Calculus I Exam 3 Review solutions 2011

Exam 3 Review solutions 1. In Section 4.2, we had the following theorem: If f 0 (x) = 0 for all x in an interval (a, b), then f is constant on (a, b). Use that theorem to prove the following: If f 0 (x) = g 0 (x) on (a, b), then f (x) = g(x) + c on (a, b). SOLUTION: Let h(x) = f (x) − g(x). Since f 0 (x) = g 0 (x), then h0 (x) = f 0 (x) − g 0 (x) = 0, and so h(x) = c on (a, b). Rewriting what h is, f (x) − g(x) = c on (a, b), so that f (x) = g(x) + c. 2. Short Answer: (a) Give the definition of a critical point for a function f : A critical point is a value of x for which f 0 (x) = 0 or does not exist (from the domain of f ). (b) State the three “Value Theorems” (don’t just name them, but also state each): • Intermediate Value Theorem: If f is continuous on [a, b] and N is any number between f (a) and f (b), then there is a c in [a, b] such that f (c) = N . • Extreme Value Theorem: If f is continuous on [a, b], then f attains an absolute maximum and minimum on [a, b]. • Mean Value Theorem: If f is continuous on [a, b] and differentiable on (a, b), then there is a c in (a, b) such that f (b) − f (a) f 0 (c) = b−a (c) What is the procedure for finding the maximum or minimum of a function y = f (x) on a closed interval, [a, b]. SOLUTION: This is the closed interval method. Find the critical points (in the interval [a, b]. Build a table using the CPs and endpoints. The largest y−value is your absolute max, the smallest y−value is your absolute min. (d) How do we determine if a function has a local maximum or minimum? SOLUTION: First we compute the critical points. Then we can use either the first or second derivative test. The first derivative test says that if the sign of the first derivative changes from positive to negative, we have a local maximum. If the sign changes from negative to positive, we have a local minimum. If there is no sign change, we do not have a local extreme point. For the second derivative test, if f 0 (c) = 0, we check f 00 (c). If that value is positive, then c is where a local minimum occurs. If that value is negative, then c is where a local maximum occurs. (e) What is meant by linearizing a function? SOLUTION: We replace f by L(x) close by some base point x = a. In this case, we say f (x) ≈ L(x) = f (a) + f 0 (a)(x − a) 3. True or False, and give a short reason: (a) If f 0 (a) = 0, then there is a local maximum or local minimum at x = a. FALSE: For example, f (x) = x3 at x = 0. √ 2 +5−3 (b) There is a vertical asymptote at x = 2 for xx2 −2x NOTE: I suspect it is FALSE because both the numerator and denominator are zero at x = 2. To show it, I need to take the limit: lim f rac x→2 p x2 + 5 − 3x2 − 2xf rac p p p 1 x2 + 5 + 3 x2 + 5 + 3 = lim f rac(x + 2)(x − 2)x(x − 2)( x2 + 5 + 3) = x→2 3 so, it is FALSE. 1 (c) If f has a global minimum at x = a, then f 0 (a) = 0. FALSE. It could be that x = a is an endpoint for an interval, or a point where f 0 (a) does not exist. In the following, “increasing” or “decreasing” will mean for all real numbers x: (d) If f (x) is increasing, and g(x) is increasing, then f (x) + g(x) is increasing. TRUE: Since f, g are increasing, f 0 , g 0 are both positive so f 0 + g 0 is positive as well. (e) If f (x) is increasing, and g(x) is increasing, then f (x)g(x) is increasing. FALSE: It might be true, but (f g)0 = f 0 g + f g 0 so we might have that g < 0 and f < 0, for example. In that case, the derivative would be negative. (f) If f (x) is increasing, and g(x) is decreasing, then f (g(x)) is decreasing. TRUE: (f (g(x)))0 = f 0 (g(x))g 0 (x) so f 0 (evaluated at g(x)) is positive, and g 0 is positive. 4. Find the global maximum and minimum of the given function on the interval provided: √ (a) f (x) = 9 − x2 , [−1, 2] √ SOLUTION: f 0 (x) = −x/ 9 − x2 , so we add the critical point x = 0. Build a table: x −1 y 2.82 0 2 3.0 2.23 So the global minimum is y = 2.23 and the global maximum is y = 3.0 (b) g(x) = x − 2 cos(x), [−π, π] SOLUTION: g 0 (x) = 1+2 sin(x). Use a triangle and/or unit circle to find the values of x = −5π/6 and x = −π/6. Now the table: x y −π −5π/6 −1.14 −0.88 −π/6 π −2.25 5.14 The global minimium is approx −2.25 and the global max is approx 5.14. 5. Find the regions where f is increasing/decreasing: f (x) = x (1+x)2 SOLUTION: Simplifying the derivative, we get: f 0 (x) = 1−x (1 + x)3 Sign chart (include x = −1 although it is a vertical asymptote): f 0 (x) − x < −1 + −1 < x < 1 − x>1 Therefore, f is decreasing if x < −1 and if x > 1, and increasing if −1 < x < 1. 6. For each function below, determine (i) where f is increasing/decreasing, (ii) where f is concave up/concave down, and (iii) find the local extrema. 2 (a) f (x) = x3 − 12x + 2 (See Exercise 33, 4.3) Hint: f 0 (x) = 3x2 − 12 = 3(x2 − 4) = 3(x − 2)(x + 2), then look at a sign chart, and f 00 (x) = 6x. √ (b) f (x) = x 6 − x (See Exercise 39, 4.3) Hints: The domain is x ≤ 6, and we can simplify f 0 (x) = f 00 (x) = 3 x−4 ·√ 2 6−x 3 x−8 · 4 (6 − x)3/2 (c) f (x) = x − sin(x), 0 < x < 4π (See Exercise 44. 4.3) SOLUTION: f 0 (x) = 1 − cos(x), and cos(x) = 1 at x = 0, 2π, 4π. Since we do not include 0, 4π in the interval, we have: + + f 0 (x) 0 < x < 2π 2π < x < 4π Therefore, f 0 (x) > 0, and f is always increasing. For concavity, f 00 (x) = sin(x). The sine is posiive (so f is increasing) on (0, π) and (2π, 3π). The sine is negative (so f is decreasing) on (π, 2π) and (3π, 4π). For local extrema, the only critical point is 2π, and from our analysis of the first derivative, we see that this is a plateau, not a local extreme point (so there are no local extrema). 7. Suppose f (3) = 2, f 0 (3) = 12 , and f 0 (x) > 0 and f 00 (x) < 0 for all x. (a) Sketch a possible graph for f . SOLUTION: At the point (3, 2), the function increases and is concave down. (b) How many roots does f have? (Explain): SOLUTION: The function f can have at most 1 root. If it had two roots, Rolle’s theorem would mean that f 0 (x) = 0 for some x between the roots- But we’re told that f 0 (x) > 0. (c) Is it possible that f 0 (2) = 1/3? Why? SOLUTION: Since f 0 (3) = 1/2 and 1/3 < 1/2, then this implies that f 0 is increasing. However, that implies that f 00 > 0, but f 00 (x) < 0 for all x. Therefore, the given value of f 0 (x) is not possible. 8. Let f (x) = 2x + ex . (a) Show that f has exactly one real root. • We see that f (−1) = −2 + e−1 < 0 and f (0) = 1 > 0. Therefore, by the IVT there is at least one real root. • The derivative is f 0 (x) = 2 + ex , which is always positive (since ex is always greater than 0). Therefore, by Rolle’s Theorem, there is at most 1 real root. • By the previous 2 items, there is exactly one (real) root. (b) Use Newton’s Method three times to approximate the root. SOLUTION: Newton’s Method is xn+1 = xn − f (xn ) f 0 (xn ) f (x) = 2x + ex , 3 f 0 (x) = 2 + ex We might use either x0 = −1 or x0 = 0 as our initial guess (from the IVT). We’ll start with x0 = 0, and the table below summarizes the computations made: n xn f (xn ) 1 0.000000 1.000000 2 −0.333333 0.049865 3 −0.351689 0.000120 4 −0.351734 6.9 × 10−10 f 0 (xn ) 3.000000 2.716531 2.703499 9. Compute ∆y and dy for the given x and dx = ∆x. Sketch a diagram and label ∆x, ∆y and dy, if f (x) = 6 − x2 , x = −2, ∆x = 1. SOLUTION: ∆y = f (x + ∆x) − f (x) = f (−1) − f (−2) = 5 − 2 = 3 dy = −2x dx = −2(−2)(1) = 4 The graph is shown below- Can you label ∆y and dy? (We’ll do this in class) 10. Linearize at x = 0: y= Use the linearization to estimate q x + 1e−x 2 3 − 14 2e SOLUTION: At x = 0, we have y = f 0 (x) = √ √ 1e0 = 1. Now for the slope: √ 2 2 1 (x + 1)−1/2 e−x + x + 1e−x (−2x) 2 so f 0 (0) = 21 , and L(x) = 1 + 12 x. Therefore, (note that the question asks you to approximate f (1/2)): f (1/2) ≈ L(1/2) = 1 + 4 1 1 5 · = 2 2 4 √ 11. Let f (x) = x − x3 on [0, 9]. Verify that the function satisfies all the hypotheses of Rolle’s Theorem, then find the values of c that satisfy its conclusion. SOLUTION: f is continuous on the given interval, and f 0 (x) does not exist at 0, but that’s OK since for Rolle’s Theorem, we need f to be differentiable on (0, 9). Finally, check that f (0) = f (9) = 0. To find the value of c, Rolle’s Theorem guarantees c so that f 0 (c) = 0: 1 1 f 0 (c) = √ − = 0 2 c 3 ⇒ c= 9 = 2.25 4 12. Let f (x) = x3 − 3x + 2 on the interval [−2, 2]. Verify that the function satisfies all the hypotheses of the Mean Value Theorem, then find the values of c that satisfy its conclusion. SOLUTION: f is a polynomial, so it is continuous and differentiable at all real numbers. Now we find c so that 4−0 f (2) − f (−2) = =1 f 0 (c) = 2 − (−2) 4 And f 0 (c) = 3c2 − 3, so: 3c2 − 3 = 1 ⇒ 2 c = ±√ 3 Both of these are in the interval [−2, 2]. Extra: The graph is below showing the tangent lines. 13. Let f (x) = tan(x). Show that f (0) = f (π), but there is no number c in (0, π) such that f 0 (c) = 0. Why does this not contradict Rolle’s Theorem? SOLUTION: f (0) = tan(0) = 0 Now, f 0 (c) = sec2 (x). But sec(x) = solution to: 1 cos(x) f (π) = tan(π) = sin(π) 0 = =0 cos(π) −1 is always greater than (or equal to) 1. That is, there is no 1 =0 cos2 (x) 5 This does not contradict Rolle’s Theorem, since tan(x) is not continuous on [0, π] (there is a vertical asymptote at x = π/2). 14. A child is flying a kite. If the kite is 90 feet above the child’s hand level and the wind is blowing it on a horizontal course at 5 feet per second, how fast is the child paying out cord when 150 feet of cord is out? (Assume that the cord forms a line- actually an unrealistic assumption). Note: As with all word problems, your notation may be different than mine First, draw a picture of a right triangle, with height 90, hypotenuse y(t), the other leg x(t). Note that these are varying until after we differentiate. So, we have that (y(t))2 = (x(t))2 + 902 and 2y(t) We want to find dy dt when y = 150 and dx dt dy dx = 2x(t) dt dt = 5. We need to know x(t), so use the first equation: 1502 − 902 = x2 ⇒ x = 120 so 2 · 150 dy dy = 2 · 120 · 5 ⇒ =4 dt dt 15. At 1:00 PM, a truck driver picked up a fare card at the entrance of a tollway. At 2:15 PM, the trucker pulled up to a toll booth 100 miles down the road. After computing the trucker’s fare, the toll booth operator summoned a highway patrol officer who issued a speeding ticket to the trucker. (The speed limit on the tollway is 65 MPH). (a) The trucker claimed that she hadn’t been speeding. Is this possible? Explain. SOLUTION: Nope. Not possible. The trucker went 100 miles in 1.25 hours, which is not possible if you go (at a maximum) of 65 miles per hour (which would only get you (at a max) 81.25 miles). In terms of the MVT: Change in Position 100 = = 80 Change in time 1.25 So we can guarantee that at some point in time, the trucker’s speedometer read exactly 80 MPH. (b) The fine for speeding is $35.00 plus $2.00 for each mph by which the speed limit is exceeded. What is the trucker’s minimum fine? By the last computation, the trucker had an average speed of 80 mph, so we can guarantee (by the MVT) that at some point, the speedometer read exactly 80. So, this gives $35.00 + $2.00 (15) = $65.00 16. Let f (x) = 1 x (a) What does the Extreme Value Theorem (EVT) say about f on the interval [0.1, 1]? SOLUTION: Since f is continuous on this closed interval, there is a global max and global min (on the interval). (b) Although f is continuous on [1, ∞), it has no minimum value on this interval. Why doesn’t this contradict the EVT? SOLUTION: The EVT was stated on an interval of the form [a, b], which implies that we cannot allow a, b to be infinite. 6 17. Let f be a function so that f (0) = 0 and You might use a value theorem to help). 1 2 ≤ f 0 (x) ≤ 1 for all x. Explain why f (2) cannot be 3 (Hint: SOLUTION: We know that: f (2) − f (0) = f 0 (c) 2−0 for some c in (0, 2). Using the restrictions on the derivative, f (2) 1 ≤ ≤1 2 2 so that 1 ≤ f (2) ≤ 2. 18. Estimate the change in the indicated quantity using differentials. (a) The volume, V = s3 of a cube, if its side length s is increased from 5 inches to 5.1 inches. dV = 3s2 ds ⇒ dV = 3 · 25 · 0.1 = 7.5 (b) The volume, V = 43 πr3 of a sphere, if the radius changes from 2 to 2.1: dV = 4πr2 · dr = 4π(2)2 · (0.1) = 1.6π (c) The volume, V = 1000 p , of a gas, if the pressure p is decreased from 100 to 99. −1000 −1000 dp ⇒ dV = (−1) = 0.1 p2 1002 q L (d) The period of oscillation, T = 2π 32 , of a pendulum, if its length L is increased from 2 to 2.2. dV = First, note that: 2π √ T =√ L 32 which makes it slightly easier to differentiate: π π dT = √ L−1/2 dL ⇒ dT = √ · 0.2 ≈ 0.0785 32 64 19. Related Rates Extra Practice: (a) The top of a 25-foot ladder, leaning against a vertical wall, is slipping down the wall at a rate of 1 foot per second. How fast is the bottom of the ladder slipping along the ground when the bottom of the ladder is 7 feet away from the base of the wall? First, make a sketch of a triangle whose hypotenuse is the ladder. Let y(t) be the height of the ladder with the vertical wall, and let x(t) be the length of the bottom of the ladder with the vertical wall. Then x2 (t) + y 2 (t) = 252 The problem can then be interpreted as: If Differentiating with respect to time: 2x(t) we have numbers for equation, and dy dt dy dt = −1, what is dx dt when x = 7? dx dy + 2y(t) =0 dt dt and x(t)- we need a number for y in order to solve for 72 + y 2 (t) = 252 ⇒ y = 24 7 dx dt . Use the original Now plug everything in and solve for 2·7· dx dt : dx dx 24 + 2 · 24 · (−1) = 0 ⇒ = dt dt 7 (b) A 5-foot girl is walking toward a 20-foot lamppost at a rate of 6 feet per second. How fast is the tip of her shadow (cast by the lamppost) moving? Let x(t) be the distance of the girl to the base of the post, and let y(t) be the distance of the tip of the shadow to the base of the post. If you’ve drawn the right setup, you should see similar triangles... Hgt of post Dist of tip of shadow to base = Hgt of girl Dist of girl to base In our setup, this means: 20 y(t) = 5 y(t) − x(t) With a little simplification, we get: 3y(t) = 4x(t) We can now interpret the question as asking what 3 dy dt is when dx dt = −6. Differentiating, we get dy dx =4 dt dt so that the final answer is dy dt = −8, which we interpret to mean that the tip of the shadow is approaching the post at a rate of 8 feet per second. (c) Under the same conditions as above, how fast is the length of the girl’s shadow changing? Let L(t) be the length of the shadow at time t. Then, by our previous setup, L(t) = y(t) − x(t) so dL dt = dy dt − dx dt = −8 − (−6) = −2. (d) A rocket is shot vertically upward with an initial velocity of 400 feet per second. Its height s after t seconds is s = 400t − 16t2 . How fast is the distance changing from the rocket to an observer on the ground 1800 feet away from the launch site, when the rocket is still rising and is 2400 feet above the ground? We can form a right triangle, where the launch site is the vertex for the right angle. The height is s(t), given in the problem, the length of the second leg is fixed at 1800 feet. Let u(t) be the length of the hypotenuse. Now we have: u2 (t) = s2 (t) + 18002 and we can interpret the question as asking what 2u(t) To solve for du dt , du dt is when s(t) = 2400. Differentiating, we get du ds du ds = 2s(t) or u(t) = s(t) dt dt dt dt we need to know s(t), u(t) and ds dt . We are given s(t) = 2400, so we can get u(t): p u(t) = 24002 − 18002 = 3000 2 Now we need ds dt . We are given that s(t) = 400t − 16t , so t. From the equation for s(t), ds dt = 400 − 32t. That means we need 2400 = 400t − 16t2 ⇒ −16t2 + 400t − 2400 = 0 8 Solve this to get t = 10 or t = 15. Our rocket is on the way up, so we choose t = 10. Finally we can compute ds dt = 400 − 32(10) = 80. Now, 3000 so du dt du = 2400(80) dt = 64 feet per second (at time 10). (e) A small funnel in the shape of a cone is being emptied of fluid at the rate of 12 cubic centimeters per second (the tip of the cone is downward). The height of the cone is 20 cm and the radius of the top is 4 cm. How fast is the fluid level dropping when the level stands 5 cm above the vertex of the cone [The volume of a cone is V = 13 πr2 h]. Draw a picture of an inverted cone. The radius at the top is 4, and the overall height is 20. Inside the cone, draw some water at a height of h(t), with radius r(t). We are given information about the rate of change of volume of water, so we are given that dV dh dt = −12. Note that the formula for volume is given in terms of r and h, but we only want dt . We need a relationship between r and h... You should see similar triangles (Draw a line right through the center of the cone. This, and the line forming the top radius are the two legs. The outer edge of the cone forms the hypotenuse). radius of top overall height 4 20 = ⇒ = radius of water level height of water r h so that r = h5 . Substituting this into the formula for the volume will give the volume in terms of h alone: 1 h 1 1 πh3 V = πr2 h = π( )2 h = 3 3 5 75 Now, dV 3π 2 dh = h dt 75 dt and we know dV dt = −12, h = 5, so dh −12 = dt π (f) A balloon is being inflated by a pump at the rate of 2 cubic inches per second. How fast is the diameter changing when the radius is 21 inch? The volume is V = 34 πr3 (this formula would be given to you on an exam/quiz). If we let h be the diameter, then we know that 2r = h, so we can make V depend on diameter instead of radius: V = Now, the question is asking for dh dt 4 h 3 π π( ) = h3 3 2 6 when h = 1, and we are given that dV dt = 2. Differentiate, and dV π dh π dh = 3h2 = h2 dt 6 dt 2 dt so that dh dt = 4 π. (g) A rectangular trough is 8 feet long, 2 feet across the top, and 4 feet deep. If water flows in at a rate of 2 cubic feet per minute, how fast is the surface rising when the water is 1 foot deep? The trough is a rectangular box. Let x(t) be the height of the water at time t. Then the volume of the water is: dV dx = 16 V = 16x ⇒ dt dt dV dx 1 Put in dt = 2 to get that dt = 8 . 9 (h) If a mothball (sphere) evaporates at a rate proportional to its surface area 4πr2 , show that its radius decreases at a constant rate. Let V (t) be the volume at time t. We are told that dV = kA(t) = k4πr2 dt We want to show that dr dt is constant. By the formula for V (t) = 34 πr3 , we know that: dV dr = 4πr2 dt dt Now compare the two formulas for proportionality! dV dt , and we see that dr dt = k, which was the constant of (i) If an object is moving along the curve y = x3 , at what point(s) is the y−coordinate changing 3 times more rapidly than the x−coordinate? Let’s differentiate: dy dx = 3x2 dt dt dx From this, we see that if we want dy dt = 3 dt , then we must have x = ±1. We also could have x = 0, y = 0, since 0 is 3 times 0. All the points on the curve are therefore: (0, 0), (−1, 1), (1, 1) 20. Graphical Exercises • 4.8: 4, 5 • 4.3: 5-6, 7-8, 31-32 • 4.2: 7 • 4.1: 3-6 10