Math 116 Exam 1 Fall 2014
Name:______This is a closed book exam. You may use a calculator and the formulas handed out with the exam. You may find that your calculator can do some of the problems. If this is so, you still need to show how to do the problem by hand. In other words, show all work and explain any reasoning which is not clear from the computations. (This is particularly important if I am to be able to give part credit.) Turn in this exam along with your answers. However, don't write your answers on the exam itself; leave them on the pages with your work. Also turn in the formulas; put them on the formula pile.
1. (10 points) Find a formula for the inverse function of the function y = f(x) = .
2. (10 points) Suppose y = g(x) is the inverse function to x = f(y) = 3 + y + tan. Find a formula for the derivative of g. Express answer as a formula involving g(x).
3. (28 points) Find for each of the following.
a. y = x2etanh x + ln( tan-1x + e1/x ) b. y =
4. (28 points) Do the following integrals. In each case, start with a substitution.
a. b.
5. You are studying a growing population of protozoa. Let N be the number of protozoa and t be the time in days since the start of the experiment. Suppose at t = 0 there are 1000 protozoa. Also suppose the population grows with the rate of growth (in protozoa per day) equal to 40% of the population size.
a. (5 points) Translate the last statement into an equation involving .
b. (9 points) Solve this equation along with the other information to get a formula for N in terms of some exponential function of the variable t and the numbers given above.
6. (10 points) Find
Solutions
1. Solve for x in terms of y. y = (1 - )/(1 + ) Û y(1 + ) = 1 - Û y + y = 1 - Û y + = 1 - y Û (y + 1) = 1 – y Û = (1 – y)/(y + 1) Þ x =f1(y) = (1 – y)2/(y + 1)2.
2. Method 1: = (4 pts) = = = = (6 pts)
Method 2: Differentiate both sides of x = 3 + y + tan(y/2) with respect to x. = (3 + y + tan(y/2)) (2 pts)
Þ 1 = + sec2(y/2) (4 pts)
Þ 1=[1 + sec2(y/2) ] Þ = = = . (4 pts)
3. a. [x2etanh x + ln(tan-1x + e1/x)] = [x2etanh x] + [ln(tan-1x + e1/x)] = x2[etanh x] + etanh x [x2] + [(tan-1x + e1/x)] (2+ 2 pts) = x2etanh x[tanh x] + 2xetanh x + [ + e1/x] (1.5 + 1 + 2 + 1.5 pts) = x2etanh xsech2x + 2xetanh x + [ - ] (2 + 2 pts)
b. = (2.5 pts) = (3 + 2.5 pts) = (3 + 3 pts)
4. a. . Let u = 2x + 1. Then du = 2x (ln 2) dx and du = 2x dx. So = (6 pts) = (5 pts) = (3 pts)
b. . Let u = 4x. Then du = 4dx and du = dx. So = (6 pts) = (5pts) = (3 pts)
5. a. = 0.4 N
b. In class we showed that if = kA, then A = Cekt where C is a constant. In our case A = N and k = 0.4, so N=Ce0.4t. (5pts) To find C we substitute t = 0 and N = 1000. This gives C = 1000, so N=1000e0.4t. (4pts)
5. Both top and bottom are zero when x = 0. Use l’Hospital’s rule. = (1 pt) = . (3 pts) Again, both top and bottom are zero when x = 0. Use l’Hospital’s rule again. = (1 pt) = . (3 pts) Plug in x = 0 to get = . (3 pts)