Section 4.8. D v(t j 1 ) t. (4.8.1) j=1


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1 Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions of integrtion to problems unrelted to the computtion of re. Depending on the context, the definite integrl of function f from to b could represent the totl mss of wire, the totl electric chrge on such wire, or the probbility tht light bulb will fil sometime in the time intervl from to b. In this section we will consider three pplictions of definite integrls: finding the distnce trveled by n object over n intervl of time if we re given its velocity s function of time, finding the position of n object t ny time if we re given its initil position nd its velocity s function of time, nd finding the length of curve. Distnce Suppose the function v is continuous on the intervl [, b] nd, for ny t b, v(t) represents the velocity t time t of n object trveling long line. Divide [, b] into n time intervls of equl length t = b n with endpoints = t < t < t < < t n = b. Then, for j =,,,..., n, v(t j ) is the speed of the object t the beginning of the jth time intervl. Hence, for smll enough t, v(t j ) t will give good pproximtion of the distnce the object will trvel during the jth time intervl. Thus if D represents the totl distnce the object trvels from time t = to time t = b, then D v(t j ) t. (.8.) j= Moreover, we expect tht s t decreses, or, equivlently, s n increses, this pproximtion should pproch the exct vlue of D. Tht is, we should hve D = lim n j= v(t j ) t. (.8.) Now the righthnd side of (.8.) is Riemnn sum (in prticulr, lefthnd rule sum) which pproximtes the definite integrl b v(t) dt. Copyright c by Dn Sloughter
2 Distnce, Position, nd the Length of Curves Section Figure.8. Grph of the velocity function v(t) = cos(πt) Hence this integrl is the vlue of the limit in (.8.), nd so we hve D = lim n j= v(t j ) t = b v(t) dt. (.8.) Exmple Suppose n object is oscillting t the end of spring so tht its velocity t time t is given by v(t) = cos(πt). Then the distnce D trveled by the object from time t = to time t = is given by Now nd Hence nd D = cos(πt) dt = cos(πt) dt. cos(πt) when t or t, cos(πt) when t. cos(πt) = cos(πt) when t or t, cos(πt) = cos(πt) when t.
3 Section.8 Distnce, Position, nd the Length of Curves Thus cos(πt) dt = = cos(πt) dt + cos(πt)dt = π sin(πt) ( ) = π = π. cos(πt) dt + cos(πt)dt + cos(πt) dt cos(πt)dt π sin(πt) + π sin(πt) ( π ) ( + + ) π π Hence D = cos(πt) dt = 8 π. Position Agin suppose v is continuous on [, b] nd v(t) represents, for t b, the velocity t time t of n object moving on line. Let x(t) be the position of the object t time t nd suppose we know the vlue of x(), the position of the object t the beginning of the time intervl. It follows tht ẋ(t) = d x(t) = v(t), (.8.) dt nd so, by the Fundmentl Theorem of Integrl Clculus, for ny t between nd b, Thus we hve v(s)ds = x(t) x(). (.8.5) x(t) = v(s)ds + x() (.8.6) for t b. In other words, if we re given the velocity of n object for every time t in the intervl [, b] nd the position of the object t time t =, then we my use (.8.6) to compute the position of the object t ny time t in [, b]. Exmple As in the previous exmple, consider n object oscillting t the end of spring so tht its velocity is given by v(t) = cos(πt). If x(t) is the position of the object t time t nd, initilly, x() =, then x(t) = cos(πs)ds + = t π sin(πs) + = sin(πt) +. π
4 Distnce, Position, nd the Length of Curves Section Figure.8. Velocity v(t) = cos(πt) nd position x(t) = π sin(πt) + You should compre the grphs of the velocity function v nd the position function x in Figure.8.. Note tht the object will oscillte between π nd + π. In prticulr, the distnce between these two extremes is π, nd so the object will trvel distnce of 8 π during complete oscilltion, in greement with our computtion in the previous exmple. Exmple Suppose the velocity of n object t time t is given by v(t) = sin(t ). If x(t) is the position of the object t time t nd its position t time is x() =, then x(t) = sin(s )ds. However, unlike the previous exmple, there does not exist simple ntiderivtive for v; hence, the best we cn do is pproximte x(t) for specified vlue of t using numericl integrtion. For exmple, we cn compute numericlly tht x() = sin(s )ds =.9, where we hve rounded the result to the third deciml plce. If we do this for enough points, we cn plot the grph of x, s shown in Figure.8.. Agin, you should compre this grph with the grph of v, lso shown in Figure Figure.8. Velocity v(t) = sin(t ) nd position x(t) = sin(s )ds
5 Section.8 Distnce, Position, nd the Length of Curves 5 x x x x x x 5 Figure.8. Approximting curve with line segments Length of curve Here we will consider the problem of finding the length of curve which is the grph of some differentible function. So suppose the function f is continuous on the closed intervl [, b] nd differentible on the open intervl (, b). Let C be the grph of f nd let L be the length of C. As we hve done previously, we will first describe method for finding good pproximtions to L. To begin, divide [, b] into n intervls of equl length x = b n with endpoints = x < x < x < < x n = b. For j =,,,..., n, we cn pproximte the length of the piece of C lying over the jth intervl by the distnce between the endpoints of this piece, s shown in Figure.8.. Tht is, since the endpoints of the jth piece re (x j, f(x j )) nd (x j, f(x j )), we cn pproximte the length of the piece of C lying over the intervl [x j, x j ] by (x j x j ) + (f(x j ) f(x j )). Since x = x j x j, (x j x j ) + (f(x j ) f(x j )) = ( x) + (f(x j ) f(x j )) ( = ( x) + (f(x ) j) f(x j )) ( x) ( ) f(xj ) f(x j ) = x +. x (.8.7)
6 6 Distnce, Position, nd the Length of Curves Section.8 Hence, when n is lrge (equivlently, when x is smll), good pproximtion for L is given by ( ) f(xj ) f(x j ) L + x. (.8.8) x Moreover, we expect tht L = lim j= n j= ( ) f(xj ) f(x j ) + x, (.8.9) x provided this limit exists. By the Men Vlue Theorem, for ech j =,,,..., there exists point c j in the intervl (x j, x j ) such tht Hence f (c j ) = f(x j) f(x j ). (.8.) x L = lim n j= Now the sum in (.8.) is Riemnn sum for the integrl b + (f (c j )) x. (.8.) + (f (x)) dx, nd so the limit, if it exists, converges to the vlue of this integrl. Thus the length of C is given by b L = + (f (x)) dx. (.8.) Exmple Let L be the length of the grph of f(x) = x on the intervl [, ], s shown in Figure.8.5. Then so L = = = 8 7 f (x) = x, + ( ) x dx + 9 x dx ( + 9 x ) = 8 =.97, 7 where we hve rounded the result to four deciml plces.
7 Section.8 Distnce, Position, nd the Length of Curves Figure.8.5 Grphs of y = x nd y = x Exmple Let L be the length of the prbol y = x from (, ) to (, ), s shown in Figure.8.5. Then dy dx = x, so L = + (x) dx = + x dx. At this point we do not hve the techniques to evlute this integrl exctly using the Fundmentl Theorem (lthough we will see such techniques in Chpter 6); however, we my use computer lgebr system to find tht L = 7 + sinh () = 7 + (log( + 7)), where log(x) is the nturl logrithm of x nd sinh (x) is the inverse hyperbolic sine of x. Since we will not study either of these functions until Chpter 6, we will use numericl pproximtion to give us L =.668 to four deciml plces, the sme nswer we would obtin by using numericl integrtion to evlute the integrl. Exmple To find the length L of one rch of the curve y = sin(x), s shown in Figure.8.6, we need to evlute π L = + cos (x) dx, n integrl which is even more difficult thn the one in the previous exmple. However, using numericl integrtion, we find tht L =.8 to four deciml plces. The lst two exmples illustrte some of the difficulties in finding the length of curve. In generl, the integrls involved in these problems require more sophisticted techniques thn we hve vilble t this time, nd frequently require the use of numericl techniques.
8 8 Distnce, Position, nd the Length of Curves Section Figure.8.6 Grph of y = sin(x) over the intervl [, π] Problems. For ech of the following, ssume tht v(t) is the velocity t time t of n object moving on line nd find the distnce trveled by the object over the given time period. () v(t) = t over t (b) v(t) = t + 6 over t (c) v(t) = t t 6 over t (d) v(t) = t t 6 over t (e) v(t) = sin(t) over t π (f) v(t) = cos(πt) over t. Suppose the velocity of flling object is given by v(t) = t feet per second. If the object is t height of feet t time t =, find the height of the object t n rbitrry time t.. Suppose x(t) nd v(t) re the position nd velocity, respectively, t time t of n object moving on line. If x() = 5 nd v(t) = t 6, find x(t).. If n object of mss m is connected to spring, pulled distnce x wy from its equilibrium position nd relesed, then, ignoring the effects of friction, the velocity of the object t time t will be given by v(t) = x k m sin ( k m t ), where k is constnt tht depends on the strength of the spring. position of the object t time t. 5. Show tht if ẋ(t) = f(t) nd f is continuous on [, b], then Find x(t), the x(t) = f(s)ds + x().
9 Section.8 Distnce, Position, nd the Length of Curves 9 6. For ech of the following, use the result from Problem 5 to find x(t). () ẋ(t) = t + 6t 7 with x() = (b) ẋ(t) = cos(6t) t with x() = (c) ẋ(t) = sin (t) with x() = (d) ẋ(t) = t sin(t) with x() = (e) ẋ(t) = + t with x() = 7. Let x(t), v(t), nd (t) be the height, velocity, nd ccelertion, respectively, t time t of n object of mss m in free fll ner the surfce of the erth. Let x nd v be the height nd velocity, respectively, of the object t time t =. If we ignore the effects of ir resistnce, the force cting on the body is mg, where g is constnt (g = 9.8 meters per second, or feet per second per second). Thus, by Newton s second lw of motion, mg = m(t), from which we obtin Using Problem 5, show tht (t) = g. x(t) = gt + v t + x. 8. Suppose n object is projected verticlly upwrd from height of feet with n initil velocity of feet per second. Use Problem 7 to nswer the following questions. () Find x(t), the height of the object t time t. (b) At wht time does the object rech its mximum height? (c) Wht is the mximum height reched by the object? (d) At wht time will the object strike the ground? 9. For ech of the following, find the length of the grph of the given function over the given intervl. [ () f(x) = x over [, ] (b) f(x) = sin(x) over, π ] [ (c) g(x) = x over [, ] (d) g(t) = tn(t) over π, π ] (e) f(t) = sin (t) over [, π] (f) g(θ) = sin(θ ) over [, π ]. A sheet of corrugted luminum is to be mde from flt sheet of luminum. Suppose cross section of the corrugted sheet, when mesured in inches, is in the shpe of the curve ( π ) y = sin t. Find the length of flt sheet tht would be needed to mke corrugted sheet tht is feet long.
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