英文版微积分考试样题3
微积分英文版3
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=
1
( x3 4cos x sin1) + x ( 3 x2 + 4sin x )
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有限次四则运算的求导法则
(u ± v)′ = u′ ± v′ (uv)′ = u′v + uv′
(Cu)′ = Cu′ ( C为常数 ) u ′ u′v uv′ (v ≠ 0) ( )= 2 v v
2. 曲线的切线斜率
曲线
在 M 点处的切线 割线 M N 的极限位置 M T (当 时)
y
y = f (x) N
C M
T
切线 MT 的斜率
o α x0
x x
= lim tan
f (x) f (x0 ) 割线 M N 的斜率 tan = x x0
→α
f (x) f (x0 ) k = lim x x0 x→x0
例.
y = x ( x 4cos x sin1) ,
3
3 解: y′ = ( x )′ ( x 4cos x sin1)
( x3 4cos x sin1)′ + x
2 x 1 y′ x=1 = (1 4cos1 sin1) + (3 + 4sin1) 2 7 7 = + sin1 2cos1 2 2
函数的可导性与连续性的关系
定理. 定理 证: 设 存在 , 因此必有 其中 故 所以函数 在点 x 连续 . 在点 x 处可导, 即
x →0
y= x
y
注意: 连续未必可导. 注意 函数在点 x 连续未必可导 反例: 反例 在 x = 0 处连续 , 但不可导.
高三英语微积分基础单选题60题(答案解析)
高三英语微积分基础单选题60题(答案解析)1.The derivative of a constant is_____.A.0B.1C.the constant itselfD.undefined答案：A。
解析：任何常数的导数都是0。
选项B，1 不是常数的导数。
选项C，常数本身不是常数的导数。
选项D，常数的导数不是未定义。
2.The integral of a constant times a function is equal to_____.A.the constant times the integral of the functionB.the integral of the function plus the constantC.the function times the constantD.the constant divided by the integral of the function答案：A。
解析：常数乘以函数的积分等于常数乘以函数的积分。
选项B，是函数积分加常数不是常数乘以函数积分的结果。
选项C，函数乘以常数不是积分的结果。
选项D，常数除以函数积分错误。
3.The derivative of a sum of two functions is_____.A.the sum of the derivatives of the two functionsB.the product of the derivatives of the two functionsC.the quotient of the derivatives of the two functionsD.the negative of the sum of the derivatives of the two functions答案：A。
解析：两个函数之和的导数等于两个函数导数之和。
选项B，不是乘积。
选项C，不是商。
Denis Auroux -MIT多变量微积分重点试卷汇编
18.02 Practice Exam 2 A
Problem 1. (10 points: 5, 5)
Let f (x, y ) = xy − x4 .
a) Find the gradient of f at P : (1, 1).
b) Give an approximate formula telling how small changes Δx and Δy produce a small change
why not. Problem 3. A ladybug is climbing on a Volkswagen Bug (= VW). In its starting position, the the surface of the VW is represented by the unit semicircle x2 + y 2 = 1, y � 0 in the xy -plane. The road is represented as the x-axis. At time t = 0 the ladybug starts at the front bumper, (1, 0), and walks counterclockwise around the VW at unit speed relative to the VW. At the same time the VW moves to the right at speed 10. a) (15) Find the parametric formula for the trajectory of the ladybug, and find its position when it reaches the rear bumper. (At t = 0, the rear bumper is at (−1, 0).) b) (10) Compute the speed of the bug, and find where it is largest and smallest. Hint: It is easier to work with the square of the speed. Problem 4. ⎠ 1 2 3 1 � M =� 3 2 2 −1 −1 (a) (5) Compute the determinant of M . b) (10) Find the numbers a and b in the formula for the matrix M −1 . x + 2y + 3z = 0 r = �x, y, z → to 3x + 2y + z = t c) (10) Find the solution � 2x − y − z = 3 d) (5) Compute Problem 5. (a) (5) Let P (t) be a point with position vector � r (t). Express the property that P (t) lies on the plane 4x − 3y − 2z = 6 in vector notation as an equation involving � r and the normal vector to the plane. r d� (b) (5) By differentiating your answer to (a), show that is perpendicular to the normal vector dt to the plane. r d� . dt as a function of t. � M −1 � ⎠ 1 1 4 1 � a 7 −8 � = 12 b −5 4
高三英语微积分基础单选题20题及答案
高三英语微积分基础单选题20题及答案1. In calculus, the derivative of a constant is _____.A. zeroB. oneC. itselfD. undefined答案：A。
常数的导数是零。
选项B“one”错误，常数的导数不是1。
选项C“itself”错误，常数的导数不是它本身。
选项D“undefined”错误，常数的导数是确定的，为零。
2. The process of finding the derivative is called _____.A. integrationB. differentiationC. summationD. multiplication答案：B。
求导数的过程叫做微分。
选项A“integration”是积分。
选项C“summation”是求和。
选项D“multiplication”是乘法。
3. If y = x², then the derivative of y with respect to x is _____.A. 2xB. x²C. 2x²D. x/2答案：A。
y = x²的导数是2x。
选项B“x²”错误，不是它本身。
选项C“2x²”错误，系数错误。
选项D“x/2”错误，计算错误。
4. The integral of a constant times a function is equal to the constant times the integral of the function. This is known as _____.A. the power ruleB. the product ruleC. the chain ruleD. the constant multiple rule答案：D。
常数乘以函数的积分等于常数乘以函数的积分，这被称为常数倍数法则。
高三英语微积分基础练习题20题
高三英语微积分基础练习题20题1.The rate of change of a function at a certain point is called _____.A.integralB.derivativeC.limitD.continuity答案：B。
本题考查微积分基本概念。
“derivative”是导数的意思，函数在某一点的变化率就是导数。
“integral”是积分；“limit”是极限；“continuity”是连续性。
都不符合题意。
2.The process of finding the area under a curve is known as _____.A.differentiationB.integrationC.limitationD.continuity checking答案：B。
“integration”是积分的意思，找到曲线下的面积的过程就是积分。
“differentiation”是求导；“limitation”是限制；“continuity checking”是检查连续性。
3.If the derivative of a function is positive at a point, then the function is _____ at that point.A.increasingB.decreasingC.constantD.none of the above答案：A。
如果函数在某一点的导数是正的，那么函数在该点是递增的。
“increasing”是递增；“decreasing”是递减；“constant”是常数。
4.The integral of a constant function is _____.A.another constant functionB.a linear functionC.a quadratic functionD.none of the above答案：A。
微积分英文版电子档
Differential CalculusNewton and Leibniz,quite independently of one another,were largely responsible for developing the ideas of integral calculus to the point where hitherto insurmountable problems could be solved by more or less routine methods.The successful accomplishments of these men were primarily due to the fact that they were able to fuse together the integral calculus with the second main branch of calculus,differential calculus.The central idea of differential calculus is the notion of derivative.Like the integral,the derivative originated from a problem in geometry—the problem of finding the tangent line at a point of a curve.Unlile the integral,however,the derivative evolved very late in the history of mathematics.The concept was not formulated until early in the 17th century when the French mathematician Pierre de Fermat,attempted to determine the maxima and minima of certain special functions.Fermat’s idea,basically very simple,can be understood if we refer to a curve and assume that at each of its points this curve has a definite direction that can be described by a tangent line.Fermat noticed that at certain points where the curve has a maximum or minimum,the tangent line must be horizontal.Thus the problem of locating such extreme values is seen to depend on the solution of another problem,that of locating the horizontal tangents.This raises the more general question of determining the direction of the tangent line at an arbitrary point of the curve.It was the attempt to solve this general problem that led Fermat to discover some of the rudimentary ideas underlying the notion of derivative.At first sight there seems to be no connection whatever between the problem of finding the area of a region lying under a curve and the problem of finding the tangent line at a point of a curve.The first person to realize that these two seemingly remote ideas are,in fact, rather intimately related appears to have been Newton’s teacher,IsaacBarrow(1630-1677).However,Newton and Leibniz were the first to understand the real importance of this relation and they exploited it to the fullest,thus inaugurating an unprecedented era in the development of mathematics.Although the derivative was originally formulated to study the problem of tangents,it was soon found that it also provides a way to calculate velocity and,more generally,the rate of change of a function.In the next section we shall consider a special problem involving the calculation of a velocity.The solution of this problem contains all the essential fcatures of the derivative concept and may help to motivate the general definition of derivative which is given below.Suppose a projectile is fired straight up from the ground with initial velocity of 144 fee t persecond.Neglect friction,and assume the projectile is influenced only by gravity so that it moves up and back along a straight line.Let f(t) denote the height in feet that the projectile attains t seconds after firing.If the force of gravity were not acting on it,the projectile would continue to move upward with a constant velocity,traveling a distance of 144 feet every second,and at time t we woule have f(t)=144 t.In actual practice,gravity causes the projectile toslow down until its velocity decreases to zero and then it drops back to earth.Physical experiments suggest that as the projectile is aloft,its height f(t) is given by the formula.The term –16t2 is due to the influence of gravity.Note that f(t)=0 when t=0 and whent=9.This means that the projectile returns to earth after 9 seconds and it is to be understood that formula (1) is valid only for 0<t<9.The problem we wish to consider is this:T o determine the velocity of the projectile at each instant of its motion.Before we can understand this problem,we must decide on what is meant by the velocity at each instant.T o do this,we introduce first the notion of average velocity during a time interval,say from time t to time t+h.This is defined to be the quotient. Change in distance during time interval =f(t+h)-f(t)/h.ength of time intervalThis quotient,called a difference quotient,is a number which may be calculated whenever both t and t+h are in the interval[0,9].The number h may be positive or negative,but not zero.We shall keep t fixed and see what happens to the difference quotient as we take values of h with smaller and smaller absolute value.The limit process by which v(t) is obtained from the difference quotient is written symbolically as follows:The equation is used to define velocity not only for this particular example but,more generally,for any particle moving along a straight line,provided the position function f is such that the differerce quotient tends to a definite limit as h approaches zero.The example describe in the foregoing section points the way to the introduction of the concept of derivative.We begin with a function f defined at least on some open interval(a,b) on the x axis.Then we choose a fixed point in this interval and introduce the differencequotient[f(x+h)-f(x)]/h.Where the number h,which may be positive or negative(but not zero),is such that x+h also lies in(a,b).The numerator of this quotient measures the change in the function when x changes from x to x+h.The quotient itself is referred to as the average rate of change of f in the interval joining x to x+h.Now we let h approach zero and see what happens to this quotient.If the quotient.If the quotient approaches some definite values as a limit(which implies that the limit is the same whether h approaches zero through positive values or through negative values),then this limit is called the derivative of f at x and is denoted by the symbol f’(x) (read as ―f prime of x‖).Thus the formal definition of f’(x) may be stated a s follows Definition of derivative.The derivative f’(x)is defined by the equation。
高三英语微积分基础单选题40题
高三英语微积分基础单选题40题1. In calculus, the derivative of a constant is _____.A.zeroB.oneC.twoD.three答案解析：A。
在微积分中，常数的导数为零。
选项B、C、D 分别为一、二、三，都不符合常数导数的定义。
2. The integral of x with respect to x is _____.A.xB.x squaredC.x cubedD.x to the fourth power答案解析：B。
对x 积分，结果是x 的平方的一半加上常数C，但这里只考虑积分结果不考虑常数项，所以答案是x 平方。
选项A、C、D 分别为x、x 的立方、x 的四次方，都不是对x 的积分结果。
3. If y = 3x^2 + 2x + 1, then the derivative of y with respect to x is _____.A.6x + 2B.6x - 2C.3x + 2D.3x - 2答案解析：A。
对y = 3x^2 + 2x + 1 求导，3x^2 的导数是6x，2x 的导数是2，1 的导数是0，所以y 的导数是6x + 2。
选项B、C、D 分别为6x - 2、3x + 2、3x - 2，都不符合求导结果。
4. The derivative of sin(x) is _____.A.cos(x)B.-cos(x)C.sin(x)D.-sin(x)答案解析：A。
sin(x)的导数是cos(x)。
选项B、C、D 分别为-cos(x)、sin(x)、-sin(x)，都不是sin(x)的导数。
5. The integral of cos(x) with respect to x is _____.A.sin(x)B.-sin(x)C.cos(x)D.-cos(x)答案解析：A。
对cos(x)积分，结果是sin(x)加上常数C，但这里只考虑积分结果不考虑常数项，所以答案是sin(x)。
英文版-微积分试卷答案
1、 (1) sin 2lim x x x→∞= 0 . (2) d(arctan )x = 1/(1+x^2) . (3) 21d sin x x =⎰ -cotx .(4).2()()x n e = 泰勒展开式（书上有。
） .(5)0x =⎰ 26/3 .2、(6) The right proposition in the following propositions is ____A____.A. If lim ()x a f x →exists and lim ()x a g x →does not exist then lim(()())x af xg x →+does not exist. B. If lim ()x a f x →,lim ()x a g x →do both not exist then lim(()())x af xg x →+does not exist. C. If lim ()x a f x →exists and lim ()x a g x →does not exist then lim ()()x af xg x →does not exist. D. If lim ()x a f x →exists and lim ()x a g x →does not exist then ()lim ()x a f x g x →does not exist. (7) The right proposition in the following propositions is __A______.A. If lim ()()x af x f a →=then ()f a 'exists. B. If lim ()()x af x f a →≠ then ()f a 'does not exist. C. If ()f a 'does not exist then lim ()()x af x f a →≠. D. If ()f a 'does not exist then the cure ()y f x =does not have tangent at (,())a f a .(8) The right statement in the following statements is __D ______. A. sin lim 1x x x→∞= B. 1lim(1)x x x e →∞+= C. 11d 1x x x C ααα+=++⎰ D. 5511d d 11bb a a x y x y =++⎰⎰ (9) For continuous function ()f x , the erroneous expression in the following expressions is __C ____. A.d (()d )()d b a f x x f b b =⎰ B. d (()d )()d b af x x f a a =-⎰ C. d (()d )0d b a f x x x =⎰ D. d (()d )()()d b af x x f b f a x =-⎰ (10) The right proposition in the following propositions is ____D____. A. If ()f x is discontinuous on [,]a b then ()f x is unbounded on [,]a b .B. If ()f x is unbounded on [,]a b then ()f x is discontinuous on [,]a b .C. If ()f x is bounded on [,]a b then ()f x is continuous on [,]a b .D. If ()f x has absolute extreme values on [,]a b then ()f x is continuous on [,]a b .3、Evaluate 2011lim()x x e x x→-- 1/24．Find 0d |x y =and (0)y ''if 20x x x y y t e +=+⎰. 隐函数求导。
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Problems 37 - 39: Solve the problem. Assume projectile is ideal, launch angle is measured from the horizontal, and launch is over a horizontal surface, unless stated otherwise. 24) An ideal projectile is launched from level ground at a launch angle of 26° and an initial speed of 48 m/sec. How far away from the launch point does the projectile hit the ground? A) ≈ 60 m B) ≈ 230 m C) ≈ 290 m D) ≈ 185 m 25) A projectile is fired with an initial speed of 528 m/sec at an angle of 45°. What is the greatest height reached by the projectile? Round answer to the nearest tenth of a meter. A) 7111.8 m B) 76.2 m C) 69,696.0 m D) 28,447.3 m Find the unit tangent vector of the given curve. 26) r(t) = (5 + 2t7 )i + (4 + 10t7 )j + (8 + 11t7 )k A) T(t) = C) T(t) = 2 2 11 i+ j+ k 15 3 15 2 2 11 i+ j+ k 225 45 225 B) T (t) = 14 14 77 i+ j+ k 15 3 15
D) 111.1°
C) 1.57
D) 1.79
B) Yes
C)
147 49 147 ij+ k 19 19 19
D)
58 58 319 i+ j+ k 15 3 15
C) - 45, 13, -36
D) 0
B) - 5/ 251, - 15/ 251, - 1/ 251 D) - 5, 15, 1
Calculate a and find the direction angle for the following vector. Give the direction angle as an angle in [0°, 360°) rounded to the nearest tenth. 11) a = - 1, 5 A) a = 26, 78.7° B) a = 26, 168.7° C) a = 26, 101.3° D) a = 26, 11.3°
D) r = - 16, - 4
C) 1, -16, - 2
D) 5, - 29, - 6
B) 74
C)
74
D) 12
4) Let v = 8 i + 6j - 9k and w = 2 i - 2j - 8k. Find v · w A) - 3 B) 12, 18, - 64
C) 76
D) - 48, 54, 72
Find a unit vector in the direction θ. 12) θ = 60° 1 1 A) 2, 2 2 2
B) -
1 1 , 2 2
3
C)
1 1 , 2 2
3
D)
1 2
3,
1 2
Solve the problem. 13) A force of magnitude 13 pounds pulling on a suitcase makes an angle of 30° with the ground. Express the force in terms of its i and j components. A) 6.500i + 11.26 j B) 11.26i + 6.500 j C) 0.8660i + + 0.5000j D) 2.005i - 12.84 j Find parametric equations for the line described below. 14) The line through the points P(- 1, - 1, - 7) and Q(3, - 6, 3) A) - 4 + t, 5 + t, - 10 - 7t C) - 1 + 4t, - 1 - 5t, - 7 + 10t
D)
13 15
For the smooth curve r(t), find the parametric equations for the line that is tangent to r at the given parameter value t = t0 . 19) r(t) = (2t2 - 3t)i + (t + 7)j + k ; t0 = 2 A) x = 2 + 5t, y = 9 + t, z = 1 C) x = 2 + 5t, y = 9 + t, z = 0 Evaluate the integral. 3 10t 4 20) i - 3t2 j + k dt 1+t (1 + t2 )2 0 9 9 A) 4 i - 27j + k B) 8i + 27j + k 2 10 B) x = 2 + t, y = t, z = t D) x = 5t, y = t, z = t
B) - t, - t, - 7t D) 1 + 4t, 1 - 5t, 7 + 10t
Find symmetric equne through P = ( - 4, 8, 3) and in the direction of v = - 7, - 4, 8 x- 7 y- 4 z+ 8 x+ 4 y-8 z- 3 A) B) = = = = 8 3 8 -4 -7 -4 C) x- 4 y+ 8 z+ 3 = = 8 -7 -4 D) x+ 7 y+4 z- 8 = = 8 3 -4
Find the angle between the given vectors in radians or degrees, as marked. 5) 3, - 3 , 4, 9 ; degrees A) 121.1° B) 55.6° C) 45.6° Find the angle between u and v in radians. 6) u = 7 i - 9 j - 10k, v = 2 i + 10j - 4 k A) 1.67 B) -0.22 Determine whether the vectors are perpendicular. 7) u = 6, 5 , v = - 7, 3 A) No Find the projection of u onto a unit vector in the direction of v. 8) u = 2 i + 10j + 11k, v = 3i - j + 3k 87 29 87 58 58 319 A) ij+ k B) i+ j+ k 19 19 19 225 45 225 Find the cross product r x v for the given vectors. 9) r = 3, - 9, - 7 , v = - 1, - 9, - 2 A) - 45, - 13, - 36 B) 45, - 13, 36 Find the indicated perpendicular vector. 10) A unit vector perpendicular to both 3, 1, 0 and - 1, 0, - 5 A) 5, - 15, -1 C) - 5/ 251, 15/ 251, 1/ 251
MATH 2415 - Sample Problems for the Final Exam Find a position vector r that is equivalent to the vector PQ defined by points P and Q. 1) P(- 7, - 6) and Q(- 9, 2) A) r = 2, - 8 B) r = - 13, 2 C) r = - 2, 8 Find the following. 2) Let u = 7, - 3, 6 , v = - 9, 2, -6 , and w = - 3, 9, 2 . Find 2v + w - 3u. A) - 4, 23, 12 B) -42, 22, - 28 3) Let w = 8 i + 1j + 3k. Find the length of - w. A) 12
Write an equation for the plane determined by the given conditions. 16) Normal vector n = 4, 5, 7 and containing the point P = (1, 0, - 7) A) 4x + 5y + 7z - 45 = 0 B) x - 7z + 45 = 0 C) 4x + 5y + 7z + 45 = 0 D) x + y + z + 6 = 0 17) Containing the points P = (3, 1, 3), Q = (3, 0, - 3), and R = (0, 2, - 1) A) - y + 6z + 19 = 0 B) 10x - 18y - 3z + 3 = 0 C) 10x + 18y - 3z = 0 D) 10x + 18y - 3z - 39 = 0 Find the perpendicular distance from the given point to the given plane. 18) P = ( -10, 3, - 3); 2x + 11y + 10z = - 4 13 13 13 A) B) C) 225 5 75