Multivariate Calculus - Bruce E. Shapiro
Multivariate Calculus - Bruce E. Shapiro Multivariate Calculus - Bruce E. Shapiro
90 LECTURE 12. THE CHAIN RULE Example 12.7 Repeat the previous example using theorem 1 instead of implicit differentiation. Solution. We had f(x, y) = x 3 + 2x 2 y − 10y 5 so that ( ∂ dy dx = −∂f/∂x ∂f/∂y = − ∂x x 3 + 2x 2 y − 10y 5) ∂ ∂y (x3 + 2x 2 y − 10y 5 ) = − 3x2 + 4xy 2x 2 − 50y 4 Example 12.8 Find dy/dx if x 2 cos y − y 2 sin x = 0. Solution.Writing f(x, y) = x 2 cos y − y 2 sin x and usingtheorem 1, ( ∂ dy dx = −∂f/∂x ∂f/∂y = − ∂x x 2 cos y − y 2 sin x ) ∂ ∂y (x2 cos y − y 2 sin x) = − 2x cos y − y2 cos x −x 2 sin y − 2y sin x Differentials Definition 12.1 The differential of a function f(x, y, z) is the quantity df = ∇f · dr where namely, dr = (dx, dy, dz) df = ∂f ∂f ∂f dx + dy + ∂x ∂y ∂z dz The differential gives an estimate of the change in the function f(x, y, z) when r is perturbed by a small amount from (x, y, z) to (x + dx, y + dy, z + dz). Example 12.9 Estimate the change in f(x, y) = x 2 + y 2 as you move from (1, 1) to (1.01, 1.01) using differentials, and compare with the exact change. Solution. We have and so that Therefore The exact change is ∆r = (0.01, 0.01) ∇f = (2x, 2y) ∇f(1, 1) = (2, 2) ∆f = ∇f · ∆r = (2, 2) · (0.01, 0.01) = .04 f(1.01, 1.01) − f(1, 1) = 1.01 2 + 1.01 2 − 1 − 1 = 0.0402 Revised December 6, 2006. Math 250, Fall 2006
LECTURE 12. THE CHAIN RULE 91 Example 12.10 The effective resistance R of two resistors R 1 , R 2 in parallel is given by the formula 1 R = 1 R 1 + 1 R 2 Suppose we have two resistors R 1 = 10 ± 1 ohms R 2 = 40 ± 2 ohms that are connected in parallel. Estimate the total resistance R and the uncertainty in R. Solution. We first calculate R, 1 R = 1 10 + 1 40 = 5 40 = 1 8 ⇒ R = 8 To calculate the uncertainty we solve for R(R 1 , R 2 ) and find the differential. Hence By the quotient rule, Thus dR = R = R 1R 2 R 1 + R 2 ∂ R 1 R 2 dR 1 + ∂ R 1 R 2 dR 2 dR 1 R 1 + R 2 dR 2 R 1 + R 2 ∂ R 1 R 2 = (R 1 + R 2 ) ∂ ∂R 1 (R 1 R 2 ) − (R 1 R 2 ) ∂ ∂R 1 (R 1 + R 2 ) dR 1 R 1 + R 2 (R 1 + R 2 ) 2 = (R 1 + R 2 )R 2 − R 1 R 2 (R 1 + R 2 ) 2 = R2 2 (R 1 + R 2 ) 2 = 40 2 ( 40 (10 + 40) 2 = 50 ) 2 ( 4 = 5 ∂ R 1 R 2 = (R 1 + R 2 ) ∂ ∂R 2 (R 1 R 2 ) − (R 1 R 2 ) ∂ ∂R 2 (R 1 + R 2 ) dR 2 R 1 + R 2 (R 1 + R 2 ) 2 dR = = (R 1 + R 2 )R 1 − R 1 R 2 (R 1 + R 2 ) 2 = R1 2 (R 1 + R 2 ) 2 = 10 2 ( 10 (10 + 40) 2 = 50 ) 2 = 16 25 ) 2 ( ) 1 2 = = 1 5 25 ∂ R 1 R 2 dR 1 + ∂ R 1 R 2 dR 2 = 16 dR 1 R 1 + R 2 dR 2 R 1 + R 2 25 (1) + 1 18 (2) = 25 25 = 0.72 and we conclude that R = 8 ± 0.72 ohms Math 250, Fall 2006 Revised December 6, 2006.
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LECTURE 12. THE CHAIN RULE 91<br />
Example 12.10 The effective resistance R of two resistors R 1 , R 2 in parallel is<br />
given by the formula<br />
1<br />
R = 1 R 1<br />
+ 1 R 2<br />
Suppose we have two resistors<br />
R 1 = 10 ± 1 ohms<br />
R 2 = 40 ± 2 ohms<br />
that are connected in parallel. Estimate the total resistance R and the uncertainty<br />
in R.<br />
Solution. We first calculate R,<br />
1<br />
R = 1 10 + 1<br />
40 = 5 40 = 1 8 ⇒ R = 8<br />
To calculate the uncertainty we solve for R(R 1 , R 2 ) and find the differential.<br />
Hence<br />
By the quotient rule,<br />
Thus<br />
dR =<br />
R = R 1R 2<br />
R 1 + R 2<br />
∂ R 1 R 2<br />
dR 1 +<br />
∂ R 1 R 2<br />
dR 2<br />
dR 1 R 1 + R 2 dR 2 R 1 + R 2<br />
∂ R 1 R 2<br />
= (R 1 + R 2 ) ∂<br />
∂R 1<br />
(R 1 R 2 ) − (R 1 R 2 ) ∂<br />
∂R 1<br />
(R 1 + R 2 )<br />
dR 1 R 1 + R 2 (R 1 + R 2 ) 2<br />
= (R 1 + R 2 )R 2 − R 1 R 2<br />
(R 1 + R 2 ) 2<br />
=<br />
R2<br />
2<br />
(R 1 + R 2 ) 2 = 40 2 ( 40<br />
(10 + 40) 2 = 50<br />
) 2 ( 4<br />
=<br />
5<br />
∂ R 1 R 2<br />
= (R 1 + R 2 ) ∂<br />
∂R 2<br />
(R 1 R 2 ) − (R 1 R 2 ) ∂<br />
∂R 2<br />
(R 1 + R 2 )<br />
dR 2 R 1 + R 2 (R 1 + R 2 ) 2<br />
dR =<br />
= (R 1 + R 2 )R 1 − R 1 R 2<br />
(R 1 + R 2 ) 2<br />
=<br />
R1<br />
2<br />
(R 1 + R 2 ) 2 = 10 2 ( 10<br />
(10 + 40) 2 = 50<br />
) 2<br />
= 16<br />
25<br />
) 2 ( ) 1 2<br />
= = 1 5 25<br />
∂ R 1 R 2<br />
dR 1 +<br />
∂ R 1 R 2<br />
dR 2 = 16<br />
dR 1 R 1 + R 2 dR 2 R 1 + R 2 25 (1) + 1 18<br />
(2) =<br />
25 25 = 0.72<br />
and we conclude that R = 8 ± 0.72 ohms <br />
Math 250, Fall 2006 Revised December 6, 2006.
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