lecture7compl-09[1]

lecture7compl-09[1] - Math 18.02(Spring 2009 Lecture 7...

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Unformatted text preview: Math 18.02 (Spring 2009): Lecture 7 Partial derivatives. Tangent plane approximation. February 20 Today: Partial derivatives. Tangent plane approximation. Last time: Velocity & acceleration. Derivatives of . and x product. Kepler’s second law. Reading Material: From Simmos 19.2, 19.3 and 19.4. From the Lecture Notes TA. . 2 Partial Derivatives We ﬁrst recall the concept of derivative for a function: Given a scalar valued function y = f(:r) we denote the derivative of f at mg as d f’{rro) = Eﬂmg) = (f) = rate of change of y when one changes a; at :ru . 10 = slope of tangent line to curve at the point (:L'g, f (30)) On the icy-plane the equation of the tangent line to the graph of y 2 f(:L') (curve) at the point Pa = (\$0,190) where ya = .1“ (1‘0) is dy (y — yo) = (—) («T — 1‘0) V‘ (LL 1'0 V (deviat. of (deviat. of y from ya) .1: from mu) Partial derivatives: Given a two variable function 2 = f (:L', y) we deﬁne o the partial derivative of f with respect to a: deﬁcit: , — — but hold y (the other variable) constant. da- dry = rate of change of 2 when one changes only .7; a the partial derivative of f with respect to y ' 2 d2 d— = — but hold :L' {the other variable) constant. 5:1; fly 2 rate of change of 2 when one changes only y Notation: We denote the partial derivative of z = f(a-, y) with respect to (r at (3:0, yo) as 82) (3f) —. = —. : fa: (3:0: HU)‘ (81' (rate) 6115 (imam) A similar notation holds for the partial derivative with respect to y: as) (Bf) — = — : f1 (34.02310)- (3y (Ithyo) all (370,310) J Exercise 1. Compute the partial derivatives of the function z = f(.’L', y) = :L'El + 2mg — y + 3 (keep for later). ‘ 2 0f r; h 2 _' = -- 2 3:1: 8:: 31 + y 2 3f — = — = .-— 1. 8y By 21 3 Tangent Planes Just like for the single variable function y : ﬁre) the derivative f’ is linked to the tangent line to its graph (curve), for the two variable function 2 2 ﬂag 3/) the partial derivatives f1- and f3, are linked to the tangent plane to the surface given by its graph. In fact given the function 2 = (1-, g), the equation for the tangent plane to its graph (surface) at the point PD : (\$0.1m, 20) for 20 = Kiowa} 13 8f . 6f 2—2 = — (._ .)+(—-) (- — ) (3.1) D ' (a\$)(\$myn) LL To By (media) y yo Handy Fact: Given a surface described as a graph of a function 2 = (my) and a point P0 = (mayo, :0) on it (that is :0 = f((.1'0, 310)), a normal vector N to this surface at P0 has coordinates N:(§) 2+(ﬂ) 3—32. ‘7‘ (\$0.1m) 5y (\$0.11“) This follows directly from the fact that the tangent plane to the surface at P0 given in (3.1) can also be written as 6f 8f —_ (m—mo)+(—) (y—yo)—(z—zn)=0 (61' ) (\$0.3m) By (\$0.31“) and we learned that the coeﬁicients of 2;, y and z are the coordinates of a vector perpendicular to the plane in question. Here we explain how one gets the equation of the tangent plane with a picture: %. . A J Exercise 2. we consider the function z = 3:3 + 23y — y + 3. Compute the tangent plane at P0 = (1, 2, 6}. solution: We computed above (“if 0 (9f _' __ .- 2 . _ _ .'— 8.7: —33. + y and By —2'L 1, hence It follows is the tangent plane we were looking for. 4 Tangent plane approximation Consider the graph of a function y = f(m). If we ﬁx the point Pa = (9:0, f (10)) and we consider the tangent line to the graph at this point '9! = f(\$0) + f’{rt'0)(:r — :50) then we notice that “near” Pu the graph of f is well approximated by the tangent line. Of course as we move away from .Pn this approximation becomes less effective. Analytically this can be recast in the following way: consider the point 3:0 + Arr, where Art is a small variation along the It—axis. Then the tangent line approximations says that for A3: small file + A5'5) E Ill-TU) 4' f’(310)A-1' or equivalently We now consider the function 3 = f(a:, y) and its graph at P0 = (mg, yo, 20), where 30 = f(.l'0, yo). Again near the point P0 this graph is well approximated by the tangent plane at the point P0 of equation 5 * 20 = (f) («T — \$0) + (—) (y n '90) 6.7, (350,110) 6y (radio) 5: ’ioujwzf than: P Fe 6 a b -.-.... I I I I X ng‘JO) Analytically this says thatWe can approximate the value of f (1-0 + Art, ya + Ay), where Arr and Ag are respectively small increments with respect to the :1; and the y— axis, by using the equation of the tangent plane. More precisely we have the following deﬁnition Deﬁnition 1. The tangent plane approximation off near (330,1;0) is deﬁned as 0 3 Az=f(:co+ArL'.yo+Ay)—f(3:o.yo)m (i) A:L-+ (El—f) Ay. ‘IL (110.110) y (Iowa) or equivalently . . a a 1m + Ass, 110 + Ag) 2: 1(m0.yo)+ (—) A2.- + (—15) Ag. 81‘ (Iowa) 83] (fringe) Exercise 3. Consider the function f(.7;,'y) I 513': + 1. Compute the ( tangent plane) approaimation to f (£1 + .02, , 2.01). At (f, 2) does f change more i‘apidty in the i or in the} direction? Solution: In this example we can take (\$0,110) = (%,2) so that ' 4 2 20 = “32) = 5‘“? +1 = §+1=1.3535. We also take £33; 2 .02 and Ag = .01. To compute the approximation we ﬁrst evaluate the partial derivatives: 6f _ cos 1' 6.1- 1’ :rr/4,'2 35 i ) af —si _ = _.:”= = —.176 . 4-2 3y 3" ma 7 i ) Then we have A2 a .3535(A\$)+{—.1767)Ay = .0071—0018 = .0053 hence 1%; +02, , 2.01) 2 :0 + A; z: 1.3535 + .0053 =1.3588. Finally we can compare (4.1) and (4.2) and deduce that the 'rate of change off is greater along i. Study Guide 1. Answer the following questions: a Is it time that any function 5 = (1', 1;) such that f1. : D is a constant function? Emplain your answer. 0 Consider the function 3 2 0:2 + ya. Does the plane 2 = U approximate well the value of the function near (0, 0) .9 What about the value of the function near (1,2)? ...
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