MA 124 (Calculus II) Lecture 2: January 24, 2019 Section A3. Professor Jennifer Balakrishnan,

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1 Wht is on tody Professor Jennifer Blkrishnn, 1 Velocity nd net chnge 1 2 Regions between curves 3 1 Velocity nd net chnge Briggs-Cochrn-Gillett 6.1 pp Suppose you re driving long stright highwy nd your position reltive to reference point (or origin) is s(t) for t. Your displcement over time intervl [, b] is the chnge in position s(b) s(). Now ssume tht v(t) is the velocity of the object t t. Recll tht v(t) = s (t), so by the FTC, we hve v(t)dt = s (t)dt = s(b) s() = displcement. We see tht the definite integrl v(t)dt is the displcement (chnge in position) between times t = nd t = b. Equivlently, the displcement over the time intervl [, b] is the net re under the velocity curve over [, b]. Not to be confused with displcement is the distnce trveled over time intervl, which is the totl distnce trveled by the object, independent of the direction of motion. If the velocity is positive, the object moves in the positive direction, nd the displcement equls the distnce trveled. However, if the velocity chnges sign, then the displcement nd the distnce trveled re not generlly equl. To compute the distnce trveled, we need the mgnitude of the velocity v(t), which is clled speed. We hve tht distnce trveled = v(t) dt. 1

2 Exmple 1 ( 6.1 Ex. 12). Let v(t) = t 2 + 5t 4 on t 5. Assume t is time mesured in seconds nd velocities hve units of m/s. 1. Grph the velocity function over the given intervl. Then determine when the motion is in the positive direction nd when it is in the negtive direction. 2. Find the displcement over the given intervl. 3. Find the distnce trveled over the given intervl. To find the displcement of n object, we do not need to know its initil position. Wht hppens if we re interested in the ctul position of the object t future time? Suppose we know the velocity of n object nd its initil position s(). The FTC gives us the nswer: Rerrnging this gives tht v(x)dx = s (x)dx = s(x) t = s(t) s(). s(t) = s() + v(x)dx. Exmple 2 ( 6.1 Ex. 24). The velocity (in mi/hr) of hiker wlking long stright tril is given by v(t) = 3 sin 2 (πt/2), for t 4. Assume tht s() = nd t is mesured in hours. 1. Determinethe position function, for t 4. (Hint: sin 2 t = 1 (1 cos 2t).) 2 2. Wht is the distnce trveled by the hiker in the first 15 min of the hike? 3. Wht is the hiker s position t t = 3? Just like we clculted position from velocity, we cn clculte velocity from ccelertion. Given the ccelertion (x), we hve (x)dx = v (x)dx = v(t) v(), 2

3 which gives v(t) = v() + (x)dx. Exmple 3 ( 6.1 Ex. 36). A cr slows down with n ccelertion of (t) = 15ft/s 2. Assume tht v() = 6ft/s, s() = nd t is mesured in seconds. 1. Determine nd grph the position function, for t. 2. How fr does the cr trvel in the time it tkes to come to rest? 2 Regions between curves Briggs-Cochrn-Gillett 6.2 pp Here we generlize the method for finding the re of region bounded by single curve to regions bounded by two or more curves. Consider two functions f nd g continuous on n intervl [, b] on which f(x) g(x). The gol is to find the re of the region bounded by the two curves nd the lines x = nd x = b. We compute the re using Riemnn sums: prtition the intervl [, b] into n subintervls using uniformly spced grid points seprted by distnce x = (b )/n. On ech subintervl, we drw rectngle extending from the lower curve to the upper curve. On the kth subintervl, we choose point x k, nd the height of the corresponding rectngle is f(x k ) g(x k ). Therefore, the re of the kth rectngle is (f(x k ) g(x k )) x. Summing 3

4 the res of the n rectngles gives n pproximtion to the re of the region between the curves: n A (f(x k) g(x k)) x. k=1 As the number of grid points increses, x pproches zero, nd these sums pproch the re of the region between the curves: Definition 4. Suppose tht f nd g re continuous functions with f(x) g(x) on the intervl [, b]. The re of the region bounded by the grphs of f nd g on [, b] is A = (f(x) g(x))dx. Exmple 5 ( 6.2 Ex. 6). Determine the re of the shded region. 4

5 Exmple 6 ( 6.2 Ex. 1). Consider the region bounded by y = cos x nd y = sin x between x = π/4 nd x = 5π/4. Sketch the region nd find its re. Exmple 7 ( 6.2 Ex. 19). Consider the region bounded by y = 2 x nd y = x 2. Sketch the region nd find its re. There re occsions when it is convenient to reverse the roles of x nd y. Consider the regions shown below tht re bounded by the grphs of x = f(y) nd x = g(y), where f(y) g(y), for c y d (which implies tht the grph of f lies to the right of the grph of g). The lower nd upper boundries of the regions re y = c nd y = d, respectively. In cses like these, we tret y s the independent vrible nd divide the intervl [c, d] into n subintervls of width y = (d c)/n. We compute the Riemnn sum nd tke the limit to obtin the following: Definition 8. Suppose tht f nd g re continuous functions with f(y) g(y) on the intervl [c, d]. The re of the region bounded by x = f(y) nd x = g(y) on [c, d] is A = d c (f(y) g(y))dy. 5

6 Exmple 9 ( 6.2 Ex. 28). Express the re of the shded region in terms of () one or more integrls with respect to x nd (b) nd one or more integrls with respect to y. Exmple 1 ( 6.2 Ex. 38). Consider the region bounded by x = y 2 4 nd y = x/3. Use ny method (including geometry) to find the re of the region. Sketch the bounding curves nd the region in question. Exmple 11 ( 6.2 Ex. 52). Find the re of the region. 6