Loyola University Chicago Midterm 2 Sample A solutions Math 161-002, Spring 2013 Name (print): Signature: Please do not start working until instructed to do so. You have 75 minutes. You must show your work to receive full credit. No calculators. You may use one one-sided 8.5 by 11 sheet of handwritten (by you) notes. Problem 1. Problem 2. Problem 3. Problem 4. Problem 5. Problem 6. Problem 7. Total. Problem 1.(25 points) Find the following derivatives. Put a box around your final answer . 0 a.(5 points) 3x5 7√x +4x . − Solution: 0 7 3x5 7√x +4x = 15x4 +ln44x − − 2√x d b.(5 points) xt3 + cos(4x) e3t dt − Solution: d xt3 + cos(4x) e3t =3xt2 3e3t dt − − dy c.(5 points) , in terms of x and y, if x2y + ln y =5 dx Solution: implicit differentiation gives 2 0 1 0 0 2 1 0 2xy 2xy + x y + y =0, y x + = 2xy, y = −2 1 y y − x + y d d.(5 points) arcsin(x5) dx Solution: d 5x4 arcsin(x5) = dx 1 (x5)2 − p d arctan y e.(5 points) dy y +1 Solution: 1 d arctan y 1+ 2 (y +1) arctan y = y − dy y +1 (y + 1)2 Problem 2.(10 points total) Use linear approximation to approximate √3 25. 3 0 1 −2/3 Solution: Notice that √27 = 3. Let f(x) = √x. Let a = 27. Then f(a)=3, f (x) = 3 x , 0 1 −2/3 1 f (a) = 3(27) = 27. The linear approximation of f(x) at x = a is then 0 1 L (x) = f(a) + f (a)(x a)=3+ (x 27) a − 27 − Plug in x = 25, get 1 25 3 + ( 2)=2 27 − 27 By the way: 2 25 2.9259 and the real value is √3 25 2.9240. 27 ≈ ≈ Problem 3.(10 points total) Simplify tan(arccos x) to an expression not involving trig functions. Solution: Consider a right triangle with one acute angle α, hypotenuse 1, adjacent side x. Then cos α = x, so α = arccos x. Then the opposite side is √1 x2, and so − √1 x2 tan (arccos x) = tan α = − x Problem 4.(10 points) At noon, ship A is 40 nautical miles due south of ship B. Ship A is sailing west at 21 knots and ship B is sailing north at 17 knots. How fast (in knots) is the distance between the ships changing at 4 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.) Solution: Using the position of ship A as the reference point, at time t measured in hours past noon, ship A is 21t miles west of this point and ship B is 40+17t north of this point. The distance between ships is then d(t) = (21t)2 +(40+17t)2. p The rate of change of this distance is dd 42t +2(40+ 17t)17 = . dt 2 (21t)2 +(40+17t)2 p Plug t =4 into this rate of change to get the answer. Problem 5.(10 points total) Write the definition of the derivative of a function f at a point x and then use this definition to find the derivative of f(x) = 5x2 7. p − Solution: The derivative of f at x is the following limit, if it exists: 0 f(x + h) f(x) f (x)= lim − h→0 h Use it with f(x) = √5x2 7: − 2 2 0 f(x + h) f(x) 5(x + h) 7 √5x 7 f (x) = lim − = lim − − − h→0 h h→0 p h 5(x + h)2 7 √5x2 7 5(x + h)2 7 + √5x2 7 = lim − − − − − h→0 p h p5(x + h)2 1 + √5x2 7 2 2 − − 5(x + h) 7 5x +7 p = lim − − h→0 h( 5(x + h)2 7 + √5x2 7) 2 − 2 2− p5x +10xh +5h 5x = lim − h→0 h( 5(x + h)2 7 + √5x2 7) − − p 10x +5h = lim h→0 5(x + h)2 7 + √5x2 7 10x − − = p 2√5x2 7 − Problem 6.(10 points) A bus will hold 60 people. the number x of people per trip who use the 2 bus is related to the fare charged (p dollars) by the law p = (3 (x/40)) . Write an expression for − the total revenue r(x) per trip received by the bus company. What number of people per trip will make the marginal revenue dr/dx equal to 0? What is the corresponding fare? (This fare is the one that maximizes the revenue.) Solution: In this case, revenue is number of people times fare. x 2 r(x) = x 3 − 40 Then marginal revenue is 2 0 x x 1 x x 2x x 3x r (x) = 3 + x2 3 = 3 3 = 3 3 − 40 − 40 −40 − 40 − 40 − 40 − 40 − 40 Then r0(x)=0 if x = 120 or x = 40. The bus will not hold 120 people, so the answer is x = 40, and for that x, p =4. Problem 7.(10 points) Find two lines tangent to the circle x2 + y2 = 4 which intersect the x-axis at the point where x = 4. Solution: First note this is about a circle centered at the origin and of radius 2. Symmetry suggests that it is enough to find the tangent line touching the circle on the upper side; the lower side one can be then deduced by symmetry: if the upper one is y = mx + b then the lower one is y = mx b. Let a be between 2 and 2. The slope of the circle at the point − − − a, 4 a2 p − is d 2 2a a 4 a = − 2 = − 2 . da − 2√4 a √4 a p − − The equation of the line tangent to the circle at the point a, 4 a2 is then p − a y 4 a2 = − (x a). − − √4 a2 − p − We want the point (4, 0) to be on that line, so plug in x =4 and y =0, get a 4 a2 = − (4 a). − − √4 a2 − p − Simplify and solve for a: (4 a2) = a(4 a), 4 + a2 = 4a + a2, 4a =4, a =1. − − − − − − Thus, the tangent line we want happens at the point corresponding to a = 1. Plug a = 1 then into the equation of the tangent line we had above: 1 y 4 12 = − (x 1). − − √4 12 − p − Simplify, obtain 1 1 y = x + + √3. −√3 √3 This is the “upper” tangent line, the lower one is 1 1 y = x √3. √3 − √3 −.