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# If x^{2} + y^{2} = 100 and dy / dt = 4, find dx/dt when y = 6.

**Solution:**

Given, x^{2} + y^{2} = 100

dy/dt = 4

We have to find dx/dt.

When y = 6,

x^{2} + (6)^{2} = 100

x^{2} + 36 = 100

x^{2} = 100 - 36

x^{2} = 64

Taking square root,

x = ±8

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,

\((\pm 8)(\frac{dx}{dt})+(6)(4)=0\)

\((\pm 8)(\frac{dx}{dt})+24=0\)

\((\pm 8)(\frac{dx}{dt})= -24\)

\(\frac{dx}{dt}=(\frac{-24}{\pm 8})\)

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

Therefore, dx/dt = ±3

## If x^{2} + y^{2} = 100 and dy / dt = 4, find dx/dt when y = 6.

**Summary:**

If x^{2} + y^{2} = 100 and dy / dt = 4, then dx/dt when y = 6 is ±3.

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