Given x^2 + y^2 = 100 and dy/dt = 8, we need to find dx/dt when y = 6.
We can differentiate both sides of x^2 + y^2 = 100 with respect to time t to obtain:
2x(dx/dt) + 2y(dy/dt) = 0
Substituting y = 6 and dy/dt = 8, we get:
2x(dx/dt) + 2(6)(8) = 0
Solving for dx/dt, we get:
dx/dt = -48/x
Using x^2 + y^2 = 100 and y = 6, we can find x:
x^2 + 6^2 = 100
x = ±8
Since we are given that y = 6, we can see that x must be negative. Therefore:
dx/dt = -48/-8 = 6
So, dx/dt = 6.
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A cylindrical tank with radius 3 m is being filled with water at a rate of 4 m^3/min. We need to find how fast the height of the water is increasing.
Let the height of the water be h. Then the volume of water in the tank is given by:
V = πr^2h
Differentiating both sides with respect to time t, we get:
dV/dt = πr^2 dh/dt
We know that dV/dt = 4 m^3/min and r = 3 m. Substituting these values, we get:
4 = 9π dh/dt
Solving for dh/dt, we get:
dh/dt = 4/(9π)
So, the height of the water is increasing at a rate of 4/(9π) m/min.