Math 210 Midterm #2 Review

Transcription

1 Math 210 Mierm #2 Review Related Rates In general, the approach to a related rates problem is to first determine which quantities in the problem you care about or have relevant information about. Then find an equation which relates these quantities. Then take the derivative of both sides to obtain a relationship between the rates of change of the variables. This equation is called the related rates equation. Plug in all the information which is known and emerge victorious over your enemies. The world is yours. Example: Suppose water is being pumped into a cylindrical tank at a rate of 3 cubic meters per second. It starts out initially empty. The radius of the tank is 5 meters and it is 10 meters high. How fast is the water level within the tank rising? Solution: Let s first assign variables. As usual, t stands for time, V is the volume of liquid in the tank, and h is the height of the liquid in the tank. The problem is asking us to compute dh, the rate of change of h with respect to t. I hope it is intuitively obvious that the water inside the cylindrical shape will itself be in the shape of a cylinder. The radius of this cylinder is the same as the radius of the tank, namely 5 meters, and the height is h. Therefore, we have a relationship between V and h given by: V = π(5) 2 h = 25πh. Now, we take the derivative of both sides of this equation with respect to t: dv = 25π dh We are told in the problem that the rate at which water is being pumped into the tank is 3 cubic meters per second. We can translate this into a mathematical statement thusly: Therefore: so dv = 3 3 = 25π dh dh = 3 25π m/s Example: A 6 foot tall woman walks at a rate of 5 feet per second away from a lamp which is 20 feet high. At what rate is her shadow lengthening when she is 10 feet from the lamp? Solution: Let s denote the length of the woman s shadow and x denote the distance the woman is from the lamppost. We are told in the problem that dx = 5 ft/sec. Note that the right triangle formed by the lamppost and the length along the ground between the lamppost and the tip of the woman s shadow is similar to the triangle formed by the woman herself and her shadow. Using the fact that ratios of corresponding lengths in similar triangles are equal, we have the following equation: We can push things around to get that Differentiating this equation with respect to t gives: x + s 20 = s 6 6x = 14s 6 dx = 14ds Plugging in dx = 5 and solving for ds gives the solution: ds = 15 7 feet/s

2 Increasing/Decreasing, Concavity The derivatives of a function f provide some important information about the graph of f. We already know that the derivative f gives the slopes of tangent lines to the graph of f. A bit of thought will indicate that this geometric interpretation of f also says something about the increasingness or decreasingness of f. We have the following facts: f(x) is increasing on the interval (a, b) whenever f (x) > 0 for all x in (a, b). f(x) is decreasing on the interval (a, b) whenever f (x) < 0 for all x in (a, b). Let s use these facts to figure out how to determine on which intervals f is increasing or decreasing: 1. Find the critical points of f. These are points where f (x) = 0 or f (x) is undefined. 2. Do a sign analysis on f. Chop up the real line into intervals whose endpoints are the critical points found above, pick a test point inside each of these intervals, evaluate f at these test points, and take note of the sign. One can infer the sign of f from just the sign of f evaluated at this test point. 3. Use the above boxes to conclude where f is increasing or decreasing. So f informs us about the increasingness or decreasingness of f. Similarly, the second derivative f informs us about the concavity of f. Recall, a function f is said to be concave up at a point x = a if around this point, the graph of f looks like a smiley face. Similarly, f is said to be concave down at a point x = a if around this point, the graph of f looks like a frowny face. We have the following statements: f(x) is concave up on the interval (a, b) whenever f (x) > 0 for all x in (a, b). f(x) is concave down on the interval (a, b) whenever f (x) < 0 for all x in (a, b). The step-by-step procedure to determine where f is concave up or down is similar to the one given to determine the intervals on which f is increasing or decreasing: 1. Find those points where f (x) = 0 or f (x) is undefined. 2. Do a sign analysis of f. Chop up the real line into inverals whose endpoints are those points found in the above step, pick a test point inside each interval, evaluate f on each of test points, and take note of the sign. 3. Use the above facts to conclude where f is concave up and where it is concave down. Extreme values of functions A point x = a is called a local maximum of a function f if there is an open interval around a on which f is the largest at x = a. The concept of local minimum is defined similarly. Local minima and maxima are collectively called local extrema. The central fact to know about local extrema is that the local extrema of a function f(x) occur at critical points, but not every critical point is a local extrema. So how can we determine if a critical point is a local extremum or not? We have two tools: The First Derivative Test: Suppose x = a is a critical point of f with f continuous at x = a: 1. If f (x) < 0 for x to the left of a, and f (x) > 0 for x to the right of a, then x = a is a local minimum. 2. If f (x) > 0 for x to the left of a, and f (x) < 0 for x to the right of a, then x = a is a local maximum. 3. If the sign of f is the same on both sides of x = a, then x = a is not a local extremum. The Second Derivative Test: Suppose f (a) = 0 and that f (a) exists. 1. If f (a) > 0, then x = a is a local minimum. 2. If f (a) < 0, then x = a is a local maximum. 3. If f (a) = 0, then this test fails and no conclusion can be reached.

3 Typically, a good strategy is to employ the second derivative test first, and if it fails, try the first derivative test. Example: Let f(x) = x x 3. Find all intervals on which f is increasing and all intervals on which it is decreasing. Identify the location of any local extrema for this function. Find all intervals on which f is concave up and all intervals on which it is concave down. Identify the location of any inflection poitns of the function. Solution: First, we find critical points by solving f (x) = 0. So we have: 12 3x 2 = 0 which we can solve as follows: 12 3x 2 = 0 3x 2 = 12 x 2 = 4 x = ±2 Next, we divvy up the real line into intervals according to these critical points. Then we test the sign of f (x) on each of these subintervals: interval test point f (test point) sign increasing/decreasing (, 2) x = 3 f ( 3) = 17 decreasing ( 2, 2) x = 0 f (0) = 12 + increasing (2, ) x = 3 f (3) = 17 decreasing At the critical point x = 2, we see that the sign of the first derivative changes from negative to positive. By the first derivative test, this implies that x = 2 is a local minimum. Also, at x = 2, the first derivative changes from positive to negative, and thus x = 2 is a local maximum. Concavity of f is investigated by examining the sign of f (x). We already calculated f (x) above, and so we see that f (x) = 6x. This function is clearly positive to the left of x = 0 and negative to the right of x = 0, hence we can conclude: f(x) is concave up on (, 0) and concave down on (0, ) Example: Consider the function f(x) = arctan(x 2 ). (a.) Compute f (x). (b.) Compute f (x). (c.) Find the critical point(s) of f(x). (d.) On what interval(s) is f(x) increasing? On what interval(s) is f(x) decreasing? (e.) Determine whether the critical point(s) you found in part (b.) are local extrema, and if so, what kind. (f.) At x = 0, is the graph of f(x) concave up, concave down, or neither? Justify your answer. Solution: Standard use of chain rule implies: and a standard use of quotient rule gives: f (x) = 2x 1 + x 4 f (x) = 2(1 + x4 ) 2x(4x 3 ) (1 + x 4 ) 2 = 2 6x4 (1 + x 4 ) 2 Now we want to find those values of x which make f (x) either 0 or undefined. Since the denominator of f is always strictly positive, we see that the only critical points occur when the numerator is 0. This only happens at x = 0. Since the denominator of f is strictly positive, we see that the sign of f (x) depends only on the sign of the numeratox. But it is obvious that 2x is positive when x is positive, and negative when x is negative. Therefore, f is increasing on the interval (0, ) and decreasing on the interval (, 0). Because f changes from negative to positive across x = 0, this is a local minimum. And at x = 0, the graph of f is concave up since f (0) > 0.

4 Applied Optimization In these problems, we apply what we have learned about finding absolute extrema to applied problems. The game here is to find a formula for the objective function, that is the quantity which is to be maximized or minimized, then employ the techniques we have to finding its absolute extrema. It is difficult to talk about this in general, so we do many examples: Example: A rectangular area of 3200 square ft. is to be fenced off. Two opposite sides will use fencing costing $1 per foot, and the remaining sides will use fencing costing $2 per foot. Find the dimensions of the rectangle of least cost. Solution: The area of the rectangle is the constraint in this problem. If we call the lengths of the sides of the rectangle x and y, then 3200 = xy. The objective function which is to be minimized is the cost of the fence. This is given by C = 4x + 2y, where x is the length of the side of the rectangle requiring the $2 per foot fencing. We use the constraint equation to solve for one variable in terms of the other: y = 3200 x then plug this into the objective function: C = 4x x With the goal of minimizing in mind, we find critical points. We begin by taking a derivative: C = x 2 Notice that x = 0 is a critical point since C is undefined there. But notice that x must be greater than 0 since it is the length of a side of a rectangle. Thus, we can ignore this critical point. Other critical points occur when C = 0. Solving, we get: 0 = x x 2 = = 4x = x 2 ±40 = x Since x denotes a length, the critical point x = 40 can be ignored. We need to figure out if x = 40 is a local maximum or minimum. If it is either, then it is also an absolute max or min since it is the only local extremum. We can figure this out using the second derivative test: so C = 12800x 3 C (40) > 0. Therefore, x = 40 is a local minimum and hence an absolute minimum. The corresponding y-value can be found by plugging x = 40 into the constraint, to give y = 80. Therefore, the rectangle of least cost is 40 by 80. Example: A closed cylindrical can is to hold 1 liter (1000 ccm) of liquid. How should we choose the height and radius of the can to minimize the amount of material needed to manufacture the can?

5 Solution: Let h, r and S denote the height, radius and surface area of the can. Note that the cylinder can be thought of as a two disks of radius r, and a rectangle of height h and wih 2πr. Therefore, S = 2π + 2πrh. This is the function we wish to minimize, and so we call it the objective function. Note that S is a function of two variables. We use the constraint in the problem to express S as a function of a single variable. The constraint is the requirement that the volume of the can be 1000 ccm. We know that the volume of a cylinder is π h, so we get the equation: We plug this into the equation for S: Now, we compute S : 1000 = π h h = 1000 π S = 2π + 2πr 1000 π = 2πr r S = 4πr 2000 Next, we find critical points. Note that r = 0 makes S undefined, but we can ignore it since r, being a length, should be positive. So we solve S = 0: 0 = 4πr = 4πr 500 π = r π = r We use the second derivative test to check that this critical point is in fact, a local min: S = 4π r Notice that when r > 0, S (r) > 0. Therefore, r = 3 π extremum, it must be an absolute minimum. must be a local minimum. Since it is the only local L Hôpital s Rule L Hôptial s Rule states that when a limit lim x a f(x) g(x) has an indeterminate form of 0 0 or ±, f(x) lim x a g(x) = lim f (x) x a g (x) Some limits which do not have these indeterminate forms can be manipulated to get them into this form. For example, 0,, 0 0, 0 and 1 are all indeterminate forms. Oftentimes, L Hôpital s rule can solve them. Example: 1 ln(x) lim x 0 e 1 x Solution: Notice that this limit has the indeterminate form (make sure you see that). So we invoke L Hôpital s Rule: 1 ln(x) 1 x lim = lim x 0 e 1 x x 0 e 1 x ( x 2 ) = lim x x 0 e 1 x 0 The form of this last limit si. This is NOT an indeterminate form. We have something which is going to 0 in the numerator while the denominator is going to. Thus, this limit is equal to 0.