Calculus Differentiation

Transcription

1 //007 Clulus Differentition Jeffrey Seguritn person in rowot miles from the nerest point on strit shoreline wishes to reh house 6 miles frther down the shore. The person n row t rte of mi/hr nd wlk t rte of 5 mi/hr. Wht route should the person tke to minimize the mount of time it tkes to reh the house? B Wht is the minimum mount of time in minutes required to reh the house? With these type of optimiztion prolems, you lwys wnt n eqution to optimize. Sometimes, you hve to e retive in how you design this eqution sed on some vrile you selet s relevnt to the prolem. Here, we know tht the shortest distne to the shore is stright line to the house BUT it s fster to wlk thn to row. 6 mi Let s mke our vrile this distne sine it reltes the oth the distne rowed nd the distne wlked mi [pge ]

2 //007 Jeffrey Seguritn Clulus Differentition person in rowot miles from the nerest point on strit shoreline wishes to reh house 6 miles frther down the shore. The person n row t rte of mi/hr nd wlk t rte of 5 mi/hr. Wht route should the person tke to minimize the mount of time it tkes to reh the house? B Wht is the minimum mount of time in minutes required to reh the house? Let s write funtion D of the distne trveled depending on the vrile we hd seleted: D But wht we relly need is to write funtion for time, using Time Distne / Rte: 6 mi T mi [pge ]

3 //007 Jeffrey Seguritn Clulus Differentition person in rowot miles from the nerest point on strit shoreline wishes to reh house 6 miles frther down the shore. The person n row t rte of mi/hr nd wlk t rte of 5 mi/hr. Wht route should the person tke to minimize the mount of time it tkes to reh the house? B Wht is the minimum mount of time in minutes required to reh the house? Let s tke look t this Time funtion to mke sure there s minimum to look for: T mi mi [pge ]

4 //007 Clulus Differentition Jeffrey Seguritn person in rowot miles from the nerest point on strit shoreline wishes to reh house 6 miles frther down the shore. The person n row t rte of mi/hr nd wlk t rte of 5 mi/hr. Wht route should the person tke to minimize the mount of time it tkes to reh the house? B Wht is the minimum mount of time in minutes required to reh the house? Now let s tke the derivtive of the Time funtion: dt d dt d dt d mi mi [pge 4]

5 //007 Jeffrey Seguritn Clulus Differentition person in rowot miles from the nerest point on strit shoreline wishes to reh house 6 miles frther down the shore. The person n row t rte of mi/hr nd wlk t rte of 5 mi/hr. Wht route should the person tke to minimize the mount of time it tkes to reh the house? B Wht is the minimum mount of time in minutes required to reh the house? Now let s find the zero of the Time funtion: dt d mi / This is ritil vlue nd should represent minimum in our Time funtion. mi [pge 5]

6 //007 Clulus Differentition Jeffrey Seguritn person in rowot miles from the nerest point on strit shoreline wishes to reh house 6 miles frther down the shore. The person n row t rte of mi/hr nd wlk t rte of 5 mi/hr. Wht route should the person tke to minimize the mount of time it tkes to reh the house? B Wht is the minimum mount of time in minutes required to reh the house? If / mi, then we n find the optiml shortest time route: 5/ mi rowing on ot, 9/ mi wlking on the shore Then the minimum mount of time in minutes required to reh the house is: rowing time 5/ mi / mi/hr 5/6 hr 50 min wlking time 9/ mi / 5 mi/hr 9/0 hr 54 min Totl time 50 min + 54 min 04 min [pge 6] 9/ 5/ mi 6 mi /

7 //007 Clulus Differentition Jeffrey Seguritn rin gutter is to e onstruted from metl sheet of width 0 m y ending up one-third of the sheet on eh side through n ngle. Wht should the mesure of the ngle e so tht the gutter will rry the mimum mount of wter? So for this prolem, let s find funtion for the enlosed re of the gutter whih depends on the ngle: 0m 0m 0m θ re of Retngle θ 00sinθ + 0osθ 0sinθ θ 00sinθ + 0osθ 0sinθ θ 00sinθ + 00osθ sinθ re of Right tringles Notie tht this re n e roken up into retngle nd right tringles [pge 7]

8 //007 Clulus Differentition Jeffrey Seguritn rin gutter is to e onstruted from metl sheet of width 0 m y ending up one-third of the sheet on eh side through n ngle. Wht should the mesure of the ngle e so tht the gutter will rry the mimum mount of wter? Let s mke sure there s mimum to the re funtion we wrote, nd we see tht there is one: θ 00sinθ + 00 osθ sinθ 0m 0m 0m θ [pge 8]

9 //007 Clulus Differentition Jeffrey Seguritn rin gutter is to e onstruted from metl sheet of width 0 m y ending up one-third of the sheet on eh side through n ngle. Wht should the mesure of the ngle e so tht the gutter will rry the mimum mount of wter? Now let s tke the derivtive of this re funtion nd find the ritil vlue: θ 00sinθ θ 00sinθ d dθ 00 osθ + 00 osθ sinθ + 50sin θ + 00os θ Trig identity: sin θ sinθ osθ 0m 0m 0m θ 00 osθ 00 osθ osθ osθ + 00os θ 00os θ os θ os θ 0 Trig identity: os θ os θ [pge 9]

10 //007 Clulus Differentition Jeffrey Seguritn rin gutter is to e onstruted from metl sheet of width 0 m y ending up one-third of the sheet on eh side through n ngle. Wht should the mesure of the ngle e so tht the gutter will rry the mimum mount of wter? Now let s tke the derivtive of this re funtion nd find the ritil vlue: osθ osθ os os os θ + osθ θ θ + 0 Now we n use the qudrti formul to solve for os θ 0m 0m 0m θ osθ ± 4 osθ osθ ± 9 4, os os [pge 0]

11 //007 Clulus Differentition Jeffrey Seguritn rin gutter is to e onstruted from metl sheet of width 0 m y ending up one-third of the sheet on eh side through n ngle. Wht should the mesure of the ngle e so tht the gutter will rry the mimum mount of wter? Sine we re looking for n ngle etween 0º nd 90º, we n only tke θ 60º. Let s go one step further nd find the mimum re: 0m 0m 0m θ θ 00sinθ + 50sin θ 60 00sin sin / + 50/ 60 50/ [pge ]

12 //007 Clulus Differentition Jeffrey Seguritn The horizontl line y intersets the urve y- in the first qudrnt on grph pper. Find so tht the res of the two shded regions re equl. The shded res inluded the tringle-like re from 0,0 to, elow y & ove the prol, the seond shded re is from, to, where the shded region is the re ove y ut within the prol., nd, re the points of intersetion. Here s n illustrtion of the res we to equlize:,, y [pge ]

13 //007 Clulus Differentition Jeffrey Seguritn The horizontl line y intersets the urve y- in the first qudrnt on grph pper. Find so tht the res of the two shded regions re equl. The shded res inluded the tringle-like re from 0,0 to, elow y nd ove the prol, the seond shded re is from, to, where the shded region is the re ove y ut within the prol., nd, re the points of intersetion. I m not sure if you ve een introdued to this onept, ut the only wy to find the re eneth urves is vi integrtion. Let me know if you hven t yet lerned integrtion.,, y [pge ]

14 //007 Clulus Differentition Jeffrey Seguritn The horizontl line y intersets the urve y- in the first qudrnt on grph pper. Find so tht the res of the two shded regions re equl. The shded res inluded the tringle-like re from 0,0 to, elow y nd ove the prol, the seond shded re is from, to, where the shded region is the re ove y ut within the prol., nd, re the points of intersetion. Now let s find formul for the re of eh shded region using integrls. [ ] d [ ] d 0,, y [pge 4]

15 //007 [pge 5] Jeffrey Seguritn Clulus Differentition The horizontl line y intersets the urve y- in the first qudrnt on grph pper. Find so tht the res of the two shded regions re equl. The shded res inluded the tringle-like re from 0,0 to, elow y nd ove the prol, the seond shded re is from, to, where the shded region is the re ove y ut within the prol., nd, re the points of intersetion. Now let s find formul for the re of eh shded region using integrls. 0 0 ] [ d + + d + + ] [

16 //007 Clulus Differentition Jeffrey Seguritn The horizontl line y intersets the urve y- in the first qudrnt on grph pper. Find so tht the res of the two shded regions re equl. The shded res inluded the tringle-like re from 0,0 to, elow y nd ove the prol, the seond shded re is from, to, where the shded region is the re ove y ut within the prol., nd, re the points of intersetion. So let s ompile ll the informtion we know: These eqution ome from the intersetion points,, nd, This eqution omes from equlizing the res. + Note tht 0 nnot e solution, sine we presume > 0 + [pge 6]

17 //007 [pge 7] Jeffrey Seguritn Clulus Differentition The horizontl line y intersets the urve y- in the first qudrnt on grph pper. Find so tht the res of the two shded regions re equl. The shded res inluded the tringle-like re from 0,0 to, elow y nd ove the prol, the seond shded re is from, to, where the shded region is the re ove y ut within the prol., nd, re the points of intersetion. Now let s try to solve: / 4 4 +, 6 0 / / 0 4 / 4 Note nnot equl ½ sine is lredy ½