Pre-Calc Homework Solutions 80

Pre-Calc Homework Solutions 80 - 80 Section 3.4 dP dV an 2...

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

View Full Document Right Arrow Icon
33. } d d P V } 5 } d d V } 1 } V n 2 R n T b } 2 } a V n 2 2 } 2 52 } d d V } ( an 2 V 2 2 ) 5 } ( 0 V 2 2 n n R b T ) 2 }1 2 an 2 V 2 3 } ( V n 2 RT nb ) 2 }1} 2 V an 3 2 } 34. } d d s t } 5 } d d t } (4.9 t 2 ) 5 9.8 t } d d 2 t 2 s } 5 } d d t } (9.8 t ) 5 9.8 35. } d d M R } 5 } d d M } 3 M 2 1 } C 2 } 2 } M 3 } 24 5 } d d M } 1 } C 2 } M 2 2 } 1 3 } M 3 2 5 CM 2 M 2 36. If the radius of a circle is changed by a very small amount D r , the change in the area can be thought of as a very thin strip with length given by the circumference, 2 p r , and width D r . Therefore, the change in the area can be thought of as (2 p r )( D r ), which means that the change in the area divided by the change in the radius is just 2 p r . 37. If the radius of a sphere is changed by a very small amount D r , the change in the volume can be thought of as a very thin layer with an area given by the surface area, 4 p r 2 , and a thickness given by D r. Therefore, the change in the volume can be thought of as (4 p r 2 )( D r ), which means that the change in the volume divided by the change in the radius is just 4 p r 2 .
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online