math53 - quiz 5 - solns

# math53 - quiz 5 - solns - + z y y s = z x 2 s + z y 2 r....

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Math 53 - Quiz #5 Solutions GSI: Santiago Canez March 1, 2006 1. The length and width of a rectangle are measured as 30 cm and 24 cm, respectively, with an error in measurement of at most 0.1 cm in each. Use diﬀerentials to estimate the maximum error in the calculated area of the rectangle. Solution. The area is given by A = lw and the maximum error is given by its diﬀerential dA = ∂A ∂l dl + ∂A ∂w dw = wdl + ldw. The values of l and w are given in the problem, and the values dw and dl are given by saying that the error of measurement is at most 0 . 1 cm in each variable. Hence we have dA = (24)(0 . 1) + (30)(0 . 1) = 5 . 4 cm 2 . 2. Use the chain rule to ﬁnd ∂z/∂s and ∂z/∂t where z = e xy tan y, x = s + 2 t, y = s/t . Solution. By the chain rule, we have ∂z ∂s = ∂z ∂x ∂x ∂s + ∂z ∂y ∂y ∂s = ( ye xy tan y )(1) + e xy ( x tan y + sec 2 y )(1 /t ) and ∂z ∂t = ∂z ∂x ∂x ∂t + ∂z ∂y ∂y ∂t = ( ye xy tan y )(2) + e xy ( x tan y + sec 2 y )( - s/t 2 ) . 3. If z = f ( x, y ), where x = r 2 + s 2 , y = 2 rs , ﬁnd 2 z/∂r∂s . Your answer should involve the ﬁrst and second partials of z with respect to x and y . Solution. First we compute the partial with respect to s : ∂z ∂s = ∂z ∂x ∂x ∂s

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Unformatted text preview: + z y y s = z x 2 s + z y 2 r. Now, we compute the partial of this with respect to r : 2 z rs = r z x 2 s + z y 2 r = r z x 2 s + r z y 2 r . The rst term on the right side is computed using the chain rule: r z x 2 s = x z x 2 s x r + y z x 2 s y r = 2 z x 2 4 rs + 2 z yx 4 s 2 . 1 The second term on the right side (two equations above) is compute using rst the product rule: r z y 2 r = r z y 2 r + z y 2 and then the chain rule on the rst part: r z y 2 r + z y 2 = 2 z xy 4 r 2 + 2 z y 2 4 rs + z y 2 . Thus we have 2 z rs = 2 z x 2 4 rs + 2 z yx 4 s 2 + 2 z xy 4 r 2 + 2 z y 2 4 rs + z y 2 . 2...
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## math53 - quiz 5 - solns - + z y y s = z x 2 s + z y 2 r....

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