MATH 220 Midterm Thurs., Sept. 20, 2007 Write your name and ID number at the top of this page. Show all your work. You may refer to one double-sided sheet of notes during the eam and nothing else. Calculators are not allowed. You have until 8:00PM to finish the eam. Stay strong, be cool, and don t forget to breathe. Begin Midterm. Find the value of each of the following epressions. For this problem, we use the fact that the functions ln and e are inverses of each other, which means ln e = and e ln =. We need to do a little bit of algebra to get epressions (a) and (c) into the first (left) and second (right) forms above, respectively. For epression (b), the numerator and denominator are already in the second (right) form above. (a) ln 3 e = ln e /3 = 3 (b) eln 7 e ln 5 = 7 5 (c) ( e ) ln 0 2 = (e ) ln 00 = e ln 00 = 00
2. (a) Determine sin ( ). We want the angle whose sine is, in the range ( sin π ) =, so sin ( ) = π 2 2. [ π 2, π ]. 2 ( (b) Determine tan tan 3π 4 ). tan 3π =, so we want the angle whose tangent is, in the 4 ( range π 2, π ). ( 2 tan π ) =, so tan ( ) = π 4 4. ( (c) Determine sin tan 5 2 Let θ = tan 5 2 ) in terms of. 5, which means tan θ =, and we need to find 2 sin θ. Draw a right triangle and label one of the angles θ. Let the side opposite θ be length 5 and the side adjacent to theta be length 2, which reflects the fact that tan θ = 5/(2) (opposite-overadjacent). Then use the pythagorean theorem to calculate the length of the hypotenuse: 25 + 4 2. 5 Then, sin θ = 25 + 4 (opposite-over-hypotenuse). 2
3. Find the inverse function of + 4 f() = 2, 4. 3 Be sure to state the domain of your new function. Let y = +4 3 2, and solve for in terms of y to get Then replace y with to get = 3(y + 2) 2 4. f () = 3( + 2) 2 4. For any inverse function f, the domain of f is the same as the range of f. Looking back at f, we see that the range is f() 2 (because the square-root portion has to be positive or zero). Therefore, the domain of f () is 2, or [ 2, )
4. Let f() = 2 + (9 4) 2. Use logarithmic differentiation to find f (). Let y = 2 + (9 4) 2. Then take the natural log of both sides, and use log properties to make things easier for differentiating later. 2 + ln y = ln (9 4) = 2 ln(2 +) 2 ln(9 4) 2 = 2 ln(2 +) 2 ln(9 4). Then differentiate with respect to, using implicit differentiation (chain rule) on ln y, Simplify, Solve for dy d, dy y d = 2 2 + 2 2 9 4 9. dy y d = 2 + 8 9 4. ( dy d = y 2 + 8 ). 9 4 Finally, replace y with f() above, ( f 2 + () = (9 4) 2 2 + 8 ). 9 4
5. A population at time t (measured in years) is given by the function P (t) = 450, 000e.02t. (a) Find the doubling time for the population. Hint: Use ln 2 0.7. Let T be the doubling time. Since P (0) = 450, 000, it must be true that P (T ) = 900, 000 (doubled from time 0 to time T ). Then, 900, 000 = 450, 000e.02T. Solve for T: 2 = e.02t ln 2 =.02T T = ln 2 35 years..02 (b) Write down the differential equation and initial condition to which P (t) above is the solution. P (t) above describes eponential growth, which we derived in class from a differential equation that assumes that the rate of increase of the population ( dp ) is proportional to the size of the population dt (P ), with proportionality constant k. In this case, k =.02. The differential equation and initial condition are dp dt =.02P, P (0) = 450, 000.
6. Calculate the following integrals. (a) e d = ln ] = ln ln e = 0 = e (b) e d Use the substitution so u = = 2, du = 2 2 = 2 d. e d = 2 e 2 d = 2 = 2e u + C = 2e + C e u du (c) sinh cosh d Use the substitution u = sinh, so du = cosh d. Then, sinh cosh d = u du = 2 u2 + C = 2 sinh2 + C Note that 2 cosh2 + C is another way to epress the answer.
7. Find dy d for each of the following: (a) y = log 3 (sin ) By the definition of log, 3 y = sin Take the natural log of both sides: ln 3 y = ln(sin ) y ln 3 = ln(sin ) Now, y = ln(sin ) ln 3 dy d = (ln 3)(sin ) cos (b) y = 3 sin Take the natural log of both sides, ln y = ln 3 sin ln y = (sin )(ln 3) Now differentiate, dy y d = (cos )(ln 3) dy d = 3sin (cos )(ln 3)
(c) y = (ln ) ln Take the natural log of both sides, ln y = ln [ ln (ln ) ] ln y = ln ln(ln ) Now differentiate, using the product rule on the right-hand side dy y d = ln(ln ) + ln ln dy d = (ln )ln ( ) ln(ln ) +
8. Find the particular solution to the differential equation given that y = 0 when =. dy d 3y = 3 + 2, First we need to put the equation into the form dy + P ()y = Q(): d dy d 3 y = 2 + 2 So P () = 3, and the integrating factor is 3 e d = e 3 ln = e ln 3 = 3 Multiply both sides by the integrating factor and proceed: 3 dy d 3 2 y = + 2 3 d d ( 3 y) = + 2 3 3 y = [ + 2 3 ] d 3 y = ln 2 + C y = 3 ln + C 3 Now use the initial condition to solve for C, 0 = 0 + C, so C = Finally, y = 3 ln + 3
9. Re-write the differential equation in the previous problem in the form suitable for using Euler s method. Then perform one step of Euler s method with stepsize h = to approimate y(2). The form suitable for Euler s method is dy = f(, y), so rearrange the d differential equation in the previous problem to get, dy d = 3 + 2 + 3y Use the initial condition in the previous problem to start off Euler s method: 0 =, y 0 = 0 With h =, = 0 + h = 2, and So y = y 0 + hf( 0, y 0 ) = 0 + f(, 0) = + 2 + 0 y(2) 3 = 3
0. Calculate the following integrals. (a) e 4 9 e 4 d Use the substitution u = 9 e 4, making du = 4e 4 d Then, e 4 d = 9 e 4 4 9 e 4 ( 4e4 ) d = 4 u du = 4 u 2 du = 4 2u 2 + C = 2 9 e 4 + C (b) e 2 9 e 4 d This time, use the substitution u = e 2, making du = 2e 2 d Then, e 2 d = 9 e 4 2 9 (e 2 ) 2 (2e2 ) d = 2 = ( ) e 2 2 sin + C 3 9 u 2 du
(c) 9 + 6 4 2 d We can rearrange the quadratic epression under the square root to put it in the form a 2 u 2 : ( 9 + 6 4 2 = 4 2 4 9 ) ( = 4 2 4 + 4 25 ) 4 4 The integral becomes, [ ] 25 = 4 ( 2)2 4 4[ 25 ( 4 2)2 ] d = 2 25 4 d ( 2)2 Use the substitution u = 2, so that du = d, and proceed, 2 25 du = ( ) 5/2 4 u2 2 sin + C u = ( ) 5 2 sin + C 2( 2)