Section 11.6: Directional Derivatives and the Gradient Vector

Size: px

Start display at page:

Download "Section 11.6: Directional Derivatives and the Gradient Vector"

Lee Parsons
6 years ago
Views:

1 Sction.6: Dirctional Drivativs and th Gradint Vctor Practic HW rom Stwart Ttbook not to hand in p. 778 # -4 p. 799 # odd Th Dirctional Drivativ Rcall that a b Slop o th tangnt lin to th surac at th point a b a b in th dirction a b a b a b Slop o th tangnt lin to th surac at th point a b a b in th dirction Instad o rstricting ourslvs to th and ais suppos w want to ind a mthod or inding th slop o th surac in an dsird dirction. Slop in dir Slop in dir D u Slop in u dir θ u < a b >

2 Lt u < a b > b th unit vctor a vctor o lngth on on th - plan which indicats th dirction w ar moving. Thn w din th ollowing: Dinition o th Dirctional Drivativ Th dirctional drivativ o a unction in th dirction o th unit vctor u < a b > dnotd b is dind th b th ollowing: D u D u a + b Nots. Gomtricall th dirctional drivativ is usd to calculat th slop o th surac. That is to calculat th slop o th surac at th point whr w comput th ollowing: Slop o Surac at point D in dirction o unit vctor u < a. b > u a + b. Th vctor u < a b > must b a unit vctor. I w want to comput th dirctional drivativ o a unction in th dirction o th vctor v and v is not a unit vctor w comput v u v. v v 3. Th dirction o th unit vctor u can b prssd in trms o th angl θ btwn th vctor u and th -ais. In this cas u < cosθ sinθ > not u is a unit vctor sinc u cos θ + sin θ and th dirctional drivativ can b prssd as D u cosθ + sinθ. 4. Computationall th dirctional drivativ rprsnts th rat o chang o th unction in th dirction o th unit vctor u.

3 3 Eampl : ind th dirctional drivativ o th unction at th π point in th dirction o th unit vctor that maks an angl o θ radians with 3 th -ais. Solution:

4 4 Eampl : ind th dirctional drivativ o th unction point -3-4 in th dirction o th vctor v i + 3j at th Solution:

5 5 Gradint o a unction Givn a unction o two variabls th gradint vctor dnotd b is a vctor in th - plan dnotd b i j + acts about Gradints. Th dirctional drivativ o th unction in th dirction o th unit vctor u < a b > can b prssd in trms o gradint using th dot product. That is D u u < a + > < a b > b. Th gradint vctor givs th dirction o maimum incras o th surac. Th lngth o th gradint vctor is th maimum valu o th dirctional drivativ th maimum rat o chang o. That is Maimum Valu o th Dirctional Drivativ D u 3. Th ngation o th gradint vctor givs th dirction o maimum dcras. o th surac. Th ngation o th lngth o th gradint vctor is th minimum valu o th dirctional drivativ. That is Minimum Valu o th Dirctional Drivativ D u

6 6 Eampl 3: Givn th unction cos. a. ind th gradint o b. Evaluat th gradint at th point P π. 3 c. Us th gradint to ind a ormula or th dirctional drivativ o in th dirction o th 3 4 vctor u < >. Us th rsult to rsult to ind th rat o chang o at P in th 5 5 dirction o th vctor u. Solution:

7 7 Dirctional Drivativ and Gradint or unctions o 3 variabls Th dirctional drivativ o a unction o 3 variabls in th dirction o th unit vctor u < a b c > dnotd b D u is dind to b th ollowing: c b a D + + u Th gradint vctor dnotd b is a vctor dnotd b k j i + +

8 8 Eampl 4: ind th gradint and dirctional drivativ o P 4 in th dirction o th point Q at Solution: W irst comput th irst ordr partial drivativs with rspct to and. Th ar as ollows Thn th ormula or th gradint is computd as ollows: i + j+ k 3 + i j + k Hnc at th point P 4 th gradint is i j + k i + j + k < > To ind th dirctional drivativ w must irst ind th unit vctor u spciing th dirction at th point P 4 in th dirction o th point Q-3. To do this w ind th vctor v PQ. This is ound to b v PQ < 3 4 >< 4 >. This must b a unit vctor so w comput th ollowing: u v v < 4 > < 4 >< 4 > Thn using th dot product ormula involving th gradint or th dirctional drivativ and th rsults or th gradint at th point P4 and u givn abov w obtain Th dirctional at th point drivativ Du 4 4 u P4 < > < >

9 9 Eampl 5: ind th maimum rat o chang o 4 and th dirction in which it occurs at th point Solution:

10 Normal Lins to Suracs Rcall that givs a 3D surac in spac. W want to orm th ollowing unctions o 3 variabls Not that th unction is obtaind b moving all trms to on sid o an quation and stting thm qual to ro. W us th ollowing basic act. act: Givn a point on a surac th gradint o at this point i + j + k is a vctor orthogonal normal to th surac.

11 Eampl 6: ind a unit normal vctor to th surac at th point Solution:

12 Tangnt Plans Using th gradint w can ind a quation o a plan tangnt to a surac and a lin normal to a surac. Considr th ollowing: Rcall that to writ quation o a plan w nd a point on th plan and a normal vctor. Sinc < > rprsnts a normal vctor to th surac and th tangnt plan its componnts can b usd to writ th quation o th tangnt plan at th point. Th quation o th tangnt plan is givn as ollows: + + Rcall to writ th quation o a lin in 3D spac w nd a point and a paralll vctor. Sinc < > is a vctor normal to th surac it would b paralll to an lin normal to th surac at. Thus th paramtric quations o th normal lin ar: t t t

13 3 W summari ths rsults as ollows. Tangnt Plan and Normal Lin Equations to a Surac Givn a surac in 3D orm th unction o thr variabls. Thn th quation o th tangnt plan to th surac at th point is givn b + + Th paramtric quations o th normal lin through th point ar givn b t t t Not: Rcall that to ind th smmtric quations o a lin tak th paramtric quations solv or t and st th rsults qual.

14 4 Eampl 7: ind th quation o th tangnt plan and th paramtric and smmtric quations or th normal lin to th surac at th point. Solution:

15 5 Not: Th ollowing graph using Mapl shows th graph o th sphr with th tangnt plan and normal lin at th point.

16 6 Eampl 8: ind th quation o th tangnt plan and th paramtric and smmtric quations or th normal lin to th surac at th point. Solution: W start b stting and computing th unction o 3 variabls Rcall that to gt an quation o an plan including a tangnt plan w nd a point and a normal vctor. W ar givn th point. Th normal vctor coms rom computing th gradint vctor o at this point. Rcall that or a givn point th gradint vctor at this point is givn b th ormula k j i + + Computing th ncssar partial drivativs w obtain Th givn point is. Thus sinc and th gradint vctor o at th point is k j k j i W us th componnts o th gradint vctor to writ th quation o th tangnt plan using th ormula continud on nt pag

17 7 + + At th point this ormula bcoms + + Using th calculations or th partial drivativs givn on th prvious pag this quation bcoms or + + W can pand this quation to gt it in gnral orm. Doing this givs + + and whn combining lik trms w hav th quation o th tangnt plan 3 +. Th paramtric quations o th normal lin through th point ar givn b t t t Using th calculations w computd abov whr that and w obtain + t + t + t which whn simpliid givs continud on nt pag

18 8 t t I w want to convrt this ths quations to smmtric orm w can tak th last two / quations o th prvious rsult and solv or t. This givs t and t. / Equation givs th smmtric quations o th normal lin. / / Th ollowing displas th graph o th unction and th normal lin at th point. th tangnt plan

2008 AP Calculus BC Multiple Choice Exam

2008 AP Calculus BC Multiple Choice Exam 008 AP Multipl Choic Eam Nam 008 AP Calculus BC Multipl Choic Eam Sction No Calculator Activ AP Calculus 008 BC Multipl Choic. At tim t 0, a particl moving in th -plan is th acclration vctor of th particl