Math 113 HW #8 Solutions

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Jeffry Anthony
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1 Math HW #8 Solutions. Exercise.8.0. A sample of tritium- decayed to 94.5% of its original amount after a year. (a) What is the half-life of tritium-? Answer: If N(t) is the amount of tritium- relative to the original amount, we know that the general form of N(t) is N(t) = Ce kt. Al, we know that N(0) =, = N(0) = Ce k 0 = C, C = and we can write Al, we know N() = 0.945, N(t) = e kt = N() = e k = e k. k = ln(0.945). N(t) = e ln(0.945)t = ( e ln(0.945)) t = (0.945) t for any t. The half-life of tritium- is the amount of time t 0 such that N(t 0 ) = 0.5. we can lve for t 0 from the equation 0.5 = N(t 0 ) = (0.945) t 0. ln(0.5) ln(0.945 t 0 ) = t 0 ln(0.945), t 0 = ln(0.5) ln(0.945).5. the half-life of tritium- is.5 years. (b) How long would it take the sample to decay to 0% of its original amount? Answer: If t is the time it takes the sample to decay to 0% of its original amount, 0. = N(t ) = (0.945) t, meaning that (if we take the natural log of both sides), ln(0.) = ln(0.945 t ) = t ln(0.945), t = ln(0.) 8.45 years. ln(0.945)

2 . Exercise.8.4. A thermometer is taken from a room where the temperature is 0 C to the outdoors, where the temperature is 5 C. After one minute the thermometer reads C. (a) What will the reading on the thermometer be after one more minute? Answer: From Newton s Law of Cooling, we know that T (t) = T s + Ce kt. The ambient temperature is T s = 5 C, whereas C =. Moreover, we know that T () =, Hence, 0 = T (0) = 5 + Ce k 0 = 5 + C, T (t) = 5 + e kt. = T () = 5 + e k = 5 + e k, e k = 7. k = ln. T (t) = 5 + e ln( )t 7 = 5 + (e ln( ) ) t 7 t = 5 +. after minutes, the temperature of the thermometer will be T () = 5 + = = 4 = (b) When will the thermometer read 6 C? Answer: The time t 0 when T (t 0 ) = 6 is given by 6 = T (t 0 ) = 5 +, ( ) 7 =. ( ) ( ) 7 ln = ln = t 0 ln. t 0 = ln ln 7.55, the thermometer will read 6 C after about and a half minutes.

3 . Exercise.8.6. A freshly brewed cup of coffee has temperature 95 C in a 0 C room. When its temperature is 70 C, it is cooling at a rate of C per minute. When does this occur? Answer: From Newton s Law of Cooling, the temperature of the coffee is given by At time t = 0, T (t) = 0 + Ce kt. 95 = T (0) = 0 + Ce k 0 = 0 + C, meaning that C = 75 and T (t) = e kt. At me time t 0, or 70 = T (t 0 ) = e kt 0, 75e kt 0 = 50, e kt 0 =. ( ) kt 0 = ln. We al know the rate of change of T at this time t 0 : In other words, Since kt 0 = ln ( ), we know that Since kt 0 = ln ( ), we know that = T (t 0 ) = 75(e kt 0 k) = 75ke kt 0. k = t 0 = ln ( k k = 75e kt 0. 75e ln( ) = 75 ) = ln ( ) = 50 ln the cup of coffee is 70 C after just over 0 minutes. 4. Exercise.0.. Find the differential of the functions (a) y = s/( + s) Answer: If f(s) = Now, 50 s +s, then, by definition, f (s) = the differential is dy = f (s)ds. = 50. ( ) 0., ( + s) s + s s ( + s) = ( + s) = dy = ds ( + s). ( + s).

4 (b) y = e u cos u Answer: If g(u) = e u cos u, then, by definition, Since dy = g (u)du. g (u) = e u cos u + e u ( sin u) = e u cos u e u sin u = e u (cos u + sin u), we have that dy = e u (cos u + sin u)du. 5. Exercise.0.8. (a) Find the differential dy of y = cos x. Answer: By definition, if f(x) = cos x, then Since f (x) = sin x, this means that (b) Evaluate dy for x = π/ and dx = dy = f (x)dx. dy = sin xdx. Answer: Given the above expression for dy and knowing that sin π =, we have that dy = (0.05) = Exercise.0.4. Use a linear approximation (or differentials) to estimate e 0.0. Answer: Let f(x) = e x. If L(x) is the linearization of f at 0, then L(x) = f(0) + f (0)(x 0) = + (x 0) = + x. Since 0.0 is close to 0, it should be the case that e 0.0 L( 0.0) = + ( 0.0) = Exercise.0.. Let f(x) = (x ), g(x) = e x, h(x) = + ln( x). (a) Find the linearizations of f, g, and h at a = 0. What do you notice? How do you explain what happened? Answer: By definition, the linearization of f is f(0) + f (0)(x 0) = f(0) + f (0)x. Since f (x) = (x ), we know that f(0) = and f (0) =, the linearization of f is x. 4

5 By definition, the linearization of g is g(0) + g (0)(x 0) = g(0) + g (0)x. Since g (x) = e x, we know that g(0) = and g (0) =, the linearization of g is By definition, the linearization of h is Since h (x) = x h(0) + h (0)(x 0) = h(0) + h (0)x. x, we know that h(0) = and h (0) =, the linearization of h is x. We notice that all three linearizations are the same. This occurs because f(0) = g(0) = h(0) and f (0) = g (0) = h (0): all three functions have the same value at 0 and their derivatives al have the same value at 0. Of course, this says nothing about the behavior of the three functions at other points. (b) Graph f, g, and h and their linear approximations. For which function is the linear approximation best? For which is it worst? Explain. Answer: Figure : Blue: f; Red: g; Purple: h; Black: linearization From the picture, the linear approximation appears to be best for f and worst for h. 5

Particular Solutions. y = Ae 4x and y = 3 at x = 0 3 = Ae 4 0 3 = A y = 3e 4x

Particular Solutions. y = Ae 4x and y = 3 at x = 0 3 = Ae 4 0 3 = A y = 3e 4x Particular Solutions If the differential equation is actually modeling something (like the cost of milk as a function of time) it is likely that you will know a specific value (like the fact that milk