must1 - Preliminary Exam II Solutions 1 . Find dy dx by...

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: Preliminary Exam II Solutions 1 . Find dy dx by implicit differentiation given that cos( y )- 3sin( x ) = 1 . Solution: Differentiation yields- sin( y ) dy dx- 3 cos( x ) = 0 hence sin( y ) dy dx =- 3cos( x ) sin( y ) . 2 . An object is moving along the circle x 2 + y 2 = 16. If dy dt = 2 when it reaches the point with coordinates x = 3 and y = 1, find dx dt at that point. Solution: Differentiating with respect to the variable t yields: 2 x dx dt + 2 y dy dt = 0 . Substituting the given information yields: 3 dx dt + 2 = 0 . Thus dx dt =- 2 3 . 3 . Find the absolute extrema of the function f ( x ) = x 3- 3 2 x 2 + 5 on the closed interval [- 1 , 2]. Solution: Differentiation yields f ( x ) = 3 x 2- 3 x = 3 x ( x- 1) hence the critical numbers are x = 0 , 1. The candidates are- 2 , , 1 , 2 and 4 f (- 1) = (- 1) 3- 3 2 (- 1) 2 + 5 =- 1- 3 2 + 5 = 5 2 f (0) = 5 f (1) = 1- 3 2 + 5 = 9 2 f (2) = 2 3- 3 2 2 2 + 5 = 8- 6 + 5 = 7 . Thus (a). max = 7 min = 5 / 2 1 2 4 . Find all of the critical numbers of the function f ( x ) = x 2 + 1 x . Solution: Differentiation yields f ( x ) = x 2- 1 x 2 . The critical numbers are obtained by setting the numerator equals zero as well as setting the denominator equals. Thus the critical numbers are x = 0 , 1 ,- 1....
View Full Document

This note was uploaded on 07/05/2010 for the course MATH 2240 taught by Professor Crespo during the Spring '09 term at SPSU.

Page1 / 5

must1 - Preliminary Exam II Solutions 1 . Find dy dx by...

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