mid1bsol

mid1bsol - Problem 1. Find the limits below or explain why...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem 1. Find the limits below or explain why they do not exist: (i) lim x,y → ( x 2 + y 2 ) 2 2 x 2 +3 y 2 We note that ≤ ( x 2 + y 2 ) 2 2 x 2 + 3 y 2 ≤ (2 x 2 + 3 y 2 ) 2 2 x 2 + 3 y 2 = 2 x 2 + 3 y 2 → . Therefore, the original limit equals as well. (ii) lim x,y → x 3 y x 4 + y 4 The limit does not exist. Indeed, approaching by keeping x = y → , the fraction equals x 3 · x x 4 + x 4 = 1 2 . On the other hand, approaching by keeping x = 0 ,y → , the fraction equals . Since the two answers are different, the limit does not exist. Problem 2 Consider the function f ( x,y ) = x 2 sin(2 y- 2 x ) and the point P (1 , π 2 + 1) . (i) Find the gradient of f at the point P . We compute the derivatives f x = 2 x sin(2 y- 2 x )- 2 x 2 cos(2 y- 2 x ) = ⇒ f x (1 , π 2 + 1) = 2sin π- 2cos π = 2 f y = 2 x 2 cos(2 y- 2 x ) = ⇒ f y (1 , π 2 + 1) = 2cos π =- 2 . Then ∇ f ( P ) = (2 ,- 2) . (ii) Calculate the directional derivative of f at P in the direction ~u = ~ 3 i + 4 ~ j 5 . We have f ~u ( P ) = ∇ f ( P ) · ~u = (2 ,- 2) · 3 5 , 4 5 = 2 · 3 5- 2 · 4 5 =- 2 5 ....
View Full Document

This note was uploaded on 03/28/2011 for the course MATH 31B taught by Professor Dragosoprea during the Winter '11 term at UCSD.

Page1 / 7

mid1bsol - Problem 1. Find the limits below or explain why...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online