Loyola University Chicago Midterm 2 Sample A solutions Math 161-002, Spring 2013

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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 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 of 1 nautical per .)

Solution: Using the position of ship A as the reference point, at time t measured in 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 −