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A point P is moving along the curve whose equation is
y=((x^3)+17)^(1/2)
When P is at (2,5), y is increasing at the rate of 2units/s. How fast is x changing?

Respuesta :

nobillionaireNobley
y = ((x^3) + 17)^(1/2)
dy/dt = (3x^2 dx/dt) / 2((x^3) + 17)^(1/2); where dy/dt = 2 units/s, (x, y) = (2, 3)
2 = (3(2)^2 dx/dt) / (2(2)^3 + 17)^1/2 = (12 dx/dt) / sqrt(33)
dx/dt = 2sqrt(33) / 12 = 0.96

Therefore, x is increasing at the rate of 0.96 units/s

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