1. (10%) Given a position vector below, determine the velocity, speed, acceleration, the tangential and normal component of acceleration, the curvature, and the unit tangent vector, unit normal vector.
 F-3ti-2j-2k

2. (10%) Find the Laplace transform of At)-[sin(t)-cos(t)]2

3. (10%) Solve the differential equation, y"-ty'+y=1, y(0)=1, y'(0)=2

4. (10%) Produce a matrix P that diagonalizes the matrix below:  

5. (10%) Find the general solutions of

6. (25%) Let u-u(x，y，t) be a function of the space (x.y) and the time t. S=S(s,f) is a function of x and t, and cl and co are constant. Determine whether the method of separation of variables is applicable to the following partial differential equations or not (each has 5 %). The rationale behind your answer needs be given; otherwise, it will not be credited.

7. (25%) Let u=u(xy) be a function of x and y. (x), g(x), f1(y), and g1(y) are given; K and L are positive constant. Solve the problem:  

Please select all the correct answer(s) to question 1 to 10. Note that there could be O to 5 correct answ wers. If non answer is correct, answer "none". If you do not wish to answer a question, leave it blank. All 10 answer rs m must be written on the first page of your answer book, and the answer to question 1 must be in the first line, the answer to question 2 must be in the second line, and so on. If you fail to follow these rules then your answers will be ignored. ne of the a 1. (3 points) Let f(n) be the number of additions in the following algorithm.  1.f(n)=2. f(n)=O(n2logn) 3. f(n)=O(n3) 4. f(n)=O(n3logn) 5. f(n)=

2. (3 points) Given two sorted list A and B (in increasing key order), the following recursive algorithn merges A and B into a sorted list C (also in increasing key order). ㆍ If either A or B is empty then the result is the other list. ㆍ If both A or B are not empty, we compare the keys of the first nodes of A and B, and select the smaller one (denoted as s), and remove it from the list. Then we recursively merge the remaining parts of A and B into a new sorted list C', then we concatenate s with C' into the iral sorted list Let f(n, m) be the minimum number of comparisons of this algorithm, where n and m are the numbers of nodes in A and B respectively. f(n,m) is? 1.Ω(n) 2. Ω(m) 3. Ω(n +m) 4. Ω(mn) 5.Ω(mar.(logm +logn))

3. (3 points) We now use the merge algorithm in the previous question to sort a set of keys. We irst make each key a list of a single node. Then we merge the first and the second lists into a sorted list of length two, the third and the fourth lists into a sorted list of length two, and so on. If we start with n keys now we have roughly n/2 sorted list of length two now. We then merge these lists of length two into roughly n/4 sorted list of length four, and so on. Finally we will have a sorted list of length n. Let f(n) denote the mazimum possible total number of comparisons of this algorithm, then f(n) is? Note that all the answers are in little-o notation. 1.o(n) 2. 3.o(nlogn) 4.o(n(logr)2) 5.o(n2)

4. (3 points) A binary minimun heap is a binary tree in which every level is completely full, except the last level, which is filled from the left to right. Here the level of a node is its distance to the root, so the root has leve! O and the children of the root will have level 1, and so on. Also the key of a parent is less than those of its children. For ease of discussion we assume that all keys in the heap are distinct. Now which of the following descriptions about a binary minimum heap of 10 nodes (see the igure below) are correct?1. The second smallest key is always in level 1. 2. The largest key is always in a leaf, ie., a node without any children. 3. The second largest key is always in a leaf. 4. Let n be the number of nodes and we consider the case of arbitrary n, then the height of the tree is θ(logn). 5. Let n be the number of nodes and we consider the case of arbitrary n, then we can find the thind stnallest key of the beap by O(1) key comparisons.

110年 - 110 國立臺灣大學_碩士班招生考試_部分系所:工程數學(E)#100594

109年 - 109國立臺灣大學_碩士班招生考試_化學工程學研究所:工程數學(E)#106052