, the tangent line is an approximation of the curve (and easier to deal with than the curve).

Size: px

Start display at page:

Download ", the tangent line is an approximation of the curve (and easier to deal with than the curve)."

Dina Ray
6 years ago
Views:

1 114 Tangent Planes and Linea Appoimations Back in-dimensions, what was the equation of the tangent line of f ( ) at point (, ) f ( )? (, ) ( )( ) = f Linea Appoimation (Tangent Line Appoimation) of f at : L ( ) = f( ) + f ( )( ) Do ou emembe as we zoom in close and close to the point of tangenc (, ), the tangent line is an appoimation of the cuve (and easie to deal with than the cuve) TEC 114 Movable Tangent Plane Now in 3-dimensions as we zoom in close and close to the point of tangenc (,, z ), the tangent plane is an appoimation of the suface Equation of the tangent plane to the suface z = f (, ) at point (, )( ) (, )( ) z z = f + f So the tangent plane to a suface is a plane that contains the two tangent lines T 1 and T (as well as all possible tangent lines at the point of tangenc), so it makes sense to see patial deivatives in thee The tangent plane at the point (,, z ) is the plane that most closel appoimates the suface at that point of tangenc Hee s wh (if ou e inteested)

2 Fom section 95, the scala equation of the plane though point (,, z ) has an equation in the fom: A( ) B( ) C( z z ) z z & divide b C + + = solve fo ( ) = ( ) ( ) C z z A B A B z z = C C ( ) ( ) let a = A/ C and b= B/ C it can be witten ( ) ( ) z z = a + b This equation epesents the tangent plane at (,, ) tangent line T 1 Setting z, so its intesection with the plane = into the above equation gives a ( ) = must be the z z =, whee = and we ecognize this as the equation of a line with slope a ling in the plane = But fom Section 113 we know the slope of the tangent T 1 is f (, ) Theefoe a = f (, ) Similal, putting = into the above equation we get b ( ) tangent line T, so b= f (, ) So, z z =, whee =, which epesents the ( ) ( ) z z = a + b becomes (, )( ) (, )( ) z z = f + f Eample: Find an equation of the tangent plane to the top half of the sphee z at the point 1 1,, + + = 1

3 Pictue the top half of a sphee and a tangent plane located at point 1 1,, With f, is constant so pictue a plane pependicula to the z-plane slicing though ou sphee ceating a cuve This tangent line is paallel to the -ais and since f = 1, if ou take a step in positive -diection ou ll go 1 With f, is constant so pictue the tangent line paallel to the -ais f =, so if ou take a step in positive -diection, ou ll go (it s like ou e walking along the idge of the sphee) Section 38 A big idea in Calculus was when f ( ) we zoom in on a point on the gaph of f ( ), the gaph becomes indistinguishable fom its tangent line and we appoimated the function b L(, ) a linea appoimation ( ) ( )( ) ( ) ( )( ) ( ) ( ) f a = f a a = f a + f a a f = L So, ( ) = ( ) + ( )( ) L f a f a a Now in 3-dimensions, we zoom in towad a point on a suface f (,, ) the suface looks moe and moe like a plane; its tangent plane see figue, page 771 We can appoimate the function b a o TEC 114 Tangent linea function of vaiables, L(, ) plane zoom What did the do to the equation of the tangent plane to make L(,? ) z z = f(, )( ) + f(, )( ) L(, ) = f( ab, ) + f ( ab, ) ( a) + f ( ab, ) ( b)

4 (, ) (, ) L f 1443 z The linea function whose gaph is this tangent plane is used to appoimate the function It s the lineaization of f at ( ab, ) Called a linea appoimation o the tangent plane appoimation It s a good appoimation of the function when ou e nea the point Use the linea appoimation of the function + + z = 1 at the point appoimate f ( 5, 1) 1 1,, to We had z = + = ( ) L, = (Remembe eal on in Calculus I, If f ( a) eists, a function f is diffeentiable at a Now: If the patial deivatives f and f eist nea ( ab, ) and ae continuous at (, ) then f is a diffeentiable at ( ab, ) ab, Eample: Eplain wh the function f (, ) = is diffeentiable at ( ) 6,

5 Back in Calculus I: The cuve is = f ( ) d is the change in height of the tangent line (the amount that the tangent line ises/falls) d Δ Δ is the amount that the cuve ises/falls d d =Δ = f ( ) d +Δ Called diffeential d d = f ( ) d change in lineaization If = f ( ), whee f is a diffeentiable function, then the diffeential d is an independent vaiable so d can be teated as a eal numbe and we d wee able to teat it as a faction f ( ) =, so d = f ( ) d d is the dependent vaiable (depends on values of and d ) We did cool poblems like If in measuing the adius of a sphee ou ma be off b as much as 1 cm, how will that effect the volume? 4 3 So we used V = π and when we had moe than one vaiable, we made a substitution to 3 get it down to 1 vaiable But since this couse is called Multi-Vaiable Calculus, we welcome man vaiables d Now in 3-dim diffeential dz (, ) (, ) dz = f d + f d z z dz = d + d dz = Total Diffeentiation dz = change in height in tangent plane So a diffeentiable function of two vaiables, z = f (, ), we define the diffeentials d and to be independent vaiables (the can be given an values) Page 775, Figue 7 d

6 The new value is ceated b a+δ, so if = a+δ, then Δ = a If we take d =Δ = a and simlal d =Δ = b, then: dz = f, d + f, d ( ) ( ) (, )( ) (, )( ) dz = f a b a + f a b b The tangent plane: z f( ab, ) = f ( ab, ) ( a) + f ( ab, ) ( b) So then the linea appoimation can be witten L(, ) = f ( a, b) + f( a, b)( a) + f( a, b)( b) (, ) (, ) f f a b + dz The function is appoimated b the tangent plane f ab, dz dz (It s eas to see geometicall how ( ) + appoimates the suface f (, ) L (, ) dz change in height of tangent plane Δz change in height of the suface = tangent plane) We can use diffeentials to estimate the lagest possible eo that ma occu in a fomula if cetain measuements ae off (incoect) b a cetain amount Eample: If z = 5 + and (, ) changes fom ( 1, ) to ( 15, 1 ), compae the values of z and Δ dz Moe inteesting tpe of poblem Eample: Use diffeentials to estimate the amount of metal in a closed clindical can that is 1 cm high and 4 cm in diamete if the metal in the top and bottom is 1 cm thick and the metal in the sides is 5 cm thick Assuming the outside can has adius and height 1

7 Impotant Note: In section 114, d and dh ae fied amounts of change, d = 5cm and dh = cm The ae set (discete) amounts of change in adius and height In section 115 these will be moving amounts of change (eample: the adius is inceasing at a ate of 3 cm / s o the height is deceasing at a ate of 7 cm / s so the d dh notation will be = 3 cm / s and 7 cm / s dt dt = ) These ae moving changes Actual Δ V Outside can, h= 1, = Inside can, h= 98, = 195 V = π h = π *1 = 4π v = π ( 195) ( 98) Δ V is 859 cm of mateial (actual amount of tin) 3 Using diffeentials, dv 88cm which is the appoimate amount of tin So we use dv to appoimate Δ V and it s suppose to be easie to compute (I m not so sue ) Diffeentials, diffeentiabilit and linea appoimation can be defined similal fo functions of moe than vaiables! Eample: V o V(,, z) = z The diffeential V V V dv = d + d + dz z

8 If we want to find an equation of the tangent plane to a paametic suface ( u, v) cetain point a ( ) + b ( ) + cz ( z) = I m sue we can find a point on the plane, but, how do we find the nomal vecto, abc,,, to the plane? What do we need fo a plane? - point on the plane - n omal vecto to the plane at a Tangent Planes to Paametic Sufaces Tangent plane to a paametic suface taced out b a vecto function ( u, v) = ( u, v), ( u, v), z( u, v) at point P with position vecto ( u, v) If we keep u constant b letting u = u, then ( u, v ) becomes a vecto function of the single paamete v and defines a famil of gid cuves The tangent vecto is found b taking the z v = v v v patial deivative of with espect to v,,, v u being held constant v P

9 Similal, b keeping v constant, ou get the othe famil of gid cuves and its tangent z vecto is,, u = u u u u v being held constant P u If vectos u and v ae on the tangent plane containing point P, how do we find the nomal vecto to this plane? Since vectos u and v ae on the tangent plane containing point P, u Χ v is the n omal vecto Fo a smooth suface the tangent plane is the plane that contains tangent vectos and the vecto u Χ v is a n omal vecto to the tangent plane u and v Eample: Find an equation of the tangent plane to the paametic suface ( ) the point u, v π = = a ( ) + b ( ) + cz ( z) = u, v = uv, usin v, vcosu What do we need fo a plane? - point on the plane - n omal vecto to the plane at

10 Paametic Repesentation of the equation of the tangent plane is: Tangent Plane at ( u, v ) : (, ) = (, ) + (, ) + (, ) t u v u v u v u u v v u v Let s stat with the paametic equation of a plane fom section 15 (, ) u v = + a u + b v Now the paametic equation of the tangent plane at ( u, v ) (, ) = (, ) + (, ) + (, ) t u v u v u v u u v v u v So ou eample would look like: t u v u v u v u u v v (, ) = (, ) + (, ) + (, ) u v t(, π) = (, π) + u(, π) u+ v(, π) v =,, π + π,, u +,,1 v = + πu+ v,+ u+ v, π + u+ 1 v = πu,, π + v So = fom ou equation and ( ) equation of the tangent plane to ( ) t, π = πu,, π + v u, v = uv, usin v, vcosu is a paametic epesentation of the at the point u =, v= π

ME 210 Applied Mathematics for Mechanical Engineers

ME 210 Applied Mathematics for Mechanical Engineers Tangent and Ac Length of a Cuve The tangent to a cuve C at a point A on it is defined as the limiting position of the staight line L though A and B, as B appoaches A along the cuve as illustated in the