# Thus f x h f x f h 1 f x 2 1 f x f h so the newtonian

• Homework Help
• 19
• 100% (7) 7 out of 7 people found this document helpful

This preview shows page 14 - 17 out of 19 pages.

Thus f ( x + h ) f ( x ) = f ( h )(1 + f ( x ) 2 ) 1 f ( x ) f ( h ) , so the Newtonian quotient for f is given by f ( x + h ) f ( x ) h = f ( h ) h braceleftBig 1 + f ( x ) 2 1 f ( x ) f ( h ) bracerightBig . Hence by the properties of limits, lim h 0 f ( x + h ) f ( x ) h = (1 + f ( x ) 2 ) braceleftBig lim h 0 f ( h ) h bracerightBig × × 1 1 lim h 0 f ( x ) f ( h ) . But then by the second set of properties for f , lim h 0 f ( h ) h = 6 , lim h 0 f ( h ) = 0 . Consequently, f ( x ) = 6(1 + f ( x ) 2 ). 028(part2of2)10.0points (ii) By solving the differential equation in Part (i) for f , find the value of f ( π 18 ). 1. f parenleftBig π 18 parenrightBig = 1 3 2. f parenleftBig π 18 parenrightBig = 3 correct 3. f parenleftBig π 18 parenrightBig = 2 4. f parenleftBig π 18 parenrightBig = 1 2 5. f parenleftBig π 18 parenrightBig = 3 2 Explanation: The differential equation for f can be writ- ten as dy dx = 6(1 + y 2 ) , setting y = f ( x ). This is a separable variables equation which in turn can be written as integraldisplay 1 1 + y 2 dy = integraldisplay 6 dx. After integration, therefore, arctan y = 6 x + C with C an arbitrary constant. Thus f ( x ) = tan(6 x + C ) . Now f (0) = 0, so f ( x ) = tan 6 x . Hence at x = π 18 , f ( π 18 ) = tan( π 3 ) = 3. 029 10.0points Solve dy dx = 7 x 13 y 1 + x 14 for y.
pacheco (jnp926) – Homework 5 – staron – (52840) 15 1. arcsin(1 + x 14 ) + C 2. arccos(1 + x 14 ) + C 3. arctan(1 + x 14 ) + C 4. None of these correct Explanation: y = 7 x 13 y 1 + x 14 y y = 7 x 13 1 + x 14 integraldisplay y y dx = integraldisplay 7 x 13 dx 1 + x 14 ln | y | = 1 2 integraldisplay 14 x 13 dx 1 + x 14 = 1 2 ln | 1 + x 14 | + C 0 y = e 1 2 ln | 1+ x 14 | + C 0 = e ln 1+ x 14 e C 0 y = C radicalbig 1 + x 14 030 10.0points Newton’s Law of Cooling asserts that the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature. A hot poker is immersed in water whose temperature is held constant at 70 F . After 6 minutes the temperature of the poker has dropped to 94 F , while after 9 minutes it has dropped to 86 F . What was the temperature of the poker initially? 5. initial temp = 109 F Explanation: If T = T ( t ) is the temperature of the poker t minutes after it is immersed in the water, Newton’s Law of Cooling ensures that dT dt = k ( T 70) . This is a separable variables differential equa- tion which becomes integraldisplay 1 T 70 dT = integraldisplay k dt after separating variables and integrating. Consequently, ln( T 70) = kt + C, i . e ., T 70 = e kt + C . Thus ( ) T ( t ) = 70 + Be kt , and so the inital temperature of the poker is given by T (0) = 70 + B . To determine B we use the conditions 031 10.0points
pacheco (jnp926) – Homework 5 – staron – (52840) 16 Diana is sick in hospital with a severe bac- terial infection. She is to be fed antibiotics intravenously at a constant infusion rate, and the doctor knows that the anti-biotic will be eliminated from the bloodstream at a rate proportional to the amount y ( t ) present in the bloodstream at time t