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Posted by on Saturday, July 17, 2010 at 4:13pm.

Both x and y denote functions of t that are related by the given equation. Use this equation and the given derivative information to find the specified derivative.

Equation:35x^2+20y^2=1

Given that dx/dt=4, find dy/dt

when (x,y)=(1/7sqrt2, 1/5sqrt2)

  • Calculus - , Saturday, July 17, 2010 at 4:58pm

    70 x dx/dt + 40 y dy/dt = 0

    280 x + 40 y dy/dy = 0

    280/7sqrt2 + 40/5sqrt2 dy/dt = 0

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