Solutions to Homework 8, Mathematics 1

Transcription

1 Solutions to Homework 8, Mathematics Problem [6 points]: Do the detailed graphing definition domain, intersections with the aes, its at ±, monotonicity, local and global etrema, conveity/concavity, inflection points, sketch of graph for the function f = arctan. Solution: We begin by looking at the maimal domain of the function f. There is only a issue with = 0, where the argument of arctan becomes undefined. So we have Domainf = R\{0}. Then it is nice to look at the its at the borders of the intervals where f is defined. In this case, we should look at f, f, and f, 0± but perhaps it would be nicer to simply realize first that f is an even function, so we can just plot the graph for the positive numbers, and then just reflect it with respect to the y-ais. We have f = arctan = arctan 0 = 0, and f = f = arctan = π Then even if the function above is not defined on = 0, we can etend it to a continuous function by letting f0 = π/. But we will not worry about it here. We see that the function is never 0, so the graph of f does not touch the -ais. Since f is not defined for = 0, its graph also does not touch the y-ais. If we etend the function as above, then the graph would have an intersection with the y-ais at 0, π/ however. So, let us study the monotonicity. To do that, since we are talking about a differentiable function, it is enough to study the derivative of f. Without further ado, we have f = = arctan + = 4 + So f is always negative for positive and positive for negative! This means that f is monotone strictly increasing on, 0 and monotone strictly decreasing on 0,. It also follows that f has no local or global etrema. Note that if we etended the function by π in 0, then f would have a maimum both local and global in 0 as f is increasing upto 0 and then decreasing after 0. Minimum of f is not achieved, infimum of f equals 0 as f 0 for ±.

2 Figure. Graph of f from Problem. Let us study the concavity of the function now, using the second derivative. We have f = 4 + = = = Therefore we know that f is positive on,,, negative on, \ {0}, and 0 on {, }. We come to this conclusion after studying the changes of sign as varies in the functions and +. We then have that f in conve on, and, and concave on, 0 and 0, after the etension, even on,. The inflection points of f are at = and =. All this information allows us to plot the function f see Figure. Problem : Let a one-dimensional displacement of a particle be given by the formula st = t + 4, t 0. Find time t0 0, 4 such that the instantenous velocity at time t 0 is equal to the average velocity over the interval [0, 4]. Identify what this t 0 is corresponding to in the Lagrange s Mean Value Theorem. Solution: By Dr. Katarina Bellová The average velocity on the interval [0, 4] is equal to s4 s v avg = = = 4 = The instantaneous velocity is equal to vt = s t = t + 4 / = t + 4 / = t + 4.

3 Figure. A sketch of the graph of f from Problem. Only [0, ] should be considered. If this is equal to v avg, we must have: On the other hand, indeed v5/ = = t + 4, t + 4 =, t + 4 = 9, t = 5/. 5/ + 4 = 9 = = v avg alternatively, one can check that the manipulations before were equivalent, namely that we can also get from the bottom to top: in particular, check that in the step where one takes square root, both sides are positive and later we don t divide by 0. Hence, the t we were looking for is t 0 = 5/. Since for this t 0 we have s t 0 = s4 s0, this t 4 0 corresponds to the ξ in Lagrange s Mean Value Theorem applied on interval [0, 4] to the function s as s satisfies the assumptions of the theorem, we are guaranteed the eistence of the ξ = t 0. Problem : A meter piece of wire is cut into two pieces and one piece is bent into a square and the other is bent into an equilateral triangle. Where should the wire be cut so that the total area enclosed by both is minimum and maimum? Solution: We cut this wire into a piece of size and a piece of size. With the piece of size we make an equilateral triangle with side length /. And with the piece of size we make a square with side length /4. Using the Pythagorean theorem, or recalling the sine of π/, we can calculate the triangle s area to be 6.

4 4 Scanned by CamScanner Figure. Sketch of the layout of the cable in Problem 4. The square s area is simply Our job is now to study the maimum and the minimum of the function f = on the interval [0, ]. We can do this if we remember how to study the quadratic function graph from school. Or we can differentiate and get f = 8 + = And we have f = 0 if and only if = [0, ]. Furthermore, since f is constant and positive, we have that the above point is a point of a global minimum. The global maimum can then be found at the etremities, but we notice that f0 = 4, and f = 9 = < 4, so the global maimum is attained at = 0 when we form only a square.

5 Problem 4 [4 points] A power house, P, is on one bank of a straight river 0m wide, and a factory, F, is on the opposite bank 40m downstream from P. The cable has to be taken across the river, under water at a cost of 0 EUR/m. On land the cost is 0 EUR/m. What path should be chosen so that the cost is minimized? Solution: Again an optimization problem. We should assume that the path taken by the cable consists of a union of straight line segments. It is easy to believe this, if you have some curved smooth path between two points A and B, its length should be bigger than the length of the straight line segment between A and B. Without loss of of generality, assume that the path looks like in Figure : first the cable crosses the river under some angle and then it continues on land. Other options first going along the river and then crossing into F, or first going along the river, then crossing along some angle and lastly going along the river on the other side will give us the same costs, as long as the angle of crossing the river will be the same. Consider R, the point directly opposed to P on the other side of the river I think the German word is gegenüber, and S, the point where the cable emerges from water on the side of the factory. That is, the path of the cable will consist of the segments P S and SF, and we have lengthp R = 0, lengthrs =, lengthsf = 40, lengthp S = 0 +. Our task is to minimize the function for [0, 40]. We have 0 lengthp S + 0 lengthsf = = f, Therefore, f = 0 if and only if f = = = = 4 = 00 = 0. This indeed lies in [0, 40]. We also notice that and f = , f 0 > 0. Therefore, 0 is a minimum a global one since f is only 0 at this point and it determines one of the optimal paths. 5

6 6 Problem 5 [4 points]: Compute the its, possibly using l Hospital rule make sure to check it s assumptions: a a a a a b, for a > 0, c cos sin, 0 d 0 cotg, tan + tan e, π sin lnln f e sin e, g ln 0 + Hint: Some material from Wednesday s lecture might be useful. Solution: a: By Dr Katarina Bellová a a 0 0 a a a ln a a a = = a a a a a = a a ln a aa a = a a ln a. Notice we used a > 0 and hence > 0 near a several times. Since the it on the right eists, l Hospital rule is justified. b: By Dr Katarina Bellová = e ln = e ln, ln 0 0 ln / = = = e = e. =, so Since the it on the right in the middle equality eists, l Hospital rule is justified. Then we used the continuity of y e y. c: We have cos sin 0 = e lncos sin 0 = ep 0 0 = ep = e 0 =. 0 0 lncos sin sin cos cos

7 7 d: 0 cotg = 0 cos sin sin cos = 0 sin 0 0 = cos cos + sin 0 sin + cos 0 0 = sin + cos 0 cos + cos sin = 0. e: π f: By Dr Katarina Bellová g: First, notice that tan + tan e sin lnln 0 0 = sin e e ln = = π =. so the bracket is positive for 0 +. We thus have ln 0 + tan cos +tan cos ln cos e = ln = 0+ = = ep 0 + ep cos e cos 0 = e. = 0, ln ln ln ln 0 + Because of, we see that the above it is also of the form 0. Therefore we may use l Hospital 0 rule to prove that the above it is equal to ep 0 + ln ln = ep = e 0 =. ln 0 + ln Here we used that 0+ ln = 0 - this result can be obtained similarly as, or one can just refer to E...0. Problem 6 [ bonus points]: Consider the function {e f = if 0, 0 if = 0. Compute f n 0 for all n N..

8 8 Solution: By Dr. Katarina Bellová First, notice that so f is continuous. Net, 0 e f +0 = f 0 = y= e 0 + e 0 = y + e y = y= = y + y= = y y e y y e y y + we used l Hospital rule... so f 0 = 0. To compute f 0, we also need to know f for 0: Then f 0 = 0 + e f = e. y= = y + e y = 0, = = 0, y + ye y = = 0, y ye y y = e y y + 4y = e y y + 4 e y = 0. To compute f n 0 for general n, we first epress f n for 0. Trying out a few more derivatives suggest that for n 0 and 0, f n has the form f n = P n e, where P n y = a n ny n + a n n y n a n y + a n 0 is a polynomial of order n actually, the lowest nonzero term will be a n n+y n+, but we will not need this. We prove statement by induction. For n = 0, this is true with P 0 y = being of order 0. We even checked the statement for n = with P y = y. Now assume holds for n = k k N 0, we will prove it for n = k +. Differentiating for n = k gives where f k+ = f k = P k e = P k e + P k e = a k k + k ak k k ak + ak 0 e + a k k + k ak k k ak + ak 0 e = ka k k k k+ ak k... k ak + a k k + k+ ak k k+ ak + 4 ak 0 = P k+ e, e e P k+ y =a k ky k+ + a k k y k+ + a k k ka k ky k a k 0 a k y a k y, which is a polynomial of order k +, so we proved for n = k + and hence for any n N 0.

9 Net, one shows that for any n N 0, 0 n e = 0. For n even, this can be done in the same way as when we computed for n = 0 and n = 4 above when checking continuity of f at 0 and computing f 0: after the substitution y =, we can use l Hospital s rule n/-times. For n odd, we need to split the it into 0 and 0+, and proceed as when we computed for n = when computing f 0. After and are proved, f n 0 = 0 follows easily. 9