{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Test3_sol

# Test3_sol - NAME Spring 2011 MAC 2313 Test 3 Section 3124...

This preview shows pages 1–2. Sign up to view the full content.

NAME: Spring 2011, MAC 2313, Test 3 UFID: 4/7/2011, Section: 3124 1. Use the Chain Rule to ﬁnd ∂z/∂s and ∂z/∂t for z = tan( u/v ), u = 2 s + 3 t , v = 3 s - 2 t . (10 points) Solution . By the Chain Rule, we have ∂z ∂s = ∂z ∂u · ∂u ∂s + ∂z ∂v · ∂v ∂s = sec 2 ± u v ² · 1 v · 2 + sec 2 ± u v ² · - u v 2 · 3 = 2 v - 3 u v 2 · sec 2 ± u v ² = - 13 t (3 s - 2 t ) 2 sec 2 ³ 2 s + 3 t 3 s - 2 t ´ , ∂z ∂t = ∂z ∂u · ∂u ∂t + ∂z ∂v · ∂v ∂t = sec 2 ± u v ² · 1 v · 3 + sec 2 ± u v ² · - u v 2 · ( - 2) = 3 v + 2 u v 2 · sec 2 ± u v ² = 13 s (3 s - 2 t ) 2 sec 2 ³ 2 s + 3 t 3 s - 2 t ´ . 2. Find the directional derivative of the function f ( x,y,z ) = xyz at (3 , 2 , 6) in the direction of v = h- 1 , - 2 , 2 i . (10 points) Solution . First we calculate f x = yz 2 xyz , f y = xz 2 xyz , f z = xy 2 xyz . Then we know that f x (3 , 2 , 6) = 2 · 6 2 36 = 1 , f y (3 , 2 , 6) = 3 · 6 2 36 = 3 2 , f z (3 , 2 , 6) = 3 · 2 2 36 = 1 2 . Also, there is

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.

{[ snackBarMessage ]}

### Page1 / 2

Test3_sol - NAME Spring 2011 MAC 2313 Test 3 Section 3124...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online