This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: are measured in miles. Since and are decreasing we take the derivatives to be negative and so we are given mi/h and mi/h. To find we use the Pythagorean Theorem, namely and differentiate with respect to time . We have and by solving for we have Now when mi and mi we have mi and so 2. Integrate: (find the antiderivative) dx x x x x x ) tan sec 4 cos 6 sec 2 ( 2 +∫ Solution C x x x x + +sec 4 sin 6 tan 2 3 2 2 3...
View
Full
Document
This note was uploaded on 05/06/2008 for the course MATH 201 taught by Professor Chuckrow during the Spring '08 term at CUNY City.
 Spring '08
 CHUCKROW
 Calculus

Click to edit the document details