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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 icyplane 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 IllTU) 4' f’(310)A1' 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 (10 + 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 ($0,110) = (%,2) so that ' 4 2
20 = “32) = 5‘“? +1 = §+1=1.3535. We also take £33; 2 .02 and Ag = .01. To compute the approximation we ﬁrst evaluate the partial
derivatives: 6f _ cos 1' 6.1 1’ :rr/4,'2 35 i )
af —si _ = _.:”= = —.176 . 42
3y 3" ma 7 i ) Then we have A2 a .3535(A$)+{—.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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