The Quick Calculus Tutorial


 Eustace Stafford
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1 The Quick Calculus Tutorial This text is a quick introuction into Calculus ieas an techniques. It is esigne to help you if you take the Calculus base course Physics 211 at the same time with Calculus I, but you o not yet have any Calculus backgroun. But we will assume that you have some algebra/trigonometry skills, of course you can refresh while you rea. This guie is accompanie by six short Youtube lectures going through the six sections (links see at the en of each section). Take a look at the vieos an rea the text below, it is not always quite the same things you will see. Do all the Exercises first yourself! Stop an think an calculate all Examples yourselves while you go along. Stop an rethink what you have learne all the time. There is also available the Calculus Concept Companion. These are notes from a course that introuces concepts with a ifferent timeline (having an eye on what s going on in your physic s class) in comparison to how the topics are covere in your section of Calculus I. Fell free to rea the companion an use its resources while you go through your Physics/Calculus courses. You can also rea the sections out of orer. 1
2 Lecture 1. What is the erivative? Calculus is the Mathematics of Change. A typical framework to introuce Calculus begins with the following question: How o we escribe the motion of a moving object? Three real numbers x(t), y(t), z(t) tell us an object s position at time t an thus tell us how the position is changing with time. Calculus begins with the question to escribe how fast the position is changing in time an to escribe the change in etail. This leas to the notion of velocity. For example x(1) = 2, y(1) = 2, z(1) = 1 means that at time t = 1 we reach the object from the origin by going two steps in the irection of the positive xaxis, then two steps in the irection of the positive yaxis an finally one step own in the irection of the negative zaxis. In this tutorial we will only consier things moving along a single axis. Often this may be a vertical motion like when you throw a ball vertically up an tell the height z(t) above groun at each point in time in some interval of time. There is a corresponing position functions z(t) (or sometimes x(t), y(t) for motions that are not vertical like a car moving along a straight line. For example imagine ropping a ball vertically own from a builing 50 feet above the groun. Suppose we place the origin of the zaxis at groun level. Then z(t) = 64 16t 2, 0 t 2 will be the height of the ball above groun for all times t in the time interval [0, 2]. When t = 2 then we have z(2) = = 0 an the ball hits the groun. This is an example of a function. We call t the inepenent variable an z the epenent variable in this case. 2
3 In Calculus books the notation is often ifferent. But this is just a matter of taste. (When we say y = 64 16x 2 for 0 x 2 an tell that x is time an y is the height above groun, we have given the same information, variable names on t matter but try to remember at all times what the variable names stan for.) Don t confuse the picture of the orbit of a moving object with the position graph. To the left you see the graph of the position graph of the object falling accoring to z(t) = 64 16t 2 uring the interval [0, 2]. Now the graph of the position function sloping own means that the object gets faster an faster on its way own, the spee is not constant. If the spee woul be constant the object woul fall equal istances in equal time intervals. If we set up a table showing the height z at times t we get t z an see that while the object only falls 4 feet within the first half secon it falls 28 feet within the fourth half secon. Let us consier another example. A car is moving along the positive xaxis. It starts from rest an accelerates within the first 10 secons. The following picture shows the graph of x(t). How fast is the car at t = 5? Imagine you take your feet from the peal for a short moment t at t = 5 an see how far you go. Note that if we stop accelerating we move with the constant velocity, which is the instantaneous spee that we ha at that moment. But how can we etermine this velocity? Suppose the position function is x(t) = 6t 2 3
4 If t is small the velocity will not change consierably in the interval 5 t 5 + t. So we can approximate the instantaneous velocity by the average velocity in this interval, which is for t = 0.1: v av = x t x(5.1) x(5) = Let us calculate this average velocity (note that you use the binomial formula) = 6 ( ) = = 6 ( ) If we replace 0.1 by t we get 6 (10 + t). Of course, when t gets smaller an smaller these number will get closer an closer to 6 10 = 60. This is the (instantaneous) velocity at t = 5. We write: v(5) = x t x = lim t=5 t 0 t Now we can o this calculation of the velocity for any time t. In this case we have the notation: v(t) = x t = lim x t 0 t You just have to be careful with the notation for x because it oes not specify for which t we consier the average velocity. x t = 6 (t + t)2 6t 2 t = 6 t2 + 2t t + ( t) 2 t 2 t = 12t + t When t get very small this quantity will be very close to 12t. Thus: v(t) = x t = 12t is the erivative of x(t). Note that v(5) = 12 5 = 60, which we calculate before. v(t) is the instantaneous velocity at time t. Now Calculus will be a collection of algebraic rules how to calculate erivatives without going through the above teious proceure of calculating a limit. But on t forget the interpretation of the numbers x (t) for a given function x(t): x (t) is the slope of the graph of x(t) for time t. In other wors: If x (t) > 0 the values of x(t) are increasing for time t: the velocity is positive. If x (t) = 0 the graph has a horizontal tangent, it usually changes from sloping up to own or vice 4
5 versa: the velocity is zero at this point. If x (t) < 0 then the values of x(t) are ecreasing at time time: the velocity is negative. Exercise: Consier x(t) = t 3. x t = t=1 x (1). Calculate x t for t = 1 an fin from this Watch the vieo Lecture 1 W hat is the Derivative? Lecture 2. How to take erivatives? In this section we will learn the basic rules of taking erivatives. I. Linearity: Imagine you are walking in a train moving along the xaxis so that x(t) is the coorinate of the center of the train, measure from some origin O. You measure how far to the right you are from the center of the train an call it y(t). Then your coorinate at time time is x(t)+y(t), which is the istance to the right from the origin O. What is your velocity with respect to groun: It obviously is the velocity of the train ae to your velocity with respect to the train. In Calculus language this means: x (x(t) + y(t)) = t t + y t. Obviously if you replace x(t) by kx(t) for a constant number k then also your velocity will be multiplie by k. Thus x (kx(t)) = k t t. 5
6 This shoul be clear from the following picture, noting that k x(t) k x(t + t) = k x. Note that if k < 0 you move in the other irection an the velocity also turns aroun. II. Power Rule: t tn = nt n 1, where n can be any real number. In particular t k = t (k 1) = t t0 = 0 for a constant k. (Get a feeling for the rule: First take the exponent as a factor own, then subtract one from the exponent.) Examples: The following examples are calculate using linearity an the power rule. 1. t (6t4 + 8t t + 4) = 6 4t t = 24t t So, for example the slope of x(t) = 6t 4 + 8t t + 4 (o you have an iea how the graph looks like?) for t = 1 is x t = = 60 t=1 2. t ( t + t 1/3 ) = t (t t 1 3 ) = 1 2 t ( 1 3 )t = 1 2 t t 4 3 = t t 4 3. Suppose you throw up a ball such that the height at time t is given by z(t) = 4 (t 2) 2 = 4 (t 2 4t + 4) = 4t t 2 = t(4 t). What is the velocity after 1 secon? (or what is the slope of the graph of the position function shown below). 6
7 We calculate: z t = t t (4t t2 ) = 4 2t, an so z t = 4 2 = 2. t=1 Does this answer look reasonable? Go from the point (1, 3) one unit to the right an 2 units up. This woul be the point you woul en up if gravity coul be cut off (which of course it can t) at t = 1. Note that z (2) = 0. This is the high point of the ball. There a few more important erivatives. t et = e t ; sin(t) = cos(t) ; t t ln(t) = 1 t cos(t) = sin(t) t Look at the graphs of the sine an cosine function an compare the slopes of sin(t) with the values of cos(t): 7
8 III. Prouct Rule an Quotient Rule: This tells how a prouct of two functions f(t) an g(t) is changing when we change the inepenent variable. f f (f(t)g(t)) = g(t) + f(t)g t t t The iea is to see the necessary quantities (fg) = ( f) g(t) + f(t) g + f g in the picture below. Here f(t) an g(t) enote the lengths of the two sies of a rectangle so that f(t) g(t) is the area. Note that for t small f g is very small so that: (fg) t = f t g(t) + f(t) g t By going over to the limit we get the prouct rule. Examples: t (t2 sin t) = ( t t2 ) sin t + t 2 ( t sin t) = 2t sin t + t2 cos t t (et sin t) = e t sin t + e t cos t = e t (sin t + cos t) Note that x(t) x(t) 1 1 = x(t) x(t) = 1 an the erivative of a constant is 0. So an application of the prouct rule gives 0 = t (x(t)x(t) 1 ) = x t x(t) 1 + x(t) t x(t) 1, 8
9 an thus by solving for t x(t) 1 we get x t t x(t) 1 = x(t) 2 Combining this with the prouct rule a simple calculation gives the quotient rule x t y = x y x y, y 2 where we have abbreviate x = x t an y = y t. Example: cos t sin t cos t t t tan t = sin t sin t t cos t = t cos 2 t cos t cos t sin t( sin t) = cos 2 t 1 = cos 2 t = sec2 t What is the erivative of e t? Let s look at the graphs of e t an e t : = cos2 t + sin 2 t cos 2 t Consier for example at the slopes of e t at t = 1. This is the negative of the slope of e t at t = 1. In fact in general we have that t e t = e t (Note that this means in particular: t e t t= 1 = e ( 1) = e 1 = t et t=1 ) 9
10 In general for each constant k: for example t e3t = 3e 3t. t ekt = ke kt, The above rule is actually a special case of the IV. Chain Rule: Consier a composition of two functions f(t) = x(u(t)) (check your Precalculus book if you forgot what this means!) What has this to o with e kt? Well recall that is is also written exp(kt), an we can consier this to be the composition of the function which multiplies by k an the exponential function, so u(t) = kt an x(u) = e u. Then the chain rule is: f t = x t = x u u t. The first term on the right han sie is more precisely x u, u(t) so you have to take the erivative an then substitute u(t) for the argument. Let s apply this to the example above: Then x u = u eu = e u an u t = t (kt) = k an the result follows. Examples: 10
11 1. Let x = e cos t. Then x(u) = e u an u(t) = cos t. Thus u eu = e u an x cos t = sin t x t = ecos t ( sin t) 2. Let x = cos(sin t). Then x(u) = cos(u) an u(t) = sin t. We get t cos(sin(t)) = u cos(u) sin t = sin(sin t) cos(t) sin t t 3. Suppose a particle moves along the zaxis an its position at time t is given by x(t) = e t2 +4t. Fin the velocity at time t = 1. This is x t t=1 We note that we have to fin the erivative of the composition of x(u) = e u with u(t) = t 2 + 4t. Note that for t = 1 we have u = = 3. Then x u = eu an u t = 2t + 4. Thus x t = x t=1 u u u=3 t = e 3 2 = 2e 3 t=1 Exercises: 1. Fin the erivatives x t : ˆ x = t 5 + t + 1 t (Hint: t = t 1/2 ) ˆ x = (t 2 + 6)e 3t ˆ x = e sin t cos(t) ˆ x = cos(cos(cos t)) ˆ x = ln cot t 2. Suppose an object moves along the yaxis an its position at time t is given by y(t) = t 2 cos(e t ). Fin the velocity at time t = ln( π 2 ). (Hint: You have to calculate t (t2 cos(e t )) t=ln( π ). The calculation of the 2 erivative requires the prouct rule, the chain rule, the power rule an the erivatives of e t an cos t). 11
12 Watch the vieo Lecture 2 How to calculate Derivatives? Lecture 3. How to calculate antierivatives? Given the function of velocities for a motion, how o we fin the position? Well certainly we nee to know aitionally an initial position, the velocity alone won t be able to tell us the position. But since the velocity is the erivative of the position, the position has to be an antierivative of the velocity. Note that such an antierivative can only be etermine up to a constant. We write v(t)t to enote the antierivative, which is a class of functions. Now what oes this mean? Let s say v(t) = t. Then v(t)t = tt = 1 2 t2 + C. Here C stans for an arbitrary constant. Note that t (1 2 t2 + C) = 1 2 t t2 + t C = 1 2 2t = t Let s say that for a motion along the xaxis we have given v(t) = 2t 2 an an initial position x(0) = 2. We first fin the antierivative of v(t): 2t 2 t = 2 3 t3 + C, so x(t) = 2 3 t3 + C for some constant C. position at time 0: an so C = 2 an the position is x(0) = C = 2 x(t) = 2 3 t This can be calculate from the But how i we know the antierivative of 2t 2. Well we can check that t (2 3 t3 + C) 2 3 3t2 = t 2. 12
13 In general taking antierivatives requires to turn aroun the rules we learne in Section 2. This is easy for some of the rules but har for others. It also requires you to get use to some conceptual unerstaning. If you take a erivative you actually perform an operation with input a function, let s say x, an output a function, then enote x t. The antierivative is an operation somehow turning this aroun. The antierivative has as input a function, let s say v(t), an as output a class of functions (a function plus all possible constants), then enote v(t)t. (We will explain the weir notation later on). Here are the basic rules: First linearity hols for integrals: (f(t) + g(t))t = f(t)t + g(t)t ; kf(t)t = k f(t)t, where k enotes a constant. t n 1 t = n + 1 tn+1 + C for n 1 (Get a feeling for the rule: First a 1 to the exponent, then take the reciprocal of the resulting number own as a factor.) Compare with the Power Rule. Note that the right han sie is not efine for n = 1. This is the exceptional rule: Examples: t t = ln t + C tt = t 1/2 t = 2 3 t3/2 + C 1 t + t5 t = ln t t6 + C Of course we have for constants k 0: e kt t = 1 k ekt + C sin(kt)t = 1 k cos(kt) + C ; cos(kt)t = 1 k sin(kt) + C 13
14 Note that you will never nee the above rules for k = 0 because e 0 = 1, sin(0) = 0 an cos(0) = 1. You will nee the special case of the Reverse Power Rule: kt = kt + C Examples: 1. Suppose that the velocity of a moving object is given by v(t) = 8 cos(2t) an the initial position is x(0) = 0. Fin the position function! We know that the position function is an antierivative for a specif value of the constant: 8 cos(2t)t = 8 cos(2t)t = 8 1 sin(2t) + C = 4 sin(2t) + C 2 Since 0 = x(0) = 4 sin(2 0) + C shows C = 0 we have x(t) = 4 sin(2t) 2. Suppose that x t = e t an x(1) = 1. Fin x(t)! We first fin the antierivative (k = 1): e t t = e t + C. Then x(1) = e 1 + C = 1 shows C = e an we get x(t) = e e t Check your answer: (1 t + 1 e e t ) = ( e t ) = e t an x(1) = e e 1 = 1 + e 1 e 1 = 1. Exercises: 1. Fin the antierivatives: ˆ t 2 + 2t 1/2 + t 1 t ˆ 2 cos t sin tt (Hint: t sin2 (t) = 2 cos t sin(t) from the chain rule.) 14
15 2. Consier the motion given by the velocity v(t) = y t = v 0 + at for a, v 0 constants an let y(0) = y 0. What is the position function y(t)? What is the meaning of a an v 0? Explain why t (v(t)) = t ( y t notation for the last function is 2 y t 2 erivative. 3. Fin 2 x t 2 for x(t) = t 4 + 3t 3 + 6t ) = a. The an is calle the secon orer Watch the vieo Lecture 3 How to calculate Anti erivatives? Lecture 4. The Funamental Theorem of Integral Calculus Suppose we are moving with constant velocity 50 miles/hr for two hours along the xaxis. How far i we get within those two hours? Of course, 100 miles. The velocity graph in this is a horizontal line at height 50. Note that we can interpret the istance covere in this case as the area of the rectangle boune by the lines x = 0, x = 50, t = 0 an t = 2. The units are in fact correct because 50 mi/hr 2 hr = 50miles. The number though can be interprete as an area. Of course this works in general: If we move with constant velocity k the position at time t will be kt, where we assume that the position at time 0 is 0. Again kt is the area of the rectangle with lengths t on the xaxis an height the velocity. Next assume that you are accelerating with constant acceleration a: v(t) = at along the xaxis. We know that if x(0) = 0. By taking the antierivative we know that: x(t) = 1 2 at2. How is this number relate with the graph of the velocity function? Is the number x(t) again the area uner the graph of x(t) > 0 an above the taxis between 0 an t. In this case we the area is that of a triangle with base length t an height at so is 1 2 at2. By subtracting areas of rectangles we see that in this case the istance covere within the interval [t 1, t 2 ] is 1 2 a(t2 2 t2 1 ). 15
16 This is the motivation for the notation t 1 t 2 t 2 v(t)t = v(t)t = x(t) t 2 t 1 = x(t 2 ) x(t 1 ) t 1 This observation is true in general. The change in position is the net area between the graph of the velocity function an the taxis. The integral a b f(t)t is calle the efinite integral of the function f(t) over the interval [a, b]. This is a number an not to be confuse with the antierivative (also calle inefinite integral), which is a class of functions. Net area means that we actually subtract the area above the graph an below the taxis. This makes sense because when the velocity is negative we have to subtract from our position. This is inicate in the picture on the below. We have that 0 t v(t)t = A 1 A 2 = x(t) x(0) is the geometric interpretation of the antierivative. Now recall that the velocity is the erivative of the position an the position function is the antierivative. The general statement a b f(t)t = F (t 2 ) F (t 1 ) for a reasonably nice function f(t) an its antierivative F (t) is often calle the Funamental Theorem of Calculus. It tells the way how we calculate for example position from a given velocity. 16
17 A relate statement is that t t f(u)u = f(t) a Note that t a f(t)t is a specific antierivative of f(t), namely the one with f(a) = 0. But of course the erivative will not epen on which antierivative we pick. If we take the erivative of the integral we get back the function. Note that taking the integral of a erivative of a function gives back the function up to a constant. In fact, when we take the erivative we lose the initial conition an the integral won t know this information. In the picture below you get the visual representation of the Funamental Theorem assertions state above. In calculating efinite integrals we use some obvious rules, which are immeiate from the net area interpretation or the Funamental Theorem. For instance a c f(t)t = for any three numbers a, b, c. Also: b a a b f(t)t = f(t)t + a b b c f(t)t f(t)t Finally recall that evaluating a efinite integral is usually one by fining the antierivative first an then taking the ifference of the values of the antierivative at the two limits of the integral. Remark: You may woner about the strange symbol for integrals. This symbol is similar to the so calle Sigma symbol Σ use to abbreviate sums. 17
18 For example or 6 k=2 5 k=1 k = = 15 k 3 = = = 440 I believe you can guess how the symbol is use in general. This integral symbol is suppose to remin you of the fact that integrals are approximate by sums in the following way: Suppose we ivie an interval [a, b] into n small equal length subintervals of length t an pick values f(t i ) in the ith interval. Then a b f(t)t n k=1 f(t k ) t Note that on the right han sie you are summing about areas of n rectangles with base lengths t an heights f(t k ). Examples: t 2 t = t3 = = 8 3. Note that this number is the area boune by the taxis, the graph of t 2 an the line t = t 3 + sin πtt = 1 4 t4 1 1 π cos πt = π ( 1) ( π 0) = π t 5 + t 1/2 + e t 4t = t t3/2 + e t 4t = e What is 5 k=0 k 2? 5 k=0 k 2 = = = 55 18
19 Exercises: 1. Calculate the following efinite integrals an interpret in terms of net areas: ˆ π 0 sin(t)t ˆ t + t2 + t 3 t ˆ 1 0 e t t ˆ 1 1 f(t)t where f(t) is any o functions (which means f(t) = f( t)). 2. Suppose an object moves along the yaxis with velocity y t = cos(πt)+1 an initial position y(0) = 0. Determine y(1) by calculating a efinite integral. Interpret the resulting number as net area for the graph of a function. Watch the vieo Lecture 4 T he F unamental T heorem of Integral Calculus Lecture 5. Working with the tool box Newton s Law is usually state as F = ma where m is the mass an a = 2 x t 2 is the acceleration. Here we assume that F is a force acting on an object moving along the xaxis. Suppose that a force F (t) = cos t is the force an m = 1. How o we fin the position function x(t) of the particle if we know that x(0) = 1 an x (0) = 2. If we let v(t) enote the velocity along the xaxis then an so by taking the antierivative v t = cos t v(t) = sin t + C Since v(0) = x (0) = 2 we get C = 2 an so v(t) = 2 + sin t. 19
20 Then since x (t) = v(t) we get x(t) by taking the antierivative again: 2t cos t + C, for another constant C. Then from x(0) = 2 0 cos 0 + C = 1 + C + 1 we calculate C = 2 an thus x(t) = 2t cos t + 2. It is easy to check whether we have foun the correct solution. We fin by ifferentiating x (t) = 2 + sin t, 2 x = cos t, t t2 an x(0) = cos = 1, x (0) = 2 + sin 0 = 2. The following coorinate system shows the graphs of x(t), x t an 2 x t 2. Do you unerstan the relations between those graphs in terms of slope an net area? Exercise: Suppose an object with mass 2 kg moves along the yaxis. A force of F (t) = 2 e 2t Newtons acts on the object. We have y(0) = 1 an y t = 0. Fin the position function of the object an graph the function. t=0 Why is the object moving at all, taking into account that it s velocity at time 0 vanishes? Discuss the motion for large t. Watch the vieo Lecture 5 W orking with the tool box Lecture 6. Some Solutions for the Exercises Exercise 1.: We calculate x x(1 + t) x(1) = t t = t + t 2 = (1 + t)3 1 t = t + 3 t2 + t 3 1 t When t gets smaller an smaller, t 2 will be even smaller. Thus Exercises 2.1.: x t x = lim t=1 t 0 t = 3 20
21 ˆ x = t 5 + t + 1 t = t 5 + t 1/2 + t 1/2, an so by linearity an power rule: x t = 5t t 1/2 1 2 t 3/2 ˆ x = (t 2 + 6)e 3t, an by using the prouct rule, linearity, the power rule an the erivative of the exponential function (incluing a constant in the exponent): x t = ( t (t2 + 6)) e 3t + (t 2 + 6) t e3t = 2te 3t + (t 2 + 6) 3e 3t = e 3t (3t 2 + 2t + 18) ˆ x = e sin t cos t, an so by using prouct rule an chain rule we get: x t = ( t esin t ) cos t + e sin t t cos t = (e sin t cos t) cos t + e sin t ( sin t) = e sin t (cos 2 t sin t) ˆ x = cos(cos(cos(t)). We have to apply the chain rule twice: cos(cos(cos t)) = t u cos u u=cos(cos t) = sin(cos(cos t)) cos(cos t) t u cos u u=cos t t cos t = sin(cos(cos t)) ( sin(cos t)) ( sin t) = sin(cos(cos t)) sin(cos t) cos t ˆ x = ln(cot t). We calculate t cot t = cos t t sin t using the quotient rule: cos t sin t sin t cos t cos t = t sin t sin 2 = 1 t sin 2 t = csc2 t Exercise 2.2: y(t) = t 2 cos(e t ). We want to calculate y t t=ln π/2 21
22 We use prouct rule an chain rule to get y t = 2t cos(et ) + t 2 ( sin(e t ))e t = 2t cos(e t ) t 2 e t sin(e t ) an by evaluating at ln π/2 an using the ientity e ln u = u for all u we get y t = 2 ln π t=ln π/2 2 cos(π 2 ) (ln π π 2 )2 2 sin(π 2 ) = π 2 (ln π 2 ) Exercise 3.1: ˆ By linearity an the power rule for antierivatives (the power rule backwars) we get t 2 + 2t 1/2 + t 1 t = 1 3 t t3/2 + ln t + C = t t2/3 + ln t + C ˆ Since from the chain rule t sin2 t = 2 sin t cos t we know that the erivatives of sin 2 t is our integran an thus Exercise 3.2: Since v(t) = y t 2 cos t sin tt = sin 2 t + C we know that y(t) is an antierivative of v(t): vt = v 0 + att = v 0 t at2 + C The constant can be etermine by evaluating at t = 0: so the constant is y 0. So we have: v a 0 + C = C = y 0, y(t) = y 0 + v 0 t at2. The velocity is not constant but changing uniformly accoring to v t = a, where a is the constant acceleration. Also v(0) = v 0 is the initial velocity. This is the uniformly accelerate motion. 22
23 Exercise 3.3: From x = t 4 + 3t 3 + 6t we get an Exercise 4.1: x t = 4t3 + 9t t 2 x t 2 = t (x t ) = t (4t3 + 9t t) = 12t t + 12 ˆ π 0 sin tt = cos t π 0 = cos π + cos 0 = ( 1) + 1 = 2 ˆ t + t2 + t 3 t = t + t2 2 + t3 3 + t = = = ˆ 1 0 e t t = e t 1 0 = e 1 ( e 0 ) = 1 1 e ˆ The efinite integral is zero. This is because the integral is the net area, an counts area below the taxis negative an the area above the taxis positive. The two contributions will cancel by the symmetry (see graph below) 23
24 In the first three cases the graph of the function is above the taxis over the corresponing interval. So the efinite integral is the blue area. In the last case the contributions on the negative axis cancel with those on the positive axis. Exercise 4.2: Let v(t) = cos(πt) + 1. For a general number τ the efinite integral y(τ) = 0 τ v(t)t is the antierivative with y(0) = 0. This is because 0 0 g(t)t = 0 for every function g(t). Thus y(1) = 0 1 cos(πt) + 1t is the efinite integral calculating y(1). Using the Funamental Theorem we get y(1) = 1 1 π sin(πt) + t = 1 sin(π) + 1 (sin(0) + 0) = 1. π Compare this with the shae area for the graph below: 0 Exercise 5.1: From Newton s law m = 2 an F = m 2 y t 2 we get 2 y t 2 = e 2t. Recall that we have also given y(0) = 1 an y t = 0. t=0 antierivative we get 2 y t t = t e 2t + C By taking the 24
25 an using the Funamental theorem an y t = 0 we get 1 t=0 4 + C = 0 an thus C = 1 4. Thus y t = t e 2t 1 4 Again we take the antierivative an get y t t = t e 2t 1 4 = t e 2t t 4 + C From y(0) = 1 we get C = 1 an thus C = 7 8. This gives y(t) = 7 8 t 4 + t e 2t You may want to check whether this y(t) satisfies the conitions: y(0) = e 2 0 = = 1 an by taking erivatives y t = t e 2t, 2 y t 2 = e 2t The object is moving because even though the velocity at t = 0 vanishes the acceleration oes not. This is like ropping a ball from rest, it starts with velocity zero but the acceleration ue to gravity makes the object move. Below you the the graph of y(t). Can you relate the graph to the given force? For large t we have F 2 is constant an the motion approaches a uniformly accelerate motion with position graph a parabola (what is the formula for the parabola?) Watch the vieo Lecture 6 Some Solutions f or the Exercises 25
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