Sction.6: Dirctional Drivativs and th Gradint Vctor Practic HW rom Stwart Ttbook not to hand in p. 778 # -4 p. 799 # 4-5 7 9 9 35 37 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 >
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 Eampl : ind th dirctional drivativ o th unction 3 4 + 6 at th π point in th dirction o th unit vctor that maks an angl o θ radians with 3 th -ais. Solution:
4 Eampl : ind th dirctional drivativ o th unction point -3-4 in th dirction o th vctor v i + 3j. 3 4 + 6 at th Solution:
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 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 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 Eampl 4: ind th gradint and dirctional drivativ o P 4 in th dirction o th point Q-3. 5 3 + at Solution: W irst comput th irst ordr partial drivativs with rspct to and. Th ar as ollows. 3 + 3 + 3 + 3 + +. Thn th ormula or th gradint is computd as ollows: i + j+ k 3 + i + 3 + j + k Hnc at th point P 4 th gradint is 4 3 + 4 i + 3 + 4 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 >< 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 < > < 4 4 + + 48 4 53.6 >
9 Eampl 5: ind th maimum rat o chang o 4 and th dirction in which it occurs. 5 3 + at th point Solution:
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.
Eampl 6: ind a unit normal vctor to th surac + + 9 at th point Solution:
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 + + +.
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.
4 Eampl 7: ind th quation o th tangnt plan and th paramtric and smmtric quations or th normal lin to th surac + + 9 at th point. Solution:
5 Not: Th ollowing graph using Mapl shows th graph o th sphr + + 9 with th tangnt plan and normal lin at th point.
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
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
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