Name: PID:Circle your section time/TA:Christine James James3pm (A01) 4pm (A02) 5pm (A03)Math 10CMidterm Examination 1Oct 18, 2010Turn off and put away your cell phone.No calculators or any other devices are allowed on this exam.You may use one page of notes, but no books or other assistance on this exam.Read each question carefully, answer each question completely, and show all of your work.Write your solutions clearly and legibly; no credit will be given for illegible solutions.If any question is not clear, ask for clarification.# Points Score1 122 93 114 8Σ 401. Suppose X represents the length of time, in hours, that a light bulb burns before itgoes out. Suppose that the cumulative distribution function for X is given byP (x) =1 − e−0.03xx ≥ 00 otherwise.(a) (4 points) Find the probability that the light bulb burns for between 1 and 2hours.Solution: Since P (x) is a cdf, the probability that the lightbulb burns forbetween 1 and 2 hours is given byP (2) − P (1) = (1 − e−0.03·2) − (1 − e−0.03)= e−0.03− e−0.06.(b) (4 points) Find the probability density function associated to X.Solution: The pdf is the derivative of the cdf, so we havepdf = p(x) = P0(x) =0.03e−0.03xx ≥ 00 otherwise.(c) (4 points) Find the mean number of hours that the light bulb burns.Solution: The mean is given by the formula mean =R∞−∞xp(x)dx. Since ourdistribution is only nonzero for x ≥ 0, we can take the integral from 0 to ∞.Using integration by parts, the mean isZ∞0x(0.03e−0.03x)dx =Z∞00.03xe−0.03xdx= −xe−0.03x∞0+Z∞0e−0.03xdx= 0 +−10.03e−0.03x∞0=10.03=10032. (a) (5 points) Find the equation of a plane through the points (1, 2, 0), (−1, 4, −8),and (2, −5, 10).Solution: This is one of many possible ways to solve this problem. Othermethods are acceptable.We know the equation looks like z = c + mx + ny. The first point gives0 = c + m + 2n, so we have c = −m − 2n. Thus, our equation isz = −m − 2n + mx + ny.Plugging the next two points into the above, we obtain the equations −8 =−2m + 2n and 10 = m − 7n. Rearranging the second equation gives m =10 + 7n. Plugging this in to the equation −8 = −2m + 2n gives −8 = −2(10 +7n) + 2n, so we can solve for n and obtain n = −1. Then m = 10 + 7n = 3,and c = −m − 2n = −3 + 2 = −1. Therefore, the equation of the plane isz = −1 + 3x − y.(b) (4 points) Suppose you have a linear function z = f (x, y). The level curve withz = 1 corresponds to the line y = 3x − 3. Explain in words what the contourdiagram of the function looks like. Give a sketch of the contour diagram.Solution: We know that the contour diagram of a linear function is a collec-tion of evenly spaced parallel lines. Thus, the contour diagram for f (x, y) willlook like evenly spaced lines of slope 3. Note that it is impossible from theproblem to tell the exact spacing of the lines.3. Let f(x) = e2x+ x2.(a) (5 points) Find the third degree Taylor polynomial for f centered at 0.Solution: The Taylor polynomial centered at 0 is of the form P3(x) = C0+C1x + C2x2+ C3x3, where Ck=f(k)(0)k!. Thus, we need to find the first threederivatives of f at 0. We have derivativesf(x) = e2x+ x2⇒ f(0) = 1f0(x) = 2e2x+ 2x ⇒ f0(0) = 2f00(x) = 4e2x+ 2 ⇒ f00(0) = 6f(3)(x) = 8e2x⇒ f(3)(0) = 8Therefore, we obtain the Taylor polynomialP3(x) = C0+ C1x + C2x2+ C3x3= 1 +21!x +62!x2+83!x3= 1 + 2x + 3x2+43x3(b) (3 points) Approximate the value of f (0.25) using the Taylor polynomial frompart (a).Solution: To approximate, we just plug 0.25 =14into P3(x). We obtainf(0.25) ≈ P3(0.25) = 1 + 214+ 3142+43143= 1 +12+316+148=8248=4124(c) (3 points) Suppose the third degree Taylor polynomial centered at 0 for a functiong is given by P3(x) = 3 + 2x − x2+ 3x3. Describe the behavior of g near 0.Solution: From the Taylor polynomial, we have that the value of g(0) is 3, gis increasing at 0 with slope 2, and g is concave down at 0.4. Let f(x, y) = (x + 1)2+ y2.(a) (4 points) Sketch a contour diagram for f(x, y) and clearly label the level curves.Show the contours for at least three levels.Solution: The contour diagram consists of circles centered at (−1, 0). Theseare the contours for z = 1, 2, 3, where z = 1 is the smallest circle and z = 3 isthe largest.(b) (4 points) Sketch the graph of z = f (x, y).Solution: The graph is a paraboloid (a bowl) with minimum point at(−1, 0,