Asked by Ke$ha
given the relationship 2x^2 + y^3 =10, with y > 0 and dy/dt = 3 units/min., find the value of dx/dt at the instant x = 1 unit.
SUPER CONFUSED DONT GET THE EQUATION!!!!!!!
SUPER CONFUSED DONT GET THE EQUATION!!!!!!!
Answers
Answered by
Steve
This is a typical exercise using the chain rule.
2x^2 + y^3 =10
4x dx/dt + 3y^2 dy/dt = 0
Now, using your values, when x=1, y=2. So
4(1) dx/dt + 3(2^2)(3) = 0
4 dx/dt + 36 = 0
dx/dt = -9
2x^2 + y^3 =10
4x dx/dt + 3y^2 dy/dt = 0
Now, using your values, when x=1, y=2. So
4(1) dx/dt + 3(2^2)(3) = 0
4 dx/dt + 36 = 0
dx/dt = -9
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