decreasing x
(C) 17/2 units/sec, increasing x
(D) 33/4 units/sec, decreasing x
(E) 8 units/sec, decreasing x
(F) 35/4 units/sec, decreasing x

Answer 1

The speed of the x-coordinate when the point is at P=(1/11, 1/11) is 17/2 units/sec, and the x-coordinate is increasing at that time.

Step-by-step explanation:

To find the speed of the x-coordinate, we need to differentiate the equation of the graph with respect to time. Differentiating both sides of the equation 5x³+6y³=xy, we get 15x²(dx/dt) + 18y²(dy/dt) = x(dy/dt) + y(dx/dt). Since the point is at P=( 1/11 , 1/11 ), we substitute x=1/11 and y=1/11 into the equation. We also know that dy/dt = 6 units/sec. By substituting these values, we can solve for dx/dt, which gives us the speed of the x-coordinate.

Plugging in the values, we get 15(1/11)²(dx/dt) + 18(1/11)²(6) = (1/11)(6) + (1/11)(dx/dt). Simplifying this equation gives us dx/dt = 17/2 units/sec. Therefore, the speed of the x-coordinate when the point is at P=( 1/11 , 1/11 ) is 17/2 units/sec.

Since dx/dt is positive, the x-coordinate is increasing at that time.