If x² + y² = 225 and dy/dt = 4, find dx/dt when y = 9.

Solution:

Given, x² + y² = 225

dy/dt = 4

We have to find dx/dt.

When y = 9,

⇒ x² + (9)² = 225

⇒ x² + 81 = 225

⇒ x² = 225 - 81

⇒ x² = 144

Taking square root,

⇒ x = ±12

Differentiating with respect to t,

\(2x(\frac{dx}{dt})+2y(\frac{dy}{dt})=0\)

\(x(\frac{dx}{dt})+y(\frac{dy}{dt})=0\)

Put dy/dt = 4 in the above expression,

\((\pm 12)(\frac{dx}{dt})+(9)(4)=0\)

\((\pm 12)(\frac{dx}{dt})+36=0\)

\((\pm 12)(\frac{dx}{dt})= -36\)

\(\frac{dx}{dt}=(\frac{-36}{\pm 12})\)

\(\frac{dx}{dt}=\pm 3\)

Therefore, dx/dt = ±3

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If x² + y² = 225 and dy/dt = 4, find dx/dt when y = 9.

Summary:

If x² + y² = 225 and dy/dt = 4, then dx/dt when y = 9 is ±3.