Xan you please tell me how you find the rate of change of Y= the square root of X when Y is increasing at 2 units per second and X=1?
y= √x
you are given dy/dt = 2 units/sec when x=1
I assume you want dx/dt, that is, how fast is x changing at that moment.
from your equation
dy/dt = 1/(2√x)*dx/dt
when x=1, y =1 from your equation, and dy/dt=2 so sub it in
2 = 1/(2√1) * dx/dt and
dx/dt = 4 units/sec
To find the rate of change of x, dx/dt, when y is increasing at 2 units per second and x=1, we can use the equation dy/dt = (1/(2√x)) * dx/dt, where y = √x.
Given that dy/dt = 2 units/sec when x=1, we can substitute these values into the equation:
2 = (1/(2√1)) * dx/dt
Since √1 = 1, the equation simplifies to:
2 = (1/2) * dx/dt
To solve for dx/dt, we can multiply both sides by 2:
4 = dx/dt
Therefore, dx/dt = 4 units/sec.