A particle moves along the curve x^2+4=y. The point on the curve at which the y coordinate changes twice as fast as the x coordinate,is
you want dy/dt = 2 dx/dt
x^2+4 = y
2x dx/dt = dy/dt
So, you want 2x dx/dt = 2 dx/dt
2x = 2
x = 1
So, at (1,5) dy/dt = 2 dx/dt
or, you want dy/dx = 2
y = x^2+4
dy/dx = 2x = 2
x = 1
To find the point on the curve where the y-coordinate changes twice as fast as the x-coordinate, we need to find the relationship between the rates of change of x and y.
Given the curve equation x^2 + 4 = y, we can differentiate both sides of the equation with respect to time (t), assuming x and y are functions of time:
d/dt (x^2 + 4) = d/dt (y)
Differentiating each term separately using the chain rule, we get:
2x * dx/dt = dy/dt
This equation represents the relationship between the rates of change of x and y. We are given that the y-coordinate changes twice as fast as the x-coordinate, which means:
dy/dt = 2 * (dx/dt)
Substituting this into the previous equation, we have:
2x * dx/dt = 2 * (dx/dt)
Canceling out the common factor of (dx/dt), we get:
2x = 2
This simplifies to:
x = 1
Now that we have the value of x, we can substitute it back into the original curve equation to find the corresponding y-coordinate:
x^2 + 4 = y
(1)^2 + 4 = y
1 + 4 = y
y = 5
Therefore, the point on the curve where the y-coordinate changes twice as fast as the x-coordinate is (1, 5).