A POINT MOVES ON A PARABOLA y²=8x IN SUCH A WAY THAT THE RATE OF CHANGE IS ALWAYS 5 units/sec HOW FAST IS THE ABSCISSA CHANGING WHEN THE ORDINATE IS 4
Question
A POINT MOVES ON A PARABOLA y²=8x IN SUCH A WAY THAT THE RATE OF CHANGE IS ALWAYS 5 units/sec HOW FAST IS THE ABSCISSA CHANGING WHEN THE ORDINATE IS 4
Solution
Given the equation of the parabola is y² = 8x.
We can differentiate both sides with respect to time (t) to find the rate of change of x with respect to time (dx/dt).
Using the chain rule, the derivative of y² with respect to t is 2y(dy/dt) and the derivative of 8x with respect to t is 8(dx/dt).
So, we have 2y(dy/dt) = 8(dx/dt).
We know that dy/dt = 5 units/sec (the rate of change of y with respect to time), so we can substitute this into the equation to get 2y(5) = 8(dx/dt).
Solving for dx/dt gives us dx/dt = (10y)/8 = 1.25y.
Finally, we substitute y = 4 (the given ordinate) into this equation to find dx/dt when y = 4.
So, dx/dt = 1.25(4) = 5 units/sec.
Therefore, the abscissa is changing at a rate of 5 units/sec when the ordinate is 4.
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