# w2014 quiz on 11.5, 11.6, and 4.8-- brief answers - WF3456...

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WF3456 11.5, 11.6, & 4.8 Diff. & Parametric Eq’ns— Brief Answers 1. The equilibrium solutions are y = – 2, y = 0, and y = 3. (Note that x = 0 is not an equilibrium solution since its graph is a vertical line rather than a horizontal one.) Using a number line and the sign of the first derivative in each interval, we see that the equilibrium solutions at y = – 2 and y = 3 are stable while the equilibrium solution at y = 0 is unstable. 2. a. We can either write dW/dt = k ( 24 W – 2800 ) or dW/dt = k ( 2800 – 24 W ) pounds/day. In the first case, k < 0, while in the second case, k > 0. In both cases, the units of k are pounds/calorie. b. dB/dt = ( 450 ) ( 12 ) + 0.009 B – ( 10 ) ( 365 ) = 1750 + 0.009 B dollars/year 3. a. dQ/dt = 0 – ( 100 grams / 450 liters ) ( 9 liters / minute ) dQ/dt = – 2 grams/minute (for 0 < t < 50) b. dQ A /dt = ( 2 grams/minute ) – ( Q A / 300 grams/liter )( 9 liters/minute ) = 2 – 9 Q A / 300 dc/dt = (1/300) ( dQ A / dt ) dc/dt = (1/300) ( 2 – 9 c ) ( grams/liter ) / minute (for 0 < t < 50) 4. a. x ( t ) = – 2 + ( 5 / 10 ) t; y ( t ) = – 3 – ( 4 / 10 ) t; 0 < t < 10 seconds. b. x ( t ) = – 3 sin ( t / 8 ) + 1; y ( t ) = – 5 cos ( t / 8 ); 0 < t < 48 seconds 5. a. dy/dx = (dy/dt)/(dx/dt)
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