{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

pm2aSolns

# pm2aSolns - Math 265(Butler Practice Midterm II —...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 265 (Butler) Practice Midterm II — A (Solutions) 1. Find the directional derivative of f ( x,y,z ) = xz 2- 3 xy + 2 xyz- 3 x + 5 y- 17 from the point (2 ,- 6 , 3) in the direction of the origin. To find a directional derivative we note that D u f = ∇ f · u . So we need to find u and ∇ f . Since we are going from the point (2 ,- 6 , 3) towards (0 , , 0) then a vector pointing in the appropriate direction is h- 2 , 6 ,- 3 i . This is not a unit vector since kh- 2 , 6 ,- 3 ik = √ 4 + 36 + 9 = √ 49 = 7, but by scaling we can make it a unit vector so that u = 1 7 h- 2 , 6 ,- 3 i . The gradient is ∇ f ( x,y,z ) = h z 2- 3 y + 2 yz- 3 ,- 3 x + 2 xz + 5 , 2 xz + 2 xy i . Evaluating at the point (2 ,- 6 , 3) we have ∇ f (2 ,- 6 , 3) = h 3 2- 3 · (- 6) + 2 · (- 6) · 3- 3 ,- 3 · 2 + 2 · 2 · 3 + 5 , 2 · 2 · 3 + 2 · 2 · (- 6) i = h- 12 , 11 ,- 12 i . Therefore the desired directional derivative will be given by ∇ f · u = h- 12 , 11 ,- 12 i · 1 7 h- 2 , 6 ,- 3 i = 24 + 66 + 36 7 = 126 7 = 18 . 2.2....
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

pm2aSolns - Math 265(Butler Practice Midterm II —...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online