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Unformatted text preview: AMS 151 (Fall, 2009) Joe Mitchell Applied Calculus I Practice Problems for Quiz # 6 Solution Notes 1. Gasoline is pouring into a cylindrical tank of radius 3 feet. When the depth of the gasoline is 4 feet, the depth is increasing at 0.2 ft/sec. How fast is the volume of gasoline changing at that instant? (give units!) Let y be the depth in feet of the gasoline in the tank; y is actually a function, y ( t ), of time. The volume, V ( y ), when the depth is y is V ( y ) = r 2 y = 9 y cubic feet. Taking the derivative with respect to time, we get dV dt = 9 dy dt . At the moment when y = 4 feet, we are told that dy dt = 0 . 2 ft/sec, at which point we get dV dt = 9 (0 . 2) = 1 . 8 cubic feet per second. 2. A voltage V across a resistance R generates a current of I = V/R . If a constant voltage of 9 volts is put across a resistance that is increasing at a rate of 0.2 ohms per second when the resistance is 5 ohms, at what rate is the current changing? (give units!) The voltage V is constant, but R ( t ) and I ( t ) are both functions of time. In particular, I ( t ) = V/R ( t ), so, taking derivatives with respect to time t , we get dI dt =- V R 2 dR dt . Thus, since V = 9 volts, the rate of change of the current, dI dt , when R ( t ) = 5 ohms and dR dt = 0 . 2 ohm/sec is dI dt =- 9 volts 5 2 ohm 2 (0 . 2)( ohm/sec ) =- 1 . 8 25 ( volt/ohm ) /sec The current is decreasing (since the rate of change is negative), and the units are (volt/ohm) per second. You may know that volt/ohm is an ampere, so the units are amperes per second....
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This note was uploaded on 02/07/2010 for the course AMS 151 taught by Professor Zhang during the Fall '08 term at SUNY Stony Brook.
- Fall '08