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1 MATH Quiz 4 Spring 8 Solutions. (5 points) Express ln() in terms of ln() and ln(3). ln() = ln( 3) = ln( ) + ln(3) = ln() + ln(3). (5 points) If g(x) is pictured in Figure and x Figure : g(x) is f(x) invertible? f(x) = x g(t) ; x, Yes. f(x) is an accumulation function and g(x) is strictly positive, so f(x) is strictly increasing. Monotonic functions have inverses. 3. (5 points) Suppose dy = f(x), f() =, and f(x) is decreasing. Suppose you use Euler s Method to approximate y(). Will your estimate be an overestimate or an underestimate? If f(x) is decreasing then y is concave down, so y curves beneath its tangent lines. Euler s Method will give an overestimate. 4. (5 points) f(x) is shown in Figure. Does f(x) have an inverse on its entire domain? Sketch a graph of f (x), restricting the domain if necessary.

2 K 3 x x K Figure : f(x) f(x) has an inverse since it is strictly increasing. To sketch a graph of the inverse, reflect diagonally. 5. (5 points) Suppose T (t) = ( 5)e t describes the temperature of an object with respect to time. Is the object getting hotter or colder? What temperature is it

3 approaching? The object is getting hotter since T (t) = 5e t is positive. T (t) is approaching since e t goes to as t increases. 6. (5 points) What is the derivative of e cosh x? Let w = cosh(x). d ( ) e cosh(x) = d (ew ) = d(ew ) dw dw = e w dw = e cosh(x) sinh(x) 7. (5 points) dy = ky, y() = and y () = 4. Find y(x). So k =. 4 = y () = dy () = k y() = k dy = y dy = y y dy = ln(y) = x + C y = e x+c = C e x y() = = C So y = e x 3

4 8. (5 points) Is sin(x) a solution to d y 4y =? No. d ( sin(x)) 4( sin(x)) = 8 sin(x) 8 sin(x) = 6 sin(x) This function is not identically zero, so sin(x) is not a solution. 9. (5 points) Calculate 4 + x 4 + x = 4( + x 4 ) substitute w = x = = 4( + w ) dw + w dw = arctan(w) w= = arctan() arctan() = π 4. (5 points) f(x) = x 3x +. Restrict the domain of f(x) so that it has an inverse, while keeping its range as large as possible. Then find f (x). Consider the derivative f (x) = x 3. This is negative for x < 3 and positive for x > 3. Thus, x = 3 is the global minimum for f(x), and if we restrict the domain to either (, 3 ] or [ 3, ) then f(x) will be strictly monotonic, hence, invertible. We will choose to restrict the domain to [ 3, ). The range of f is y

5 f(x) = y = x 3x + = x 3x + ( y) x = 3 ± 9 4( y) (by quadratic formula) x = ( y) (since x 3 ) x = y f (x) = x ; x 5 4. (5 points) A batch of brownies cooks in a 35 oven. The temperature in the room is 7 and the temperature inside the refrigerator is 38. The brownies are taken out of the oven and placed on the counter. After 5 minutes you try a brownie but burn yourself, they are still. How much longer must you wait until the brownies reach and are safe to eat? Newton s law of cooling gives us the formula: T (t) = T A + (T T A )e kt T A = 7 and T = 35. We also know that T (5) = = 7 + (35 7)e 5k 4 8 = e 5k k = 5 ln(4 8 ) = ln() 5 Knowing k we can now solve for the time when the brownies will be. T (t) = = 7 + (35 7)e kt 4 8 = e kt t = k ln( 7 ) = ln(7) k So the brownies will be cool in another 9 minutes. 5 = 5 ln(7) ln() 4

6 . (5 points) A tank initially contains 5 gallons of brine, with 3 lbs of salt in solution. Fresh water runs into the tank at 3 gallons per minute, and the well stirred solution runs out at gallons per minute. How long will it take until there are 5 lbs of salt in the tank? Let A(t) be the amount of salt in pounds in the tank after t minutes. Let V (t) be the volume of water in the tank. A() = 3. V () = 5. The volume in the tank is not constant, more water enters the tank than leaves. dv = rate in rate out = 3 gal/min gal/min = gal/min V (t) = t + V = t + 5 da(t) da(t) = rate in rate out A(t) lbs = (3 gal/min)( lbs/gal) ( gal/min) V (t) gal = A(t) t + 5 A(t) da(t) = t + 5 A(t) da(t) = t + 5 ln(a(t)) = ln(t + 5) + C A(t) = e ln(t+5)+c ln(t+5) = Be A(t) = B(t + 5) A() = 3 = B 5 B = 3 5 A(t) = 3 5 (t + 5) 6

7 Now if A(t) = 5 then A(t) = 5 = 3 5 (t + 5) (t + 5) = 3 t = min 3. ( points) In 4 the population of the world was people and was increasing at a rate of people per year. Assume that the maximum population that the world can support is L =.78. Using the logistic model for population growth, determine the year in which the rate of population growth begins to slow. where dp (t) = hp (t)(l P (t)). P (t) = LP P + (L P )e Lht dp () = hp (L P ) h = dp () P (L P ) The rate of population growth will be greatest when the second derivative of population is zero. d P (t) dp (t) dp (t) = hp (t)( ) + h (L P (t)) dp (t) = h(l P (t)) So the population grows at the highest rate when P (t) = L. 7

8 L = P (t) = LP P + (L P )e Lht P + (L P )e Lht = P e Lht = P L P ( ) P Lht = ln L P t = ( ) L Lh ln P 5.3 P So the rate of population growth begins to slow in 9. 8