calculus

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

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asked by john

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