{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Calculus 2 - Assignment 5 Solutions

Calculus 2 - Assignment 5 Solutions - MAT1332 Spring/Summer...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
MAT1332 Spring/Summer 2009 Assignment 5 Solutions 1. Consider the function f ( x , y ) : = x 2 y 2 x - 3 xy 3 (a) What is the gradient of f ? (b) What is 2 f yx ? (c) What is the tangent plane to f at the point (3,5) ? First, note that f ( x , y ) simplifies to f ( x , y ) = xy 2 - 3 xy 3 . For (a), we observe that the gradient f of f is given by f ( x , y ) = £ y 2 - 3 y 3 x 2 - 9 xy 2 / For (b), we have that, since f x = y 2 - 3 y 3 , 2 f yx = 1 2 - 9 y 2 . For (c), we have that the tangent plane at (3,5) is given by z - f (3,5) = ( 5 2 - 375)( x - 3) + ( 3 2 - 675)( y - 5) = - 745 2 ( x - 3) + 1347 2 ( y - 5). Since f (3,5) = - 2235 2 we have z = ( 5 2 - 375)( x - 3) + ( 3 2 - 675)( y - 5) - 2235 2 = - 745 2 ( x - 3) + 1347 2 ( y - 5) - 2235 2 2. Solve the following linear system of differential equations dv dt = 6 v + w dw dt = 9 v - 2 w with intial conditions v 0 = 7 and w 0 = 2 . First we observe that the the coefficient matrix A = 6 1 9 - 2 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
has characteristic polynomial ( A - λ I ) equal to λ 2 - 4 λ - 21. This factors as ( λ - 7)( λ + 3) and therefore the eigenvalues are λ = 7, - 3. We now solve for the corresponding eigenvectors. Since when λ = 7 we have ( A - λ I ) = - 1 1 9 - 9 it follows (taking 9 * R 1 + R
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}