y = x^2 + 1
dy/dt = 2x dx/dt
when x=1 and dx/dt = 2
dy/dt = 2(1)(2) = 4
Also, when x = 1, y = 2
Distance = (x^2 + y^2)^(1/2)
d(distance)/dt = (1/2)(x^2 + y^2)^(-1/2) (2x dx/dt + 2y dy/dt)
= (1/2)(1/√2) (2(2) + 2(2)(4))
= 10/√2 = 5√2
Let the height of the water be h ft
let the width of the water level be 2x ft
2x/h = 3/3
h = 2x or x = h/2
V = area of triangle x 12
= 12(h/2)(h) = 6 h^2
dV/dt = 12h dh/dt
a) when h = 1 and dV/dt = 2
2 = 12(1)(dh/dt)
dh/dt = 1/6 ft/min
b) You do it.
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