exam2_fa04_answer

exam2_fa04_answer - Math 20B Second Midterm Solutions...

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: Math 20B Second Midterm Solutions November 15, 2004 1. The curve goes from x = 0 to x = 1. Solving for y(x), we have y = y ′ = −2x3 (1 − x4 )−1/2 . 1 1 1 + (y ′ )2 dx = (a) 1+ 0 0 4x6 dx = 1 − x4 2πy(x) 1 + (y ′ )2 dx = 2π 0 (1 − x4 ) 1 + 0 0 1 = 2π 0 1 − x4 . Thus 1 − x4 + 4x6 dx. 1 − x4 1 1 (b) 1 √ 4x6 1 − x4 dx 1 − x4 + 4x6 dx. 2. (a) 0.08/42 = 0.005, since error is roughly C/n2 and we have multiplied n by 4. (b) The error in the Midpoint Rule is about half that of the Trapezoidal Rule and opposite in sign. Hence the answer is −0.04. (You will get partial credit for 0.04.) √ √ 3. (a) 2 cos(3π/4) + 2i sin(3π/4) = − 2 + 2 i. (b) r = 21/3 . The values of θ are 3π/4 π = , 3 4 3π/4 + 2π 11π = , 3 12 3π/4 + 4π 19π = . 3 12 (You need not do the arithmetic to simplify the angles.) 4. Separating variables: dy = 2x dx and so ln y = x2 + C. Using y(0) = 2, we have y ln 2 = 02 + C and so C = ln 2. At x = 3 we have ln y = 32 + C = 9 + ln 2 and so y = e9+ln 2 = 2e9 . 5. We have 2x2 2 1 1 = 2+ = 2+ − , 2−1 x (x − 1)(x + 1) x−1 x+1 where it is up to you how you find the partial fractions. Integrating gives 2x + ln x−1 +C. x+1 ...
View Full Document

This note was uploaded on 04/29/2010 for the course MATH MATH 20B taught by Professor Takeda during the Spring '07 term at UCSD.

Ask a homework question - tutors are online